N-ary tree or K-way tree data structure

Hello people! In this post, we will talk about a generic tree data structure which is N-ary tree or also known as k-way tree. N-ary tree is defined as a rooted tree which has at most N children for any node. So, a binary tree is a special case of the N-ary tree, where N = 2. Example –

N-ary tree Theory of ProgrammingThe above example is a N-ary tree with N = 3. You can observe that each node has no more than 3 children.

In many scenarios, having a binary tree many not be enough as a situation can often lead to more than 2 situations. In such cases, we must form a N-ary tree of sufficiently large N.

Implementing N-ary tree using structures

Implementing N-ary tree is very simple. It is even more easy if you already know how to implement a Trie Tree. Now, in the case of binary tree, the node structure was like this –

class TreeNode {
    String label;
    Node leftChild, rightChild;

Having two references was enough for us then because we knew we would have at most 2 children. Now, we can have at most N children. How will you store a collection of N references? By using an array! In the case of Java, I’d like to use an ArrayList rather than an array. In the case of C++, I’d suggest you go for a C++ STL vector. So now our node structure will look like this –

class NaryTreeNode {
    String label;
    ArrayList<NaryTreeNode> children;
    int n;

Storing the value of N inside the node will enable us to put a restriction on the number of children. So we will add methods inside our newly formed NaryTreeNode class to manipulate the member variables. Some important methods would be –

  • addChild(child) – to add a child node.
  • getChildren() – to get all the child nodes.
  • getChild(index) – to get child at a certain index.
  • print(root) – to print the tree. (This can be a static method, where we just supply the root node).

Also, don’t forget to add a constructor to your class! Now you are all set to write the code for N-ary tree. Try to code it, this is something you should be able to code in 1-3 attempts. You can refer to my code below if you get stuck –

         Carbon dioxide

I have added a helper method for printing the tree so that the tree can be printed more legibly based on the depth of nodes. I have also modified the addChild() method a little. It only accepts the new child’s label as it’s N value can be deduced from the parent node. The code should be easy to understand. Do leave a comment if you have any doubts 🙂

Implementing N-ary tree using arrays

Remember you can implement a binary tree using arrays. Similarly, you can also implement an N-ary tree using arrays. It is very similar to how you would do it in the case of binary trees, so here the formula for computing child node and parent node changes.

Parent node Child node
Binary tree Parent node in binary tree Left child –
Left child in binary tree
Right Child –
Right child in binary tree
N-ary Tree Parent node in N-ary tree Cth child –
Child in an N-ary tree

Using the above table, you can find the parent and child node for a node at index i. Generally, I like to implement the tree using structures, but if you have a complete n-ary tree, then the array representation will you a better compact storage.

I hope this article was able to give you an idea of how to implement an N-ary tree. Keep practicing! Happy coding! 😀

Compressed Trie Tree

Hello, people! In this post, we will discuss a commonly used data structure to store strings, the Compress Trie Tree, also known as Radix Tree or Patricia (Practical Algorithm to Retrieve Information Coded in Alphanumeric) Tree. If you remember, the problem with a Trie Tree is that it consumes a lot of memory. So, due to its memory consumption, developers prefer using a compressed trie tree to get the same functionality at the same runtime complexity in memory-critical situations such as in android apps.

“Compress”-ing the tree

Now, the idea of the compressed trie tree is to convert long chains of single-child edges to one single edge. Example, suppose we have two words “facebook” and “facepalm” in our regular trie tree, it would look like –

Trie Tree

It’s a pretty long one, isn’t it? Now, how about this one below?

compressed trie tree

This one surely looks a lot compact! This is the compressed trie tree. As you can see what we do here is to make long chains of single-child edges into one edge. Unlike a regular trie tree, in a compressed trie tree, the edge labels are strings.

Node in Compressed Trie Tree

For a regular trie tree, our tree node looked something like this –

class Node {
    Node[] children = new Node[26];
    boolean isWordEnd;

So, in every node, we had an array of references. The first references corresponded to ‘a’, the second to ‘b’, and so on. So, essentially, we had a mapping of alphabets to references. We had a way of saying, “An edge ‘a’ is denoted by this particular element in the array of references”. Now, in a compressed trie tree, the edges are labeled by strings. Now, we need a way of saying, “An edge ‘face’ is denoted by this particular element in the array of references”. To accomplish this, we re-design our tree node as follows –

class Node {
    Node[] children = new Node[26];
    StringBuilder[] edgeLabel = new StringBuilder[26];
    boolean isEnd;

So, what we did is that we added an additional array of Strings along with the array of references. Now edgeLabel[0] will denote the string starting with ‘a’, edgeLabel[1] will denote the string starting with ‘b’, and correspondingly, children[0] will denote the edge with the label edgeLabel[0].

Example, in the above diagram, the root node will have edgeLabel[5] = “face” and children[5] will point to the internal node. The internal node will have edgeLabel[1] = “book” and children[1] will point to the leaf node which will denote the occurrence of the word “facebook”. The same internal node will also have edgeLabel[15] = “palm” and children[15] will point to the leaf node which will denote the occurrence of the word “facepalm”. The rest of the values of edgeLabel and children in the internal node will be null.

