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Disjiont Set Union-Find Data Structure | Code In C++

 Introduction:  In this tutorial we are going to write full program of disjoint set union find advance data structure in c++.  Problem Description: Disjoint Set Union (DSU) is an advance data structure, which basically uses in graph algorithms to find cycles. Codes:  Method1: Brute Force #include<bits/stdc++.h> using namespace std; int find(int f,vector<int>&dsuf){     if(dsuf[f]==-1)         return f;     return find(dsuf[f],dsuf); } void union_op(int from, int to, vector<int>& dsuf){     dsuf[from]=to; } bool isCycle(vector<pair<int,int> >&edge_list, vector<int>&dsuf){     for(int i=0;i<edge_list.size();i++){         int parent1=find(edge_list[i].first,dsuf);         int parent2=find(edge_list[i].second,dsuf);         if(parent1==pare...
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Construct Binary Tree from preorder and inorder | Data Structure

Introduction: In this tutorial we are going to see how we can construct the binary tree from given preorder and inorder. Prerequisites: you should know about binary tree traversal and on paper you can draw binary tree from given preorder and inorder traversal. Inorder:left->root->right; Preorder:root->left->right; Problem Statement: we have given two arrays. one for preorder and another for inorder. by using these two array we have to built a binary tree. eg: preorder = [3,9,20,15,7] inorder = [9,3,15,20,7] solution: Solution: We will follow recursive approach to solve this question.let's discuss how we can solve it. Trick: In the given preorder the very first element will be the root of the tree. then we will find root element in inorder also. and we know in inorder traversal we have left part then root and then right part of the tree. by using preorder we can get the root of the main tree and by using inorder and root we can get the left part and right part of the...
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Group Anagrams solution with explanation:Leetcode solution

what is Anagrams? suppose we have two strings and the characters present in both of the strings are same. then both will be called anagram of each other. eg:" rat" and "art" both are anagram of each other. the characters present in both string are same. eg: "raat" and "art" these are not anagrams of each other. and the reason is that the frequency of "a" in "raat" is 2. while the frequency of "a" in "art" is 1. so these are not anagrams of each other. in short two strings will be anagram of each other if they have same characters and the count(frequency) of each and every character must be same in both strings. Solution: There are different ways to solve this question, so here we are going to use Hashmap to solve this question. cause whenever there is need to search we should go with the data structure which use balanced binary tress or hashing. and we know that Hashmap  uses balanced binary trees. so using...
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