LeetCode 1031: Maximum Sum of Two Non-Overlapping Subarrays Solution

Master LeetCode problem 1031 (Maximum Sum of Two Non-Overlapping Subarrays), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

1031. Maximum Sum of Two Non-Overlapping Subarrays

Medium

Array Dynamic Programming Sliding Window

Problem Explanation

Explanation

To solve this problem, we can use a dynamic programming approach. We will iterate over the input array nums and calculate the prefix sums. Then, we will find the maximum sum of subarrays with lengths firstLen and secondLen by considering all possible combinations where the two subarrays do not overlap. We will keep track of the maximum sum found so far.

Calculate the prefix sums of the input array nums.
Iterate over the array and consider all possible combinations of non-overlapping subarrays with lengths firstLen and secondLen.
Update the maximum sum if a new maximum is found.
Return the maximum sum.

Time Complexity: O(n) where n is the size of the input array nums. Space Complexity: O(n) for storing the prefix sums.

Solution Code

class Solution {
    public int maxSumTwoNoOverlap(int[] nums, int firstLen, int secondLen) {
        int n = nums.length;
        int[] prefixSums = new int[n + 1];
        for (int i = 0; i < n; i++) {
            prefixSums[i + 1] = prefixSums[i] + nums[i];
        }
        
        int result = 0;
        for (int i = 0; i <= n - firstLen; i++) {
            int sum1 = prefixSums[i + firstLen] - prefixSums[i];
            for (int j = i + firstLen; j <= n - secondLen; j++) {
                int sum2 = prefixSums[j + secondLen] - prefixSums[j];
                result = Math.max(result, sum1 + sum2);
            }
        }
        for (int i = 0; i <= n - secondLen; i++) {
            int sum1 = prefixSums[i + secondLen] - prefixSums[i];
            for (int j = i + secondLen; j <= n - firstLen; j++) {
                int sum2 = prefixSums[j + firstLen] - prefixSums[j];
                result = Math.max(result, sum1 + sum2);
            }
        }
        
        return result;
    }
}

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