The above code is written in Java. For Java, it is much better to use StringBuilder rather than String because we would be doing a lot of string manipulation. Using String will heavily slow down your program. If you are not familiar with StringBuilder, you can refer to my post.

insert() operation

All operations in the compressed trie tree are similar to what we would do in a regular trie tree. Insert operation is the one which will differ the most. In the insert operation, we need to take care of a few cases, they are –

  1. Inserting new node for a new word – This occurs when the starting character of the word is new and there’s no edge corresponding to it. This may occur at root, or after traversing to an internal node.compressed-trie-tree-4
  2. Inserting a prefix of an existing word – Inserting prefix into compressed trie tree
  3. Inserting a word which has a partial match to an existing edge – This occurs when we are trying to insert “this” when “there” is already inserted into the tree. Remember that “there” can have further children too, like if “thereafter” and “therein” are already inserted.breaking words during compressed trie tree insertion

So, for these cases, we would have to break the existing word or the newly inserted word accordingly. The faster we perform these string operations, the faster the insert operation will be.

search() operation

Searching in a compressed trie tree is much like searching. Here, instead of comparing a single character, we compare strings. The following cases will arise –

  • The string we want to search does not exist. Example, searching for “owl” when the tree only has “crow” in it.
  • The string we want to search exists as a prefix. Example, searching for “face” when the tree only has “facebook”.
  • Only the prefix of the target string exists. Converse of the previous case. Searching for “facebook” when the tree only has “face”.
  • The string we want to search matches partially with an existing string. Example, searching for “this” where the tree only has “there”.
  • Another case is when the edge label fully matches to the starting portion of the string. Example, searching for “thereafter” when “thereafter” and “therein” exist in the tree. For this, after a full match with “there”, we traverse to the node which corresponds to that label and then resume our search (searching for “after”).

If we are able to fully traverse the tree via the target string and arrive on a tree node, we check if that node is a word ending or not. If yes, we return true or, we return false. For rest of the cases, return false.

startsWith() operation

The startsWith() operation is a popular operation performed on the compressed trie tree which checks if there’s any word in the tree which starts with the target word. This method would be exactly as the searching method. The minor change with the search operation would be, in this operation, we will just check if we are able to fully traverse the tree via the target string and arrive on a node (which may be the root). If we can we return true, regardless of whether the current node is a word ending or not. This is because, even if it is not a word ending, its children will lead to a node which would be a word ending.

Printing the compressed trie tree

For each edge traversed, add its label to a variable and recursively do this for the child node. While traversing another edge, remove the previously added label and add the new label of the new edge traversing. Print the variable only if the current node is a word ending.

This recursive method should print all the words in the compressed trie tree in a lexicographical order.


Start with your existing trie tree code, modify the node definition and then work on the insert method cases. Once you get the insert correctly, then the rest will work out easily. For the insert cases, you just have to do the string operations and change the references carefully. Try to code those cases. Come back and have a look at the diagrams if you need to.

You can check your code’s correctness with LeetCode’s Implement Trie problem. Try to solve that question using a compressed trie tree. Once you solve it, try to reduce the runtime of your program.

You can refer to my code if you get stuck. 🙂


This is the Java implementation. I will update this post with the C/C++ implementation soon.

In terms of runtime complexity, compressed trie tree is same as that of a regular trie tree. In terms of memory, a compressed trie tree uses very few amount of nodes which gives you a huge memory advantage especially for long strings with long common prefixes. In terms of speed, a regular trie tree would be slightly faster because its operations don’t involve any string operations, they are simple loops.

I hope my post has demystified everything regarding a compressed trie tree. Tutorial and code for a compressed trie tree are hard to find. I hope my post saved you the effort of finding further tutorials. Do comment below if you have any doubts. Keep practising! Happy Coding!! 😀

Binary Heaps and Heapsort Algorithm

Hello people…! In this post I will talk about the Binary Heaps and Heapsort Algorithm implemented using structures and not arrays. This is post is the successor to my previous post on Binary Heaps. Using arrays to code Binary Heaps is very comfortable for the programmer. But it is not fit for the day-to-day applications. This is because the Binary Heap is used to implement the Priority Queue and it is meant to grow, not stay of fixed size. But given that it is meant to grow, we cannot use STL Vector for this. This is because the Vector keeps re-allocating memory as it grows. So, it is slow. Here we will learn the proper structure based implementation of the Binary Heaps.

Now, currently we have the knowledge of the Binary Heap, its structure, working and other properties. But we implemented this using arrays. Now if we were to implement using a structure, there is one challenge that we will face. That is to ensure the Structural Property of the heap. Why? For insertion into an array, we created a node at the end which itself ensured the Structural Property is not violated and then we bubbled upwards to ensure that the Heap Property is not violated at any point. In the array, we knew exactly where we are supposed to insert the node. But in a tree coded using structures how will you find where to insert the new element such that the tree remains a complete binary tree..? The same goes when we want to delete the root. How do we know where the last node is. For this, we will use the binary representation of the size of the heap to traverse to the location we want.

