LeetCode 3381: Maximum Subarray Sum With Length Divisible by K Solution

Master LeetCode problem 3381 (Maximum Subarray Sum With Length Divisible by K), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

3381. Maximum Subarray Sum With Length Divisible by K

Medium

Array Hash Table Prefix Sum

Problem Explanation

Explanation

To solve this problem, we can use a prefix sum technique to keep track of the sum of elements up to each index. We iterate through the array and calculate the prefix sum at each index. Then, for each subarray whose length is divisible by k, we calculate the sum of that subarray using the prefix sum array. We keep track of the maximum sum encountered so far.

Algorithm

Initialize a prefix sum array and set its first element to 0.
Iterate through the input array, calculating the prefix sum at each index and storing it in the prefix sum array.
Initialize a variable maxSum to store the maximum sum of subarrays with length divisible by k.
Iterate through the prefix sum array and for each index i, iterate again from the beginning of the array to index i-k (inclusive).
Check if the difference between the prefix sum at index i and the prefix sum at index j is divisible by k. If it is, update maxSum if the current sum is greater.
Return maxSum.

Time Complexity

The time complexity of this algorithm is O(n^2), where n is the number of elements in the input array.

Space Complexity

The space complexity of this algorithm is O(n) for storing the prefix sum array.

Solution Code

class Solution {
    public int maxSumDivK(int[] nums, int k) {
        int n = nums.length;
        int[] prefixSum = new int[n + 1];
        for (int i = 1; i <= n; i++) {
            prefixSum[i] = prefixSum[i - 1] + nums[i - 1];
        }
        int maxSum = 0;
        for (int i = 1; i <= n; i++) {
            for (int j = 0; j <= i - k; j++) {
                if ((prefixSum[i] - prefixSum[j]) % k == 0) {
                    maxSum = Math.max(maxSum, prefixSum[i] - prefixSum[j]);
                }
            }
        }
        return maxSum;
    }
}

Try It Yourself

Loading code editor...

Frequently Asked Questions

How to solve LeetCode 3381 (Maximum Subarray Sum With Length Divisible by K)?

This page provides optimized solutions for LeetCode problem 3381 (Maximum Subarray Sum With Length Divisible by K) in Java, C++, and Python, along with a detailed explanation and an interactive code editor to test your code.

What is the time complexity of LeetCode 3381 (Maximum Subarray Sum With Length Divisible by K)?

The time complexity for LeetCode 3381 (Maximum Subarray Sum With Length Divisible by K) varies by solution. Check the detailed explanation section for specific complexities in Java, C++, and Python implementations.

Can I run code for LeetCode 3381 on DevExCode?

Yes, DevExCode provides an interactive code editor where you can write, test, and run your code for LeetCode 3381 in Java, C++, or Python.

Back to LeetCode Solutions