# HackerEarth The smallest permutation problem solution

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.

`#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)))`