Size does matter !

While inserting itself we could keep a track of the heap size. So, this takes constant time. While inserting, we must search for the appropriate location. For this, we will convert the size into its binary representation and begin the traversal from the root. How? Keep reading the bits of the binary representation you just generated, if you encounter ‘0’ go to the left child of the parent, if it is a ‘1’ go to the right child of the parent. I know it sounds really weird, but that’s the best thing we can do. Let’s take up an example to understand this better. Consider the Binary Heap below –

Binary Heap Insertion - 1

Binary Heap Insertion – 1

Now consider the binary representation of the number (HeapSize + 1) = 8, which is “1000”. We omit the first bit (left-to-right) because that bit says “come to the root”.

  • The Binary String left is “000”. The leftmost bit is ‘0’. So, go to the Left Child. You come to Node 4.
  • The Binary String left is “00”. The leftmost bit is ‘0’. So, go to the Left Child. You come to Node 1.
  • The Binary String left is “0”. There is only one bit left. This bit doesn’t tell you to “go left” or “go right”, it tells you to “insert it to left” or “insert it to right”. As this bit is “0”, we place the new node as the left child of Node 1.

This is how you must use the binary representation of the (HeapSize + 1) to get to the location you want to insert the node. This is process is depicted in the sketch below –

Binary Heap Insertion - 2

Binary Heap Insertion – 2

Every time you insert a node, you will go all the way down the tree, you will not perform any insertions in the middle, so the traversal always takes O(log n), once inserted, you travel up, ensuring that the heap property holds good, so you have to travel all the way up again, so it is another O(log n) operation. So, the over all process takes O(log n), which is same as in the case of an array implementation.

But for deletion, we are supposed to find something else. We are supposed to find out the last node. In the sketch above, before the insertion of 8, we are supposed to somehow find out that the last node is Node 5. There are a couple of methods by which we can accomplish this –

  1. BFS or DFS – By using the Breadth First Search or the Depth First Search we can find out which is the last node in the Binary Heap by searching the whole tree for a node at the maximum depth which is encountered as last as possible. The time complexity of BFS and DFS is O(|E| + |V|). We would have N vertices and what about the number the edges? We would have N – 1 edges, because every node is pointed by a child pointer, except the root. So, the overall complexity becomes O(N + N), which is O(N).
  2. Binary Traversal – This is the same traversal as we did in the case of Insertion. I just called it Binary Traversal. But the method which we discussed about gave us the Node to be inserted but not the last node. Can you think of a way of modifying the procedure such that it would give us the last node directly…? Think about this for a couple of minutes and you will figure out the simple logic behind it. For those who can’t, don’t worry I will discuss it, when I do, try to figure how silly a thing you missed…! The logic is simple, you used the binary representation of (HeapSize + 1) to get to the location where you were supposed to insert a node, if you use the binary representation of (HeapSize) itself for traversal, you will end up at the last node…! Try working out this on paper and you will realise the logic behind it. As, we would traverse the whole tree from root to leaf, this takes O(log N), which is way faster than BFS and DFS which was an O(N) operation. This kind of Binary Traversal is depicted in the sketch below –
Binary Heap Insertion - 3

Binary Heap Insertion – 3

The next thing that we must know about the structure implementation is that unlike the array implementation, where we could construct the heap using build heap procedure, we cannot do that here. The only way to construct the heap from a given collection of numbers is to keep inserting them one-by-one. As we discussed that in insertion, we would have to go all the way down and come up, inserting a node in the structure implementation will be slower than inserting into a heap implemented using arrays.

So, now let us plan what we would be writing as our code. First things first..! The structure of our nodes in the heap would be –

struct node
	int value;
	struct node * parent;
	struct node * leftChild;
	struct node * rightChild;

Apart from the standard variables that we see in many structure related codes, we would be needing the parent pointer here. This is because, without that, we cannot travel up the tree, which is needed in insertion and deletion.

Heapsort – If you have understood the concepts of constructing the Binary Heap, finding the Last Node and Extracting maximum element, then Heapsort becomes obvious. You just have to keep extracting the maximum element and store it in an array or a list. Then, you will have a sorted array. This essentially destroys the entire heap.

