HackerEarth The smallest permutation problem solution

YASH PAL, 31 July 202416 February 2026

In this HackerEarth The smallest permutation problem solution You are given an array A of N elements which is a permutation. You are required to perform the following operation exactly one time:

Select two different indices i and j (1 <= i < j <= n) and swap elements at these indices.

Your task is to find the lexicographically smallest array A you can achieve.

HackerEarth The smallest permutation problem solution

HackerEarth The smallest permutation 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; cin>>n;
      vector<ll>a(n);
      vector<ll>b(n+1);
      for(auto &it:a) cin>>it;
      ll first=-1;
      for(ll i=0;i<n;i++){
          if(a[i]!=(i+1)){
              first=i;
              i=n+3;
          }
      }
      if(first==-1){
          swap(a[n-1],a[n-2]);
      }
      else{
          ll secondval=first+1;
          ll second;
          for(ll i=0;i<n;i++){
              if(a[i]==secondval){
                  second=i;
              }
          }
          swap(a[first],a[second]);
      }
      for(auto &it:a){
          cout<<it<<" ";
      }
      cout<<endl;
    }
    return 0;
}

Second solution

t = int(input())
while t > 0:
    t -= 1
    n = int(input())
    a = list(map(int, input().split()))
    i = 0
    while i < n - 2 and a[i] == i + 1:
        i += 1
    j = a.index(min(a[i + 1:]))
    a[i], a[j] = a[j], a[i]
    print(' '.join(map(str, a)))

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