Jump Trip Score#13064

Jump Trip Score (Editorial)

Hint : performing a jump on index $i$ will skip the elements $i, i + 1, ..., i + d - 1$ i , i+1 , ... , i+d-1.

Solution: jump operation will skip a continuous subarray from the array $a$ , so the final score will be .

$c = i = 1 \sum n a_{i}$ c = \sum\limits_{i=1}^{n} a_i $-$ the sum of skipped subarray.

since we need to maximize the score it's optimal to skip the subarray which has the smallest sum, defining the sum of an empty subarray $= 0$ .

you can find the min subarray sum using kadane's algorithm.

Time complexity : $O (n)$ .

#include<bits/stdc++.h>
#define endl '\n'
#define int long long
#define F first
#define S second
using namespace std;
const int N = 2e5 + 1;
const int M = 1e9 + 7;
signed main()
{
    ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
    int tc = 1;
    cin>>tc;
    while(tc--)
    {
        int n;
        cin>>n;
        int a[n+1] ,mn = 0 , mx = 0 , sum = 0;
        for(int i=1;i<=n;i++)
        {
            cin>>a[i];
            sum += a[i];
            mx = max(mx , sum);
            mn = min(mn , sum - mx);
        }
        cout<<sum - mn<<endl;
    }
}