Take a break and come back to code the binary heap. If you have already coded the binary search tree or a binary tree before this should be relatively easy for you. If you have not, then this will be challenging, and I’m sure you’ll be proud of your efforts once you have successfully coded this data structure. You will spend 80% of the time debugging your code, and trust me that’s the way programming is done. I can’t discuss the issues that you might encounter while coding but you can always comment them if you need help. Once you are able to code Insertion and Deletion, Heap Sort should be easy. If you get stuck, you can always refer to my code below –


The code is big and that is how it should be and moreover, it is well commented with information. If you still have doubts regarding the code or any concept, feel free to comment them. I hope my post has helped you to get a complete idea about the Binary Heap. If it did, let me know by commenting…! Keep practising… Happy Coding…! 🙂

Segment Trees

Hello people…! In this post I will talk about one of the most versatile data structure in competitive programming, the Segment Tree. Almost in every contest that you do in the major coding websites, at least one question is asked which uses a Segment Tree in its solution. This is because it is easy to code, and does some useful operations quickly, in logarithmic time. Segment Trees are used where the problem demands these two queries on a given array arr[1 … n] –

  1. Sum of elements in the range arr[L … R]. Where, 1 <= L <= R <= n.
  2. Updating an element arr[i].

You can think of doing this in the naive approach where you directly operate on the input array itself. You could do the first operation in O(n) time by running a loop calculate the sum. The second operation could be done in O(1) time, because, it’s only simply updating. This is good if the number of sum queries asked are very less and the update queries are more.

The next best thing you could think of is, Prefix Sum Array. The Prefix Sum Array is an array where the element arr[i] holds the sum of values from arr[1] to arr[i]. You can construct the Prefix Sum Array in O(n) time and solve the first query in O(1) time. Prefix Sum Array is one of my favourite techniques to find out the sum in a given range. But the update query takes O(n) time. If you want to update the ith element, you’d have to update all the elements from i to n, which takes O(n) in the worst case.

How about doing both of them in O(log n) time…? That’s really quick..! This is the speciality of the Segment Tree. We can get the sum query and the update query done in logarithmic time. The Segment Tree works on the principle of partial sums. If you have read my post on Fenwick Trees (or) Binary Indexed Trees, or if you already know BIT, Segment Tree should be easy for you. For those who don’t know, don’t worry, you’ll have an easier time learning the Binary Indexed Trees than compared to the others. So, let’s get started..!

As I mentioned earlier, the Segment Tree works on the principle of partial sums. What this principle is that we somewhere hold the sum of a few elements and keep building on it. In a Segment Tree, this “partial” is actually “half”. The Segment Tree structurally is a Binary Tree. Every internal node is associated with a range of numbers, whose sum is stored in that node. Now, what we do is partition the array into two halves. The left child will hold the sum of elements in the left partition and the right child will hold the sum of elements in the right partition. You may not get the idea of Segment Tree in the first reading. That’s okay. Give another reading and ponder upon the sketch below for a better understanding.

Structure of Segment Tree

Structure of Segment Tree

I hope it is clear from the picture how the Segment Tree works on the partial sums, which here is the left partition sum and the right partition sum. The leaf nodes represent the individual elements of the input array.

Note that the array we will not store the array in the nodes of our Segment Tree as was depicted in the sketch. This was done in the sketch only to give you the idea. The array will not be stored in the nodes. However, we will store the range limits on which the sum is defined, i.e, we will store the start index and the end index of the span of elements in the input array whose sum is there in the node. We will use a struct for the node which looks like this –

struct node
	int value, low, high;
	struct node * leftChild;
	struct node * rightChild;

The variable ‘value’ was to store the sum, ‘low’ and ‘high’ is to store the range. The variables ‘leftChild’ and the ‘rightChild’ are to point to the children.

Building the Segment Tree –

The Segment Tree is built in a recursive manner. We could write it in a step-by-step procedure –

  • Allocate memory.
  • Start with the root.
  • Partition the array you are given into two halves.
  • Create nodes for each half.
  • The value of the parent node is the sum of values on its children.
  • This is calculated by recursively calling the procedure on the children.
  • When the array given reduces to one element. The procedure must create a node and return the value.
  • When the parent node has acquired the sum of its children, it must return its current value to carry on the recursion.

The step-by-step procedure may not be clear in the first reading. You will know what I was talking about when you understand the pseudo-code below –

int buildSegmentTree(Array, low, high, SegmentTree)
    if low == high
        SegmentTree->value = Array[low]
        SegmentTree->low = SegmentTree->high = low
        return SegmentTree->value
    mid = (low + high) / 2
    SegmentTree->low = low
    SegmentTree->high = high
    SegmentTree->leftChild = allocate memory using calloc()
    SegmentTree->rightChild = allocate memory using calloc()
    SegmentTree->value = buildSegmentTree(Array, low, mid, SegmentTree)
                       + buildSegmentTree(Array, mid + 1, high, SegmentTree)
struct node * getSegmentTree(Array, ArraySize)
    if ArraySize <= 0
        return NULL
    struct node * SegmentTree = allocate memory using calloc()
    buildSegmentTree(Array, 1, ArraySize, SegmentTree)
    return SegmentTree

I made the pseudo-code a little extra-descriptive so that you can understand the step-by-step procedure and the code simultaneously. If you have any doubts, feel free to comment them.

The time complexity of the code is O(n). This is because in the worst case we would end up building a complete binary tree which will have 2 ✗ n – 1 nodes, and we preform calculations that take O(1) time for every node.

