HackerEarth Xsquare And Number List problem solution YASH PAL, 31 July 202415 February 2026 In this HackerEarth Xsquare And Number List problem solution Xsquare loves to play with numbers a lot. Today, he has a multi set S consisting of N integer elements. At first, he has listed all the subsets of his multi set S and later replaced all the subsets with the maximum element present in the respective subset. Now, Xsquare wonders that given an integer K how many elements in the final list are greater than ( > ) , less than ( < ) or equals to ( == ) K. To make this problem a bit more interesting, Xsquare decided to add Q queries to this problem. Each of the queries has one of the following type. > K : Count the number of elements X in the final list such that X > K.< K : Count the number of elements X in the final list such that X < K.= K : Count the number of elements X in the final list such that X == K. HackerEarth Xsquare And Number List problem solution.#include <bits/stdc++.h>using namespace std ;#define LL long long int#define ft first#define sd second#define PII pair<int,int>#define MAXN 1000001#define mp make_pair#define f_in(st) freopen(st,"r",stdin)#define f_out(st) freopen(st,"w",stdout)#define sc(x) scanf("%d",&x)#define scll(x) scanf("%lld",&x)#define pr(x) printf("%dn",x)#define pb push_back#define MOD 1000000007#define inf INT_MAX/2#define ASST(X,L,R) assert(X >= L && X <= R)int N,Q ;vector<int> A ;int power(int a,int b){ int res = 1 ; while(b){ if(b%2){ res = (1LL * res * a) % MOD ; } b /= 2 ; a = (1LL * a * a) % MOD ; } return res ;}int less_than_X(int X){ X -- ; int c = upper_bound(A.begin(),A.end(),X)-A.begin() ; return power(2,c) ;}int less_than_equals_to_X(int X){ int c = upper_bound(A.begin(),A.end(),X)-A.begin() ; return power(2,c) ;}int main(){ sc(N) ; sc(Q) ; ASST(N,1,500000) ; ASST(Q,1,500000) ; A.resize(N) ; for(int i=0;i<N;i++){ sc(A[i]) ; ASST(A[i],1,1000000000) ; } sort(A.begin(),A.end()) ; char ch ; int X; int total_subsets = power(2,N) ; while(Q--){ cin >> ch ; sc(X) ; assert(ch == '>' || ch == '<' || ch == '=') ; ASST(X,1,1000000000) ; int ans = 0 ; switch(ch){ case '<' : pr(less_than_X(X)) ; break ; case '>' : ans = less_than_equals_to_X(X) ; ans = total_subsets - ans ; if( ans < 0 ) ans += MOD ; pr(ans) ; break ; case '=' : ans = less_than_equals_to_X(X) ; ans -= less_than_X(X) ; if( ans < 0 ) ans += MOD ; pr(ans) ; break ; } } return 0 ;} Second solution#include <bits/stdc++.h>#define MAX 500005#define lli long long#define M 1000000007using namespace std;int A[500005];lli pow2[500005];void pre(){ pow2[0] = 1; for ( int i = 1; i <= 500001; i++ ) { pow2[i] = pow2[i-1] + pow2[i-1]; if ( pow2[i] >= M ) pow2[i] -= M; } return;}int main(){ pre(); int n,q,x; char s[2]; scanf("%d%d", &n, &q); assert(n<=500000); assert(q<=500000); for ( int i = 0; i < n; i++ ) { scanf("%d", &A[i]); assert(A[i] >= 1 && A[i]<=1000000000); } sort(A,A+n); while ( q-- ) { scanf("%s%d", &s, &x); assert(x>=1&&x<=1000000000); if ( s[0] == '=' ) { int idx2 = upper_bound(A,A+n,x) - A; idx2--; int idx1 = lower_bound(A,A+n,x) - A; if ( idx1 == n || A[idx1] > x ) { printf("0n"); continue; } lli ans1 = pow2[idx1]; lli ans2 = pow2[idx2-idx1+1] - 1; if ( ans2 < 0 ) ans2 += M; printf("%lldn", (ans1*ans2)%M); } else if ( s[0] == '<' ) { int idx = lower_bound(A, A+n, x) - A; idx--; lli ans1; else ans1 = pow2[idx+1]; printf("%lldn", ans1); } else { int idx = upper_bound(A,A+n,x) - A; lli ans1 = pow2[idx]; lli ans2 = pow2[n-idx] - 1; if ( ans2 < 0 ) ans2 += M; printf("%lldn", (ans1*ans2)%M); } } return 0;} coding problems solutions HackerEarth HackerEarth