programming for beginners

Introduction: In this tutorial, we are going to implement DFS  to solve another important problem. this is a good question for interview point of view. Problem Statement: Leetcode Input: [[0,0,1,0,0,0,0,1,0,0,0,0,0], [0,0,0,0,0,0,0,1,1,1,0,0,0], [0,1,1,0,1,0,0,0,0,0,0,0,0], [0,1,0,0,1,1,0,0, 1 ,0, 1 ,0,0], [0,1,0,0,1,1,0,0, 1 , 1 , 1 ,0,0], [0,0,0,0,0,0,0,0,0,0, 1 ,0,0], [0,0,0,0,0,0,0,1,1,1,0,0,0], [0,0,0,0,0,0,0,1,1,0,0,0,0]] Output: 6 Solution: So basically we are going to use DFS to solve this problem, and we will use a 2D vector to track the already visited element. let's see the code: Full Code: int helper(vector<vector<int>>& grid,int x, int y, vector<vector<int>>&visited){ if(x<0 || x>=grid.size() || y<0 || y>=grid[x].size()) return 0; if(grid[x][y]==0){ return 0; } if(visited[x][y]==1){ return 0; } visited[x][y]=1; // cout<<x...

Intoduction: In this tutorial we are going to solve a problem from leetcode, which is a good problem for applying recursion on 2D matrix or how can we apply DFS on 2D matrix. Question Link : Leetcode Input: nums = [ [ 9 ,9,4], [ 6 ,6,8], [ 2 , 1 ,1] ] Output: 4 Explanation: The longest increasing path is [1, 2, 6, 9] . Solution: So basically what we are going to do is making recursive calls in all four direction. Approach1: Recursion(this will fail in last three test cases) and that's why we will use approach2. Full Code:     int helper(vector<vector<int>>& matrix, int x, int y, int val, int flag, int temp){         if(x<0 || x>=matrix.size() || y<0 || y>=matrix[x].size()){             return temp;         }         if( flag!=0 && matrix[x][y]>val){   ...

Introduction: In this tutorial we are going to solve a problem "Target Sum" which is from leetcode. and believe me it's really a good problem to understand Backtracking(Recursion). and if you try to understand the problem as well as code you will get a clear picture of Backtracking. Problem Statement: Link To Problem You are given a list of non-negative integers, a1, a2, ..., an, and a target, S. Now you have 2 symbols + and - . For each integer, you should choose one from + and - as its new symbol. Find out how many ways to assign symbols to make sum of integers equal to target S. Input: nums is [1, 1, 1, 1, 1], S is 3. Output: 5 Explanation: -1+1+1+1+1 = 3 +1-1+1+1+1 = 3 +1+1-1+1+1 = 3 +1+1+1-1+1 = 3 +1+1+1+1-1 = 3 There are 5 ways to assign symbols to make the sum of nums be target 3. Solution: As given in question, we have two operation + and -. so we will make recursive call for + and - . and if we have sum as target and we have reached upto the last ind...

Introduction: In this tutorial we are going to solve a good question which will clear your doubts on how BFS or level order traversal is useful to solve binary tree problems. this is basically a question from Leetcode, and it is really very good problem to practice BFS. Problem Statement: Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center). For example, this binary tree [1,2,2,3,4,4,3] is symmetric: But the following [1,2,2,null,3,null,3] is not: Link : Leetcode Link please make binary tree on paper by the given array above. and you will get a clear picture of the problem. what we are going to do in this problem. Solution: As we know we are going to apply BFS to solve this problem. actually you have to just get one thing to solve this problem. we will perform BFS, but with little change or modification. So first we will store the root->val in a vector or array of the left subtree. and during this we will first insert left child and then ...

Introduction: In this tutorial we are going to solve a very famous interview problem "first unique character in the string". Problem Statement: We have a string. and then we have to find very first character present in the string which is unique. It means we have to return very first character of the string, whose frequency(count) is 1 in the string. eg: string="hehemon"      Output: m Solution: In this type of problem we prefer hashmap to easily solve the problem. so what we are going to do is, first of all we are going to create a hashmap. and we will store the character as key and index of character as value. Trick: whenever the frequency of character is greater than 1. then we will make the value part of that character -1 in the hashmap. it means now this character is not unique. it's frequency is more than 1. After iterating whole string, we will have a hashmap. In which we have characters and it's index. and if the frequency of character is more than ...