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Merge pull request Kvaibhav01#5 from CruiseDevice/master
some of the solutions for algorithms problem from hackerearth
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/*
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You have been given an undirected graph consisting of N nodes and M
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edges. This graph can consist of self-loops as well as multiple edges.
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In addition , you have also been given Q queries. For each query , you
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shall be given 2 integers A and B. You just need to find if there exists
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an edge between node A and node B. If yes, print "YES" (without quotes)
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else , print "NO"(without quotes).
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*/
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#include <iostream>
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using namespace std;
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int arr[1000][1000];
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void initialize(){
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for(int i = 0; i < 1000; i++){
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for(int j = 0; j < 1000; j++){
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arr[i][j] = false;
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}
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}
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}
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int main(){
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int x,y,n,m,q;
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initialize();
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cin >> n >> m;
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for(int i = 0 ; i < m; i++){
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cin >> x >> y;
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arr[x][y] = true;
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}
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cin >> q;
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for(int i = 0; i < q; i ++){
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cin >> x >> y;
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if(arr[x][y]== true){
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cout << "YES"<<"\n";
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}else{
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cout << "NO"<<"\n";
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}
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}
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return 0;
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}
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/*
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You are given container full of water. Container can have limited amount
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of water. You also have N bottles to fill. You need to find the maximum
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numbers of bottles you can fill.
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*/
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#include <iostream>
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#include <algorithm>
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using namespace std;
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const int MAX = 100000;
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int A[MAX];
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int main(){
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int T;
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cin >> T;
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while(T--){
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int N,Capacity;
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cin >> N >> Capacity;
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int numberBottles = 0, numberOfBottles = 0;
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for(int i = 0; i < N; i++){
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cin >> A[i];
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}
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sort(A,A+N);
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for(int i = 0; i < N; i++){
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numberBottles += A[i];
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if(numberBottles > Capacity){
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break;
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}else{
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numberOfBottles++;
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}
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}
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cout << numberOfBottles << "\n";
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}
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return 0;
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}

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