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)))