Help me in solving REMONE problem

help
,
1

My issue

let’s say we ignored Ai; lets say sum of all elements in A as a, and B as b respectively. a-Ai gives sum of all elements except Ai; now add x to each element(i.e. n-1 times) so now the sum becomes (a-Ai)+ (n-1)X = b; so X= (b-a+A[i])/(n-1). For this to be possible, we must ensure b-a+A[i] is divisible by n-1. we have to find the least value possible for x by iteration. But for some reason my approach is giving wrong answer. Some one please explain.

My code

#include <iostream>
#include<bits/stdc++.h>
#define ll long long
using namespace std;

int main() {
	// your code goes here
	int t;cin>>t;
	while(t--){
	       int n;   ll  a=0;  ll b=0;int min=INT_MAX;
	    cin>>n;
	    vector<int>A;
	    for(int i=0;i<n;i++){ int x;cin>>x;a+=x;A.push_back(x); }
	    for(int i=0;i<n-1;i++){int x;cin>>x;b+=x; }
	   // sort(A.begin(),A.end());
	    for(int i=0;i<n;i++){
	            int d= b-a+A[i];
	        if(d%(n-1) == 0 && d>0){
	            if(min>d/(n-1)){min=d/(n-1);}
	        }
	    }
	    cout<<min<<endl;
	    
	}
	return 0;
}

Learning course: Jump from 2* to 3*
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2

@gowtham78444
plzz refer my c++ code .
I have implemented it in much simpler way .
Hope u will get it. Let me know in case u get stuck at any point.

#include <bits/stdc++.h>
using namespace std;

int main() {
	// your code goes here
	int t;
	cin>>t;
	while(t--)
	{
	    long long int n,ans=1e15;
	    cin>>n;
	    long long int a[n],b[n-1];
	    for(int i=0;i<n;i++)
	    {
	        cin>>a[i];
	    }
	    for(int i=0;i<n-1;i++)
	    {
	        cin>>b[i];
	    }
	    sort(a,a+n);
	    sort(b,b+n-1);
	    long long int x=b[n-2]-a[n-1];
	    long long int x1=b[n-2]-a[n-2];
	    int j=0;
	    for(int i=0;i<n;i++)
	    {
	        if(a[i]+x==b[j])
	        {
	            j++;
	        }
	    }
	    if(j==n-1&&x>0)
	    ans=min(ans,x);
	    j=0;
	    for(int i=0;i<n;i++)
	    {
	        if(a[i]+x1==b[j])
	        {
	            j++;
	        }
	    }
	    if(j==n-1&&x1>0)
	    ans=min(ans,x1);
	    cout<<ans<<endl;
	}
	return 0;
}