Sum Query –

Now, it is time to discuss the two important queries that we started our discussion with. We will begin with the sum query. We are supposed to tell the sum of all the elements in a given range. Now we will use our ‘low’ and ‘high’ variables in each node of the Segment Tree. The way we do this query mimics the Binary Search. We would be given a well defined range and we will keep comparing that with the range in each node of the Tree. If it fits perfectly, then we simply return the value stored in that node. If it doesn’t we have a few cases to handle. To understand the cases, consider the following notations –

  • NodeLeft – This is the left bound of the range whose sum is stored in the node. We actually named this as ‘low’ inn our struct.
  • NodeRight – This is the right bound of the range whose sum is stored in the node. We actually named this as ‘high’ inn our struct.
  • QueryLeft – This is the left bound of the range whose sum we are supposed to calculate.
  • QueryRight – This is the right bound of the range whose sum we are supposed to calculate.
  • Mid – This is the index upon which the partition occurs to assign values to the children nodes. It is essentially the mean of NodeLeft and NodeRight.
  • Value – This is the value that the current node is holding. Same as the ‘value’ variable in our struct.

Now, let’s talk about the cases we would encounter –

  1. (NodeLeft == QueryLeft && NodeRight == QueryRight) – This is the best thing that can happen. As we discussed a moment ago, we simply return the Value.
  2. (NodeLeft >= QueryLeft && NodeRight >= QueryRight) – This is the situation where among the elements that the current node holds the sum of, a few of the elements are not needed. The range that the current node has covers more than that’s needed. Now, to proceed with the recursion, we use Mid –
    • (QueryLeft <= Mid) → getSumQuery(SegmentTree->leftChild, QueryLeft, Mid) – This condition checks if the query demands the sum of elements which are in the left partition. So, we carry on the recursion with query range concerning only the left partition.
    • (QueryRight > Mid + 1) → getSumQuery(SegmentTree->leftChild, QueryLeft, Mid) – This checks if the query wants elements of the right partition. So, we carry on the recursion with query range concerning only the right partition.

    Note that both the conditions must be checked for the sum query to give accurate results. A special case arises when (QueryLeft == QueryRight), which means that we are looking for one element, which would obviously be either in the left or the right partition. In that case, we don’t have the necessity of  checking the second condition if the first condition is already met. This is because if we know the query asks for an element in the left partition, we have nothing to do with the right partition in this case.

This will be more clear once you put this idea to work by pen-and-paper. Coding this is simple too. Once you are able to code it, you will realise its resemblance to Binary Search. In the worst case, we would have to travel all the way down the tree, which takes O(log n).

Update Query –

What do we do if we were to update an element initially given in the input. How would we have to modify our Segment Tree so that it is still correct? We know that the initial input elements can be found at the leaves of the Segment Tree. So, we should be updating that. How do we traverse down to that leaf? We do this similarly as we did it in Sum Query, where we took a range and went down the tree based on it. If we put it in Sum Query terms, QueryLeft is equal to QueryRight. We search for it in a similar fashion. Then, we update the node which we came searching for. We also keep a note of the difference, and we add the difference to all those nodes which we came by. This is because, when you update a leaf of the Segment Tree, the internal nodes which would get affected are those which contain the partial sum in which the leaf is involved. We already met those nodes in our recursive traversal. When we are “coming out” of recursion, we update them by the difference that was caused by the new value. So, basically we go down the tree and come back. So, this operation takes O(2 log n) which is O(log n).

Take a break for a while and come back to start coding the Segment Tree. You will need to do a lot of pen-and-paper work while coding this data structure. And it is totally worth it. To print your Segment Tree, you could use the Tree Walks, In-order or Pre-order or whichever pleases you. If you are getting stuck, you can always refer to my code below –


There are many varieties of questions that can be asked based on this Data Structure. I’m sure you’ll start liking and appreciating this data structure as you keep solving many questions on it. If you have any doubts, feel free to comment them. I hope my post has helped you in understanding the Segment Tree. If it did, let me know by commenting…! Keep practising… Happy Coding…! 🙂

Trie Tree Implementation

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Hello people…! In this post we will talk about the Trie Tree Implementation. Trie Trees are are used to search for all occurrences of a word in a given text very quickly. To be precise, if the length of the word is “L“, the trie tree searches for all occurrences of this data structure in O(L) time, which is very very fast in comparison to many pattern matching algorithms.

But I must mention, this data structure is not exactly used for “pattern matching”, it is used to search for the occurrences of the word in the given text. How these both functionalities differ…? We’ll get to know that shortly. The Trie Tree has many applications. Your browser could be using a trie tree internally to search for words when you press Ctrl + F. So, let’s get started with this data structure…!

