1096. Maximum Sum Of Two Non Overlapping Subarrays¶

Difficulty: Medium

1096. Maximum Sum of Two Non-Overlapping Subarrays

Medium

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

Solution¶

class Solution {
    public int maxSumTwoNoOverlap(int[] nums, int firstLen, int secondLen) {
        int n = nums.length;
        int current_maxi = 0, current_sum = 0, start = 0;
        for (int i = 0; i < firstLen; i++) current_sum += nums[i];
        if (firstLen + secondLen < n) current_maxi = Math.max(current_maxi, current_sum + solve(nums, firstLen + 1, n - 1, secondLen));
        for (int i = firstLen; i < n; i++) {
            current_sum += nums[i];
            current_sum -= nums[start++];
            if (start >= secondLen) current_maxi = Math.max(current_maxi, current_sum + solve(nums, 0, start - 1, secondLen));
            if (i + 1 + secondLen < n) current_maxi = Math.max(current_maxi, current_sum + solve(nums, i + 1, n - 1, secondLen));
        }
        return current_maxi;
    }
    private int solve(int arr[] , int low, int high, int len) {
        int maxi_sum = 0, current_sum = 0, start = low;
        for (int i = low; i < low + len; i++) current_sum += arr[i];
        maxi_sum = Math.max(maxi_sum , current_sum);
        for (int i = low + len; i <= high; i++) {
            current_sum += arr[i];
            current_sum -= arr[start++];
            maxi_sum = Math.max(maxi_sum, current_sum);
        }
        return maxi_sum;
    }
}

Complexity Analysis¶

Time Complexity: O(?)
Space Complexity: O(?)

Approach¶

Detailed explanation of the approach will be added here