1031. Maximum Sum of Two Non-Overlapping Subarrays (Medium)

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:

1. L >= 1
2. M >= 1
3. L + M <= A.length <= 1000
4. 0 <= A[i] <= 1000

Related Topics:
Array

Solution 1.

// OJ: https://leetcode.com/problems/maximum-sum-of-two-non-overlapping-subarrays/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(N)
class Solution {
private:
    int get(vector<int> &v, int i) {
        return (i >= 0 && i < v.size()) ? v[i] : 0;
    }
public:
    int maxSumTwoNoOverlap(vector<int>& A, int L, int M) {
        int N = A.size(), ans = 0;
        partial_sum(A.begin(), A.end(), A.begin());
        vector<int> maxLeft(N, 0), maxRight(N, 0);
        for (int i = L - 1; i < N; ++i) maxLeft[i] = max(get(maxLeft, i - 1), A[i] - get(A, i - L));
        for (int i = N - L; i >= 0; --i) maxRight[i] = max(get(maxRight, i + 1), A[i + L - 1] - get(A, i - 1));
        for (int i = M - 1; i < N; ++i) {
            int sum = A[i] - get(A, i - M)
                + max(get(maxLeft, i - M), get(maxRight, i + 1));
            ans = max(ans, sum);
        }
        return ans;
    }
};

Solution 2.

// OJ: https://leetcode.com/problems/maximum-sum-of-two-non-overlapping-subarrays/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
// Ref: https://leetcode.com/problems/maximum-sum-of-two-non-overlapping-subarrays/discuss/278251/JavaC%2B%2BPython-O(N)Time-O(1)-Space
class Solution {
public:
    int maxSumTwoNoOverlap(vector<int>& A, int L, int M) {
        partial_sum(A.begin(), A.end(), A.begin());
        int ans = A[L + M - 1], Lmax = A[L - 1], Mmax = A[M - 1];
        for (int i = L + M; i < A.size(); ++i) {
            Lmax = max(Lmax, A[i - M] - A[i - L - M]);
            Mmax = max(Mmax, A[i - L] - A[i - L - M]);
            ans = max(ans, max(Lmax + A[i] - A[i - M], Mmax + A[i] - A[i - L]));
        }
        return ans;
    }
};