3381. Maximum Subarray Sum With Length Divisible by K LeetCode Solution
In this guide, you will get 3381. Maximum Subarray Sum With Length Divisible by K LeetCode Solution with the best time and space complexity. The solution to Maximum Subarray Sum With Length Divisible by K problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.
Problem Statement of Maximum Subarray Sum With Length Divisible by K
You are given an array of integers nums and an integer k.
Return the maximum sum of a subarray of nums, such that the size of the subarray is divisible by k.
Example 1:
Input: nums = [1,2], k = 1
Output: 3
Explanation:
The subarray [1, 2] with sum 3 has length equal to 2 which is divisible by 1.
Example 2:
Input: nums = [-1,-2,-3,-4,-5], k = 4
Output: -10
Explanation:
The maximum sum subarray is [-1, -2, -3, -4] which has length equal to 4 which is divisible by 4.
Example 3:
Input: nums = [-5,1,2,-3,4], k = 2
Output: 4
Explanation:
The maximum sum subarray is [1, 2, -3, 4] which has length equal to 4 which is divisible by 2.
3381. Maximum Subarray Sum With Length Divisible by K LeetCode Solution in C++
class Solution {
public:
long long maxSubarraySum(std::vector<int>& nums, int k) {
long ans = LONG_MIN;
long prefix = 0;
// minPrefix[i % k] := the minimum prefix sum of the first i numbers
vector<long> minPrefix(k, LONG_MAX / 2);
minPrefix[k - 1] = 0;
for (int i = 0; i < nums.size(); ++i) {
prefix += nums[i];
ans = max(ans, prefix - minPrefix[i % k]);
minPrefix[i % k] = min(minPrefix[i % k], prefix);
}
return ans;
}
}; /* code provided by PROGIEZ */
3381. Maximum Subarray Sum With Length Divisible by K LeetCode Solution in Java
class Solution {
public long maxSubarraySum(int[] nums, int k) {
long ans = Long.MIN_VALUE;
long prefix = 0;
// minPrefix[i % k] := the minimum prefix sum of the first i numbers
long[] minPrefix = new long[k];
Arrays.fill(minPrefix, Long.MAX_VALUE / 2);
minPrefix[k - 1] = 0;
for (int i = 0; i < nums.length; ++i) {
prefix += nums[i];
ans = Math.max(ans, prefix - minPrefix[i % k]);
minPrefix[i % k] = Math.min(minPrefix[i % k], prefix);
}
return ans;
}
} // code provided by PROGIEZ
3381. Maximum Subarray Sum With Length Divisible by K LeetCode Solution in Python
class Solution:
def maxSubarraySum(self, nums: list[int], k: int) -> int:
ans = -math.inf
prefix = 0
# minPrefix[i % k] := the minimum prefix sum of the first i numbers
minPrefix = [math.inf] * k
minPrefix[k - 1] = 0
for i, num in enumerate(nums):
prefix += num
ans = max(ans, prefix - minPrefix[i % k])
minPrefix[i % k] = min(minPrefix[i % k], prefix)
return ans # code by PROGIEZ
