LeetCode 1031 Maximum Sum Of Two Non-Overlapping Subarrays - Medium

// 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:
 
// L >= 1
// M >= 1
// L + M <= A.length <= 1000
// 0 <= A[i] <= 1000
 
 
 
 
class Solution {
public:
    int maxSumTwoNoOverlap(vector<int>& A, int L, int M) {
        int prefix [A.size()];
        for (int i = 0; i < A.size(); ++i){
            if (i == 0){
                prefix[i] = A.at(i);
                continue;
            }
            prefix[i] = prefix[i - 1] + A.at(i);
        }        
        int res = prefix[L + M - 1], maxL = prefix[L - 1], maxM = prefix[M - 1];
        for (int i = L + M ; i < A.size(); ++ i ){
            maxL = max(maxL, prefix[i - M] - prefix[i - L - M]);
            maxM = max(maxM, prefix[i - L] - prefix[i - M - L]);
            res = max(res, max(maxL + prefix[i] - prefix[i - M], maxM + prefix[i] - prefix[i - L]));
        }
        return res;
    }
};

class Solution:
    def maxSumTwoNoOverlap(self, A: List[int], L: int, M: int) -> int:
        prefix = [A[0]]
        for i in range(1, len(A)):
            prefix.append(prefix[i - 1] + A[i])
        res = prefix[L + M - 1]
        maxL = prefix[L - 1]
        maxM = prefix[M - 1]
        for i in range(L + M, len(A)):
            maxL = max(maxL, prefix[i - M] - prefix[i - M - L])
            maxM = max(maxM, prefix[i - L] - prefix[i - M - L])
            res = max(res, max(maxL + prefix[i] - prefix[i - M], maxM + prefix[i] - prefix[i - L]))    
        return res