HackerEarth A Needle in the Haystack problem solution YASH PAL, 31 July 202414 February 2026 In this HackerEarth A Needle in the Haystack problem solution Our hacker, Little Stuart lately has been fascinated by ancient puzzles. One day going through some really old books he finds something scribbled on the corner of a page. Now Little Stuart believes that the scribbled text is more mysterious than it originally looks, so he decides to find every occurrence of all the permutations of the scribbled text in the entire book. Since this is a huge task, Little Stuart needs your help, he needs you to only figure out if any permutation of the scribbled text exists in the given text string, so he can save time and analyze only those text strings where a valid permutation is present. HackerEarth A Needle in the Haystack problem solution.#include <iostream>#include <cstring>#include <vector>using namespace std;int main(){ int t; cin>>t; char c; cin.get(c); while(t--) { char p[10002],hay[101000]; cin>>p>>hay; int lp = strlen(p); int lh = strlen(hay); int p_abc[26] = {0}; int f = 0; for(int i=0;i<lp;i++) p_abc[p[i]-'a']++; int checker_abc[26] = {0}; for(int i=0;i<lp;i++) { checker_abc[hay[i]-'a']++; } for(int i=lp;i<lh;i++) { int found = 1; for(int k=0;k<26;k++) { if(p_abc[k] != checker_abc[k]) { found = 0;break; } } if(found) { f=1;break; } checker_abc[hay[i-lp]-'a']--; checker_abc[hay[i]-'a']++; } int found = 1; for(int k=0;k<26 && !f;k++) { if(p_abc[k] != checker_abc[k]) { found = 0;break; } } if(found) f=1; if(!f) cout<<"NO"<<endl; else cout<<"YES"<<endl; } return 0;} coding problems solutions HackerEarth HackerEarth