#3381

Medium

auto_awesomeArray scanning plus hash lookup

LeetCode Problem Workspace

Maximum Subarray Sum With Length Divisible by K

Find the maximum sum of a subarray where the length of the subarray is divisible by k.

Problem Statement

You are given an array of integers nums and an integer k. Your task is to return the maximum sum of a subarray of nums such that the length of the subarray is divisible by k.

For example, given nums = [1, 2] and k = 1, the sum of the entire array is valid because its length is divisible by 1. Other edge cases may involve negative numbers or larger subarrays.

Examples

Example 1

Input: nums = [1,2], k = 1

Output: 3

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

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

The maximum sum subarray is [1, 2, -3, 4] which has length equal to 4 which is divisible by 2.

Constraints

1 <= k <= nums.length <= 2 * 105
-109 <= nums[i] <= 109

Solution Approach

Array Scanning and Hash Lookup

To solve this problem, iterate through the array while keeping track of prefix sums. Use a hash table to store the minimum prefix sum ending at each index % k. This allows us to efficiently calculate the maximum sum of valid subarrays.

Modulo-Based Optimization

Given the condition on subarray lengths, maintain a running prefix sum and use the modulo operation to check if the current subarray length is divisible by k. If it is, compute and update the potential maximum sum.

Sliding Window Strategy

By tracking the prefix sums and using a sliding window approach, you can minimize unnecessary checks and focus on the subarrays that could yield the maximum sum. This reduces the time complexity by eliminating redundant calculations.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity depends on the final implementation. A solution using array scanning with hash table lookups can achieve a time complexity of O(n), where n is the length of the array. The space complexity is O(k) due to the hash table storing the prefix sums modulo k.

What Interviewers Usually Probe

Candidate shows an understanding of prefix sums and hash table usage.
Candidate successfully applies the modulo-based optimization technique.
Candidate is able to implement an efficient sliding window approach.

Common Pitfalls or Variants

Common pitfalls

Forgetting to account for negative numbers in prefix sums.
Incorrectly updating the hash table, leading to wrong results.
Not using the modulo operation to efficiently reduce unnecessary checks.

Follow-up variants

Consider variations where the subarray length must be divisible by other values, not just k.
Try solving the problem with different array types (e.g., large integers or large negative values).
Explore variations that require returning the minimum or product of the subarray instead of the sum.

FAQ

How do I approach the Maximum Subarray Sum With Length Divisible by K problem?

Start by understanding the prefix sum technique. Use hash lookups to efficiently track potential subarrays and compute the maximum sum when the length is divisible by k.

What is the primary pattern for solving the Maximum Subarray Sum With Length Divisible by K?

The primary pattern is array scanning combined with hash table lookups, where you track the minimum prefix sum modulo k to efficiently find valid subarrays.

What is the time complexity of this problem?

The time complexity is O(n), where n is the length of the array, using array scanning with hash table lookups.

How do I handle negative numbers in this problem?

When working with negative numbers, make sure to correctly compute prefix sums, as they will affect the overall sum of subarrays. The algorithm should handle this case naturally if implemented correctly.

Can I modify the problem to find the subarray with the maximum product instead of the sum?

Yes, the problem can be adapted to find the subarray with the maximum product by modifying the way sums are calculated, but this requires a different approach.

terminal

Solution

Solution 1: Prefix Sum + Enumeration

According to the problem description, for a subarray's length to be divisible by $k$, it is equivalent to requiring that for subarray $\textit{nums}[i+1 \ldots j]$, we have $i \bmod k = j \bmod k$.

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class Solution:
    def maxSubarraySum(self, nums: List[int], k: int) -> int:
        f = [inf] * k
        ans = -inf
        s = f[-1] = 0
        for i, x in enumerate(nums):
            s += x
            ans = max(ans, s - f[i % k])
            f[i % k] = min(f[i % k], s)
        return ans

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