The Trie Tree is a very straight forward data structure. It is a simple tree where the nodes have an alphabet associated with them. And the way these nodes are arranged is the best part of the Trie Tree. To get an idea take a close look at the sketch below –

Structure of Trie Tree

Structure of Trie Tree

The arrows in the sketch above indicate how to traverse the trie tree in order to tell if a word exists or not. We travel through the root node, down the tree until we hit a leaf. Picking up a character at every edge, we construct our word. Now, you can easily tell why the time to search for a node in the text will be in the order of length of the word to be searched. It’s obvious…! One would have to go down till the leaf. In the worst case, one would have to travel the height of the tree which would be the length of the longest word in the whole text..! 😉

So, as we got a little idea about working with a Trie Tree, let us move on to the serious part. Let us talk about how this is implemented and then we will talk about three fundamental operations done on the Trie Tree –

  • Insert
  • Delete
  • Search

In C, the Trie Tree is implemented using structures. But the C implementation of this data structure gets a little dirty. So we will switch to C++ but we will keep it as C-ish as possible. As we keep discussing about the implementation, you will notice how many advantages we have when we use C++. This will be our structure of each node in the tree –

struct node
    struct node * parent;
    struct node * children[ALPHABETS];
    vector<int> occurrences;
  • parent – This points to the node which is the parent of the current node in the tree hierarchy. It may seem useless at the first, but we need this to travel up the tree when we want to delete any word. We’ll get to the deletion operation in a moment.
  • children – It points to the children nodes. It is made an array so that we can have O(1) accessing time. But why is the size 26…? Its simple. Consider a word, “th”, now, what could the third letter possibly be, if it had one…? One among the 26 english alphabets…! So that’s why the size is made 26. If you want to make a trie tree for another language, replace the number 26 with the number of letters in your language.
  • occurrences – This will store the starting indices of the word occurring in the given text. Now, why this is made a vector is that, vector is as good as a Linked List with random access. It is one of the most handy ready made data structures available in the C++ STL Library. If you are not familiar with vectors this is a good place to start.
    If this were C, we would have to give a fixed array size, and we would have a problem if the occurrences of a particular node are more. We could avoid this by putting a Linked List there. But we sacrifice random access and a whole lot of operations get time taking. Moreover, the code will get really really cumbersome to manage if you have a tree and a linked list.

Having got a picture of the implementation, let us look at how the operations are done in a Trie Tree.

Insert Operation

When we do an insert operation, there are a few cases –

  1. The word to be inserted does not exist.
  2. The word to be inserted already exists in the tree.
  3. The word to be inserted does not exists, but as the suffix of a word.

The first case is simple. One would have to traverse till the alphabets of the words have nodes in the trie tree or else create new nodes one-after-the-other. And at the end of the word, i.e., the node for the last alphabet, we will mark it as a leaf and push the starting index into the vector indicating the occurrence of the newly inserted word.

During this course of traversal, we will be cutting off the string of the word we have one-by-one as they are processed. This is done by putting using a vector of characters and popping off one character after the other. This is less code to handle and more efficient as we can use a vector as a queue. This is another advantage of using C++.

After having learnt what to do with the first case, you can guess what we would have to do in the second case. We simply have to add a new value to the occurrences vector at the node corresponding to the last alphabet of the word. We can also know the number of occurrences in constant time, we simply return the size of the vector. This is another advantage of using C++.

To understand the challenge in the third case, let’s take a simple example. What would you do with your trie tree if you wanted to insert the word “face” if the word “facebook” is already there in your tree…? This is the third case. The answer to this is the occurrence vector itself. We simply push the starting index of the word into the vector of that node which corresponds to the last alphabet of the word to be inserted, in the above example, this would be the node of “e”. So, what really tells if there’s a word ending with the alphabet corresponding to the current node is the size of the vector.

So I hope you understand how important our vector is. A lot depends on it…!

Delete Operation

The deletion of a word in the trie tree is similar to the insertion, we have a few cases –

  • Error 404 : Word not found…!
  • Word exists as a stand-alone word.
  • Word exists as a prefix of another word.

If the word is not there at all, we don’t have to do anything. We just have to make sure that we don’t mess up the existing data structure…!

The second case is a little tricky. We would have to delete the word bottom-up. That is, we will delete that part of the word which is not a part of any other word. For example, consider the sketch above. If we were to delete “this”, we would delete the letters ‘i’ and ‘s’ as, ‘h’ is a part of another word. This keeps us away from distorting the data structure. If the word were existing multiple number of times we will simply remove the occurrence from the vector of the concerned node.

In the third case too, we will simply delete the occurrence of the word from the vector. We needn’t write a lot of code as we can use the functions in algorithm header file. This is another advantage of using C++.

Note – When we delete the occurrence of a word, we are not concerned about the validity of the indices stored as occurrences of other words. What I mean to say is, suppose we have 10 words. If we delete the 3rd word, the 5th word or the 9th word is supposed to become the 4rth and the 8th word as far as the original text is concerned. But we will not consider this. The data stored in the trie tree is not meant to me deleted or inserted. The Trie Tree is meant for processing the given text not to manipulate the given text.

Search Operation

The search operation is simple and is as we discussed when we began our discussion about the Trie Tree. We go down the tree traversing the nodes and keep “picking up characters” as we go. And the occurrences vector tells us if a word exists that ends with the alphabet associated with the current node, and if so, it gives us the indices of occurrences and also the number of occurrences.

