Formatted question description: https://leetcode.ca/all/1031.html

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/

// 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/

// 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;
    }
};

Java

class Solution {
    public int maxSumTwoNoOverlap(int[] A, int L, int M) {
        int length = A.length;
        if (L == M) {
            int[] rightMaxSums = new int[length - M + 1];
            int rightSum = 0;
            for (int i = 0; i < M; i++)
                rightSum += A[length - i - 1];
            rightMaxSums[length - M] = rightSum;
            for (int i = length - M - 1; i >= L; i--) {
                rightSum -= A[i + M];
                rightSum += A[i];
                rightMaxSums[i] = Math.max(rightMaxSums[i + 1], rightSum);
            }
            int maxSum = 0;
            int leftSum = 0;
            for (int i = 0; i < L; i++)
                leftSum += A[i];
            maxSum = leftSum + rightMaxSums[L];
            int end = length - M;
            for (int i = L; i < end; i++) {
                leftSum -= A[i - L];
                leftSum += A[i];
                maxSum = Math.max(maxSum, leftSum + rightMaxSums[i + 1]);
            }
            return maxSum;
        } else {
            int[] rightMaxSums1 = new int[length - M + 1];
            int rightSum1 = 0;
            for (int i = 0; i < M; i++)
                rightSum1 += A[length - i - 1];
            rightMaxSums1[length - M] = rightSum1;
            for (int i = length - M - 1; i >= L; i--) {
                rightSum1 -= A[i + M];
                rightSum1 += A[i];
                rightMaxSums1[i] = Math.max(rightMaxSums1[i + 1], rightSum1);
            }
            int maxSum1 = 0;
            int leftSum1 = 0;
            for (int i = 0; i < L; i++)
                leftSum1 += A[i];
            maxSum1 = leftSum1 + rightMaxSums1[L];
            int end1 = length - M;
            for (int i = L; i < end1; i++) {
                leftSum1 -= A[i - L];
                leftSum1 += A[i];
                maxSum1 = Math.max(maxSum1, leftSum1 + rightMaxSums1[i + 1]);
            }
            int[] rightMaxSums2 = new int[length - L + 1];
            int rightSum2 = 0;
            for (int i = 0; i < L; i++)
                rightSum2 += A[length - i - 1];
            rightMaxSums2[length - L] = rightSum2;
            for (int i = length - L - 1; i >= M; i--) {
                rightSum2 -= A[i + L];
                rightSum2 += A[i];
                rightMaxSums2[i] = Math.max(rightMaxSums2[i + 1], rightSum2);
            }
            int maxSum2 = 0;
            int leftSum2 = 0;
            for (int i = 0; i < M; i++)
                leftSum2 += A[i];
            maxSum2 = leftSum2 + rightMaxSums2[M];
            int end2 = length - L;
            for (int i = M; i < end2; i++) {
                leftSum2 -= A[i - M];
                leftSum2 += A[i];
                maxSum2 = Math.max(maxSum2, leftSum2 + rightMaxSums2[i + 1]);
            }
            return Math.max(maxSum1, maxSum2);
        }
    }
}

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