Help me in solving FINALSALE problem

help
,
1

My issue

exlpain me the approach of this question

My code

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

int main() 
{
	int t;
	cin >> t;
	while(t--)
	{
	    int N, sum = 0, max = 0;
	    long long a[300000];
	    cin >> N;
	    for(int i = 0; i < N; i++)
	    {
	        cin >> a[i];
	        if(a[i] > max){
	            max = a[i];
	        }
	    }
	    
	    for (int i = 0; i < N; i++)
	    {
	        if(a[i] == max)
	        {
	            sum = sum + (2*max);
	            break;
	        }
	        else{
	            sum = sum + a[i];
	        }
	    }
	    cout<<sum<<endl;
	}
}

Problem Link: Sale Practice Coding Problem - CodeChef

2

You are assuming that the day of max sales is the best day to break down the sales. I’ll say this counter example:

A = [1,1,1]
max is 1

Then, when you get into the loop,
i = 0
A[i] = 1
A[i] == max
sum = 0 + 2*1
break

cout << 2 << endl

But that is not true. The max amount of sales is 4, because you can close the last day, which would be 1+1+2

You don’t know what day is the best to close the store, so just try each day like this:

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

int main() 
{
	int t;
	cin >> t;
	while(t--)
	{
	    long long N, sum = 0, maxSales = 0;     // Modified
	    //long long a[300000];
	    cin >> N;
	    vector<long long> a(N);                 // Added
	    for(int i = 0; i < N; i++)
	    {
	        cin >> a[i];
	        //if(a[i] > max){
	        //    max = a[i];
	        //}
	    }
	    
	    long long prevDays = 0;                 // Added
	    for (int i = 0; i < N; i++)
	    {
	        maxSales = max(maxSales, prevDays + a[i]*2);    // Added
	        prevDays += a[i];                               // Added
	    }
	    
	    cout << maxSales << endl;       // Modified
	}
}