1031.Maximum-Sum-of-Two-Non-Overlapping-Subarrays

1031. Maximum Sum of Two Non Overlapping Subarrays

题目地址

https://leetcode.com/problems/maximum-sum-of-two-non-overlapping-subarrays/

题目描述

Given an array A of non-negative integers, return the maximum sum of elements in two non-overlapping (contiguous) subarrays, which have lengths L and M.  (For clarification, the L-length subarray could occur before or after the M-length subarray.)

Formally, return the largest V for which V = (A[i] + A[i+1] + ... + A[i+L-1]) + (A[j] + A[j+1] + ... + A[j+M-1]) and either:

0 <= i < i + L - 1 < j < j + M - 1 < A.length, or
0 <= j < j + M - 1 < i < i + L - 1 < A.length.

Example 1:
Input: A = [0,6,5,2,2,5,1,9,4], L = 1, M = 2
Output: 20
Explanation: One choice of subarrays is [9] with length 1, and [6,5] with length 2.

Example 2:
Input: A = [3,8,1,3,2,1,8,9,0], L = 3, M = 2
Output: 29
Explanation: One choice of subarrays is [3,8,1] with length 3, and [8,9] with length 2.

Example 3:
Input: A = [2,1,5,6,0,9,5,0,3,8], L = 4, M = 3
Output: 31
Explanation: One choice of subarrays is [5,6,0,9] with length 4, and [3,8] with length 3.

Note:
L >= 1
M >= 1
L + M <= A.length <= 1000
0 <= A[i] <= 1000

代码

Approach #1

class Solution {
  public int maxSumTwoNoOverlap(int[] A, int L, int M) {
        int[] prefixSum = new int[A.length + 1];
    for (int i = 0; i < A.length; i++) {
      prefixSum[i + 1] = prefixSum[i] + A[i];
    }
    return Math.max(maxSum(prefixSum, L, M), maxSum(prefixSum, M, L));
  }

  private int maxSum(int[] p, int L, int M) {
    int ans = 0;
    int maxL = 0;
    for (int i = L + M; i < p.length; i++) {
      maxL = Math.max(maxL, p[i - M] - p[i - M - L]);
      int m = p[i] - p[i - M];         // m = p[i] - p[i - M]
            ans = Math.max(ans, maxL + m);
    }
    return ans;
  }
}

Approach #2 Sliding Window

class Solution {
  public int maxSumTwoNoOverlap(int[] A, int L, int M) {
        return Math.max(maxSum(A, L, M), maxSum(A, M, L));
  }

  private int maxSum(int[] A, int L, int M) {
    int sumL = 0, sumM = 0;
    for (int i = 0; i < L + M; i++) {
      if (i < L) {
        sumL += A[i];
      } else {
        sumM += A[i];
      }
    }
    int ans = sumM + sumL;
    int maxL = sumL;
    for (int i = L + M; i < A.length; i++) {
      sumM += A[i] - A[i - M];
      sumL += A[i - M] - A[i - L - M];
      maxL = Math.max(maxL, sumL);
      ans = Math.max(ans, maxL + sumM);
    }

    return ans;
  }
}