1031. Maximum Sum of Two Non-Overlapping Subarrays

1031. Maximum Sum of Two Non-Overlapping Subarrays | Leetcode Solution

Given an integer array nums and two integers firstLen and secondLen, return the maximum sum of elements in two non-overlapping subarrays with lengths firstLen and secondLen.

The array with length firstLen could occur before or after the array with length secondLen, but they have to be non-overlapping.

A subarray is a contiguous part of an array.

Example 1:

Input: nums = [0,6,5,2,2,5,1,9,4], firstLen = 1, secondLen = 2
Output: 20
Explanation: One choice of subarrays is [9] with length 1, and [6,5] with length 2.

Example 2:

Input: nums = [3,8,1,3,2,1,8,9,0], firstLen = 3, secondLen = 2
Output: 29
Explanation: One choice of subarrays is [3,8,1] with length 3, and [8,9] with length 2.

Example 3:

Input: nums = [2,1,5,6,0,9,5,0,3,8], firstLen = 4, secondLen = 3
Output: 31
Explanation: One choice of subarrays is [5,6,0,9] with length 4, and [0,3,8] with length 3.

Constraints:

1 <= firstLen, secondLen <= 1000
2 <= firstLen + secondLen <= 1000
firstLen + secondLen <= nums.length <= 1000
0 <= nums[i] <= 1000

class Solution {
public:
    int helper(vector<int>&nums,int x,int y){
        int n=nums.size();
        vector<int>pre(n);
        vector<int>next(n);
        int sum=0;
        for(int i=0; i<n; i++){
            if(i<x){
                sum+=nums[i];
                pre[i]=sum;
            }
            else{
                sum+=nums[i]-nums[i-x];
                pre[i]=max(sum,pre[i-1]);
            }
        }
        sum=0;
        for(int i=n-1; i>=0; i--){
            if(i+y>=n){
                sum+=nums[i];
                next[i]=sum;
            }
            else{
                sum+=nums[i]-nums[i+y];
                next[i]=max(sum,next[i+1]);
            }
        }
        int ans=0;
        for(int i=0; i<n-1; i++){
            ans=max(ans,pre[i]+next[i+1]);
        }
        return ans;
    }
    
    
    
    int maxSumTwoNoOverlap(vector<int>& nums, int firstLen, int secondLen) {
        return max(helper(nums,firstLen,secondLen),helper(nums,secondLen,firstLen));
    }
};