Besides these basic operations, there is another very interesting operation that is done with the Trie Tree –

  • Lexicographical Sort – If we want to print all the words processed into the trie tree lexicographically, all we have to do is do a Preorder Walk on the tree. This will automatically print all the words in the lexicographical order or the dictionary order. This is due to the very structure and arrangement of nodes in a Trie Tree. Now, I’d like to share another interesting thing about pre-order walk in trees… The Pre-order walk works exactly as a Depth First Search (DFS) in graphs, i.e., the sequence in which both the algorithms visit the nodes is the same. Think about this for a while and word out the two algorithms on an example (you could take mine in the sketch above), and you can see why it is so. You will also notice why the printed words would be lexicographically sorted.

Now, having learned a lot about the trie tree, try coding it in C++. If you are uneasy with C++, you can try it in C, but make sure you try at least 3 times. Trying is very important. I don’t know if you are new to reading my posts, but I insist a lot on trying in every post of mine…! If you have succeeded, you’re fabulous…! If not, check out my code below any try figuring out how just close you were…!!


The code is highly commented with explanation. It is well described, but if you have any doubts regarding the data structure or the code, feel free to comment them. I have used a few macros. The macro CASE indicates for which case the Trie Tree works. If we mention ‘A’ as the macro, the Trie Tree will work for upper case words only.

Other Implementations

The code is well tested against fairly large input. You can download the test case file here – Trie Tree Input (PDF). You can just clock your code for the insert operations. My code took 1.236 seconds to execute that loop which reads a word and inserts it into the Trie Tree. There are 5000 words in total. The last word in the input file is the word to be deleted.

If you think you can give a suggestion to make my code better, please do comment them too. I appreciate your suggestions. For those there who are struggling to get their code right, “Keep the effort going guys”…! Remember, you won’t learn anything if you keep writing Hello World program again and again. So, keep practising…! I hope my post has helped you in getting to know about the Trie Tree. If it did, let me know by commenting! Happy Coding…! 😀

Binary Indexed Tree (or) Fenwick Tree

Every programmer needs a suitable data structure to compute some complex function or to do a time-taking task quickly. In this post, I will explain the Binary Indexed Tree (BIT) data structure also known as the Fenwick Tree in post-soviet countries. This data structure is very easy to code in competitions. So, a programmer who knows BIT will have a massive edge over others. But the problem with BIT is that it is NOT easy to understand! A beginner may go nuts in trying to understand the working of a Binary Indexed Tree. I will explain this data structure step-by-step using as many images, tables and examples as I can.

Binary Indexed Tree becomes handy in manipulating Cumulative Frequency Tables. Now, I assume you know what Frequency and Cumulative Frequency is! Now, consider we have 12 objects with us of the respective frequencies –

Item No. Frequency Cumulative Frequency
1 4 4
2 2 6
3 7 13
4 5 18
5 1 19
6 3 22
7 6 28
8 4 32
9 6 38
10 6 44
11 3 47
12 3 50

Now, if we were to increase the quantity of Item 5 by 1. We would have to change the whole table. The change would be like,

Item No. Frequency Cumulative Frequency
1 4 4
2 2 6
3 7 13
4 5 18
5 1 + 1 19 + 1
6 3 22 + 1
7 6 28 + 1
8 4 32 + 1
9 6 38 + 1
10 6 44 + 1
11 3 47 + 1
12 3 50 + 1

If we were to code this manipulation using an array of integers, the implementation would go like,

Array Element Element Value
arr[0] 4
arr[1] 6
arr[2] 13
arr[3] 18
arr[4] 19 + 1
arr[5] 22 + 1
arr[6] 28 + 1
arr[7] 32 + 1
arr[8] 38 + 1
arr[9] 44 + 1
arr[10] 47 + 1
arr[11] 50 + 1

Clearly, implementing the simple increment would cost O(n). This is because in Cumulative Frequency, every element holds the “prefix-sum” of the elements, i.e., the sum of elements till that index. Now what if we hold only partial sums? Say, there are three variables that hold the partial sums and the final sums can be evaluated by summing up the partial sum. Now, the question arises is, “How does this help the increment manipulation?”. This is explained by the following diagram.

Partial Sums

Partial Sums

So, we can see how by using the Partial Sum Technique, we managed to decrease the number of operations to perform. This is the raw underlying idea of the Binary Indexed Tree. But the Binary Indexed Tree uses this partial sum technique far more efficiently and quite amazingly actually! The BIT is constructed on the principle that “all numbers can be represented as the sums of powers of 2”. This is more or less similar to the principle behind the binary numbers. Now the working of the BIT is that, at indices
of the values of powers of 2, they contain the total sum of the frequencies till that index. Which means if we have an element of an array, arr[7], it will have the sum of elements from arr[0] to arr[7] (including the value of arr[7], i.e, frequency of the 8th element). Thats ok! But what will the other elements contain…?! The Partial Sums…! I make this clear with the help of my next sketch where I show a simple Binary Search Tree (BST) where the nodes have the values of the frequencies and the nodes are constructed in such a way that the value of the “index” is taken into consideration while constructing the BST but the nodes hold the frequencies as the values. Then I will give a super simple step-by-step explanation.

