HarshKumarChoudary

class Solution{
    public:
    int min_sprinklers(int gallery[], int n)
    {
        int ans = 0;
        vector<pair<int,int>>v;
        for(int i = 0; i < n; ++ i){
            if(gallery[i] == -1)continue;
            int left = max(0,i-gallery[i]);
            int right = min(n-1,i+gallery[i]);

HarshKumarChoudary

int findDeterminant(vector<vector<int>>&Matrix){
        // Variable to store the determinant value
        int det = 0;
        if (Matrix.size() == 1)
        {
            return Matrix[0][0];
        }

        else if (Matrix.size() == 2)
        {

HarshKumarChoudary

/*
   // Difficulty : Hard  Prerequisite:  Inclusion-Exclusion Principles, Sieve of Eratosthenes  Time complexity: O(sqrt(n)*log(n))   If any integer has a perfect square as factor, then it is guaranteed that it has a square of prime as a factor too. So we need to count the integers less than or equal to N which have square of primes as a factor.  Let us look at the number of integers less than N which has 2*2=4 as factor. Every 4th number has 4 as factor(4,8,12,16..) so there are total [N/4] numbers who has 4 as factor, similarly there are [N/9] numbers who has 9 as factor. (Here [N] is greatest integer of N).  Now to calculate the number of integer who has both 4,9 as factor, we cant simply add [N/4] and [N/9] as we are counting numbers with 36 as factor two times. To avoid this we use Inclusion-Exclusion Principle. To avoid the double counting of number with 36 as factor ,we subtract [N/36] from our answer.  So number of integers with both 4,9 as factor=[N/4]+[N/9]-[N/36]  Similarly, the number of integers

HarshKumarChoudary

See description for link    

HarshKumarChoudary

class Solution {
    public:
    vector<int>par;
    vector<int>rank;
    int find(int a){
        return par[a] == a?a:par[a] = find(par[a]);
    }
    void unin(int a,int b)
    {
        if(a == b)return;

HarshKumarChoudary

//Geek collects the balls :- 
// https://practice.geeksforgeeks.org/problems/geek-collects-the-balls5515/1/?page=1&status[]=unsolved&curated[]=7&sortBy=submissions#
class Solution{
public:
    int maxBalls(int n, int m, vector<int> a, vector<int> b){
       int i=0,j=0;
        int s1=0,s2=0;
        sort(a.begin(),a.end());
        sort(b.begin(),b.end());
        while(i<n || j<m){

HarshKumarChoudary

class Solution {
public:
    int permutation(int m, int n) {
        // m number with len n
        // m * (m - 1) * (m - 2) .. * (m - n + 1) 
        int res = 1;
        while (n) {
            res *= m;
            m--;
            n--;

HarshKumarChoudary

int Solution::jump(vector<int> &a) {
    int jmp = 0;
    int cur = 0;
    int next = 0;
    for(int i = 0; i <= cur; ++ i){
        if(i == a.size()-1)return jmp;
        next = max(next,a[i]+i);
        if(i == cur){
            ++jmp;
            cur = next;

HarshKumarChoudary

class Solution
{
  public:
    vector<int> findOrder(int n, int m, vector<vector<int>> pre) 
    {
        vector<int>ans;
        vector<int>indeg(n,0);
        vector<int>adj[n];
        queue<int>q;
        for(auto a:pre){

HarshKumarChoudary

class Solution {
public:
    vector<int>ans;
    vector<int>cnt;
    vector<vector<int>>adj;
    void dfs(int i,int cur){
        for(auto a:adj[i]){
            if(a!=cur){
                dfs(a,i);
                cnt[i]+=cnt[a];