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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, or0 <= 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 >= 1M >= 1L + M <= A.length <= 10000 <= A[i] <= 1000
Related Topics:
Array
Solution 1.
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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); } } } ############ class Solution { public int maxSumTwoNoOverlap(int[] nums, int firstLen, int secondLen) { int n = nums.length; int[] s = new int[n + 1]; for (int i = 0; i < n; ++i) { s[i + 1] = s[i] + nums[i]; } int ans = 0; for (int i = firstLen, t = 0; i + secondLen - 1 < n; ++i) { t = Math.max(t, s[i] - s[i - firstLen]); ans = Math.max(ans, t + s[i + secondLen] - s[i]); } for (int i = secondLen, t = 0; i + firstLen - 1 < n; ++i) { t = Math.max(t, s[i] - s[i - secondLen]); ans = Math.max(ans, t + s[i + firstLen] - s[i]); } return ans; } } -
// 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; } }; -
class Solution: def maxSumTwoNoOverlap(self, nums: List[int], firstLen: int, secondLen: int) -> int: n = len(nums) s = list(accumulate(nums, initial=0)) ans = t = 0 i = firstLen while i + secondLen - 1 < n: t = max(t, s[i] - s[i - firstLen]) ans = max(ans, t + s[i + secondLen] - s[i]) i += 1 t = 0 i = secondLen while i + firstLen - 1 < n: t = max(t, s[i] - s[i - secondLen]) ans = max(ans, t + s[i + firstLen] - s[i]) i += 1 return ans ############ # 1031. Maximum Sum of Two Non-Overlapping Subarrays # https://leetcode.com/problems/maximum-sum-of-two-non-overlapping-subarrays/ class Solution: def maxSumTwoNoOverlap(self, nums: List[int], firstLen: int, secondLen: int) -> int: n = len(nums) prefix = [0] + list(accumulate(nums)) res = 0 for i in range(firstLen, n + 1): first = prefix[i] - prefix[i - firstLen] second = 0 for j in range(secondLen, i - firstLen + 1): second = max(second, prefix[j] - prefix[j - secondLen]) for j in range(secondLen + i + 1, n + 1): second = max(second, prefix[j] - prefix[j - secondLen]) res = max(res, first + second) return res -
func maxSumTwoNoOverlap(nums []int, firstLen int, secondLen int) (ans int) { n := len(nums) s := make([]int, n+1) for i, x := range nums { s[i+1] = s[i] + x } for i, t := firstLen, 0; i+secondLen-1 < n; i++ { t = max(t, s[i]-s[i-firstLen]) ans = max(ans, t+s[i+secondLen]-s[i]) } for i, t := secondLen, 0; i+firstLen-1 < n; i++ { t = max(t, s[i]-s[i-secondLen]) ans = max(ans, t+s[i+firstLen]-s[i]) } return } func max(a, b int) int { if a > b { return a } return b }