The Initial Binary Search Tree

The Initial Binary Search Tree

We all know that the Binary Search Tree is such that the left-subtree holds all the values less than the current node. Now, the Binary Indexed tree is such that, the current node holds the sum of all values in it’s left-subtree + the value of the current node. For example, in the first tree in the above image,

Value of Node 10 = Value of Node 9 + Value of Node 10

Value of Node 12 = Values of Node 12 + Value of Node 10 (= Value of Node 9 + Value of Node 10…!)


Value of Node 4 = Values of Node 4 + Value of Node 3 + Value of Node 2 (= Value of Node 2 + Value of Node 1…!)

So, now we see that the Node 4 will be the sum of Nodes 1 to 4, similar is the case with all the nodes that are powers of 2. These nodes are somewhat special, as they contain the “prefix-sum”. Hence, I marked them in my previous image. Now we will apply the concept to our second tree in the above image.

The Binary Indexed Tree !

The Binary Indexed Tree !

Wait…! What…?! The “array” is the “tree”…?! Yes…! The array we obtained is called the Binary Indexed Tree. For whatever applications a BIT is used for, it is used as this array. Pretty cool, uh…! So, this means you don’t have to get messed up creating structures, pointers and doing tree travels! All you’ll need is this simple array. Now, before we get into the coding part of the BIT lets observe this BIT “array” a little more closely so that we can understand it more clearly. Again, I use an image because we all know, “A picture speaks more than a thousand words!”.

Partial Sums in a Binary Indexed Tree

Partial Sums in a Binary Indexed Tree

The picture clearly shows how the Fenwick Tree (or BIT) stores the partial sums and that at indices of powers of two it will have the complete sum of all the elements behind it. What really allows the Fenwick Tree to have this feature is the very structure of the Binary Search Tree (BST). In a BST, the nodes with index in powers of two lie at the left-most part of the tree (recall our very first tree of the first image), now, we know for a Binary Indexed Tree, the value of a node is the sum of the components of the left-subtree + the value that is supposed to be in that node. But what will the left-subtree of a Binary Search Tree have?… “Some” elements surely less than it. Now, what will the left-subtree of the left-most wing of a tree have? “All” the elements surely less than it. If you ponder on this statement for a while you will understand why the indices of the powers of 2 have the whole “prefix-sum”. Now I will depict you the effect of increment of an Item in the Binary Indexed Tree, for this I use the tree representation once again.

Incrementing an item in BIT

Incrementing an item in BIT

As we can clearly see, in a worst case scenario, such as the above, we’d have to update all the nodes beginning from the lowest level and go up till the root node. So, in a worst case scenario, we’d have to update the “height of the tree” number of nodes, which is log n as we know it. So, the increment operation here costs only O(log n), which is very fast! Recall that our brute force or the naive approach costed us O(n), so now we know how using BIT speeds up the manipulation. But how do we do this in an array? For a given frequency table, how do we construct the BIT? We will see this coding part now, which is also difficult to understand but easy to implement.

For a given frequency array freq[] of size ‘n’, create another array bit[] of size ‘n’ and initialize bit[] to zero. Now, run a loop where we take every element of freq[] one-by-one and put it into bit[]. What actually are we doing is that we are doing the increment operation over and over again. Look over the code closely –


You might feel uncomfortable with the “j = j + (j & (-j))” thing, for now, forget that, we’ll come back to it in a minute. Look what is logically happening in the code. Being able to read raw code is a gift, if you can, you’re awesome! Be proud of it! For those who can’t, here is what happening. Listen carefully!… First we take the collection of items to be empty, so, all items have zero frequency. Then, we take the first item, it’s frequency is 4, so we “increment” the frequency of that item from 0 → 4. Now, recall the picture of the effect of increment in Fenwick Trees. I drew a Binary Search Tree and incremented the leaf node and up till the root by 1. Here, we will increment by 4, all the corresponding log n nodes. For better understanding watch the picture below closely –

Working of a Fenwick Tree

Working of a Fenwick Tree

It is clear from the explanation how the Fenwick Tree gets going. But how does one traverse so strangely, I mean, the first Item we had to go like, 1 → 2 → 4 → 8, then for the next Item we had to go like, 2 → 4 → 8, and then like 3 → 4 → 8…! How does one put it in a code? This is exactly what “j = j + (j & (-j))” does! It takes you to the next location where you are to increment the value. It is a standard expression for traversing a Fenwick Tree. If you are really eager to understand the logic behind the standard expression, you have to read this wonderful blog by Devendra Agarwal. It was where I was able to start my BIT journey. Lastly, I will just show you an output of the same above code but with little modification so that it prints the values of variables and BIT array step-by-step.

If you have understood till here, you will be able to understand the output without breaking a sweat. I will conclude this post here. It has become too large already, and I guess my reader’s would also like a break. Thanks a lot for being so patient with me! If you really are eager to do some questions on this topic, go for Insertion Sort Advanced Analysis of HackerRank. I enjoyed solving that question, and I hope you will too! Feel free to comment your doubts..! Keep practising…! Happy Coding…! 🙂