In this HackerEarth Swap, It problem solution Bob loves sorting very much. He is always thinking of new ways to sort an array.His friend Ram gives him a challenging task.He gives Bob an array and an integer K .The challenge is to produce the lexicographical minimal array after at most K-swaps.Only consecutive pairs of elements can be swapped.Help Bob in returning the lexicographical minimal array possible after at most K-swaps.


HackerEarth Swap It problem solution


HackerEarth Swap It problem solution.

#include <bits/stdc++.h>
#define _ ios_base::sync_with_stdio(false);cin.tie(0);
using namespace std;
#define pb push_back
#define pob pop_back
#define pf push_front
#define pof pop_front
#define mp make_pair
#define all(a) a.begin(),a.end()
#define bitcnt(x) __builtin_popcountll(x)
#define MOD 5000000007
#define total 500005
#define M 1000000007
#define NIL 0
#define INF (1<<28)
typedef unsigned long long int uint64;
typedef long long int int64;

int main(){
freopen("input1.txt","r",stdin);
freopen("output.txt","w",stdout);
int t,n,i,k;
cin>>t;
while(t--){
cin>>n>>k;
int a[n];
for(i=0;i<n;i++)
cin>>a[i];
for(i=0;i<n;i++){
int pos=i;
for(int j=i+1;j<n;j++){
if((j-i)>k)
break;
if(a[j]<=a[pos])
pos=j;
}
for(int j=pos;j>i;j--)
a[j]=(a[j]+a[j-1])-(a[j-1]=a[j]);
k-=(pos-i);
}
for(i=0;i<n;i++)
cout<<a[i]<<" ";
cout<<endl;
}
fclose(stdout);
return 0;
}

Second solution

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<vector>
#include<cassert>
#include<set>
#include<queue>
#include<map>

using namespace std;

#define vi vector < int >
#define pb push_back
#define mp make_pair
#define ll long long
#define llu unsigned long long
#define MOD 1000000007
#define INF 1000000000
#define dbg(x) { cout<< #x << ": " << (x) << endl; }
#define all(x) x.begin(),x.end()

int a[1003];

int main()
{
int t,n,k,i,j;
scanf("%d",&t);
assert(1<=t && t<=10);
while(t--)
{
scanf("%d%d",&n,&k);
assert(1<=n && n<=1000);
assert(1<=k && k<=1000);
for(i=0;i<n;i++)
{
scanf("%d",&a[i]);
assert(1<= a[i] && a[i] <= 1000000);
}
j = 0;
while(k>0 || j<n)
{
int mn = INF,id;
for(i=j;i<n;i++)
{
if(i-j <= k && mn > a[i])
mn = a[i] , id = i;
}
for(int x=id;x-1>=j;x--)
{
swap(a[x],a[x-1]);
}
k -= (id-j);
j++;
}
for(i=0;i<n;i++)
{
printf("%d ",a[i]);
}
printf("\n");
}
return 0;
}