HackerEarth The smallest string problem solution

In this HackerEarth The smallest string problem solution You are given a string S which consists of lower case Latin letters and you need to perform the following operation exactly K times:

Select any character and replace it with its next character [‘a’ with ‘b’, ‘b’ with ‘c’…. ‘z’ with ‘a’].

You need to find the lexicographically smallest string after performing exactly K operations on string S.

HackerEarth The smallest string problem solution.

#include<bits/stdc++.h>
using namespace std;
#define FIO ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0)
#define mod 1000000007
#define endl "n"
#define test ll txtc; cin>>txtc; while(txtc--)
typedef long long int ll;
typedef long double ld;
int main() {
    FIO;
    test
    {
      ll n,k,ops; cin>>n>>k;
      string s; cin>>s;
      for(ll i=0;i+1<n;i++){
          ops=('z'-s[i]+1)%26;
          if(k>=ops){
              s[i]='a';
              k-=ops;
          }
      }
      k%=26;
      ops=(s[n-1]-'a');
      ops+=k;
      ops%=26;
      s[n-1]=char('a'+ops);
      cout<<s<<endl;
    }
    return 0;
}

Second solution

t = int(input())
while t > 0:
    t -= 1
    n, k = map(int, input().split())
    s = list(input())
    for i in range(n):
        if (ord('z') - ord(s[i]) + 1) % 26 <= k:
            k -= (ord('z') - ord(s[i]) + 1) % 26
            s[i] = 'a'
    k %= 26
    s[-1] = chr(ord(s[-1]) + k if ord(s[-1]) + k <= ord('z') else ord(s[-1]) + k - 26)
    print(''.join(s))

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