HackerEarth ICPC Team Management problem solution YASH PAL, 31 July 202414 February 2026 In this HackerEarth ICPC Team Management problem solution Little Chandan is an exceptional manager – apart from his role in HackerEarth – as the person who has to bug everyone, in general… and if possible, try to get some work done. He’s also offered a job as the coach of the best Russian teams participating in the ACM-ICPC World Finals. Now, Chandan is an extremely good coach, too. But he’s a weird person who thrives on patterns in life, in general. So, he has decided that if there are n number of students in total, and he is supposed to divide them into camps of k students – he wants them to be arranged in such a way that the length of names of all the students in a camp is equal. I know, totally weird, right? HackerEarth ICPC Team Management problem solution.tc = int(raw_input())assert(tc>0 and tc<51)for i in xrange(tc): n, k = map(int, raw_input().split()) l = [0]*1000 ok = True assert(n>0 and n<1001) assert(k>0 and k<1001) assert(n>=k) assert(n%k==0) for j in xrange(n): a = raw_input() l[len(a)] +=1 for z in l: if z%k!=0: ok = False break if (ok): print "Possible" else: print "Not possible" Second solution#include<bits/stdc++.h>using namespace std;#define ll long long int#define vi vector<int>#define vl vector<ll>#define pii pair<int,int>#define pil pair<int, ll>#define pll pair<ll, ll>#define pli pair<ll, int>#define pb(v, a) v.push_back(a)#define mp(a, b) make_pair(a, b)#define MOD 1000000007#define rep(i, a, b) for(i=a; i<=b; ++i)#define rrep(i, a, b) for(i=a; i>=b; --i)#define si(a) scanf("%d", &a)#define sl(a) scanf("%lld", &a)#define pi(a) printf("%d", a)#define pl(a) printf("%lld", a)#define pn printf("n")ll pow_mod(ll a, ll b){ ll res = 1; while(b) { if(b & 1) res = (res * a) % MOD; a = (a * a) % MOD; b >>= 1; } return res;}int cnt[105];int main(){ int t, i; si(t); rep(i, 1, t) { memset(cnt, 0, sizeof(cnt)); int n, k, j; si(n); si(k); string str; rep(j, 1, n) { cin>>str; cnt[str.length()]++; } bool flag = true; rep(j, 1, 100) { if(cnt[j] % k) { flag = false; break; } } if(flag) cout<<"Possiblen"; else cout<<"Not possiblen"; } return 0;} coding problems solutions HackerEarth HackerEarth