How do you solve Maximum Subarray Sum With Length Divisible by K on LeetCode?

Maximum Subarray Sum With Length Divisible by K (LeetCode #3381) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Using prefix sums allows us to efficiently calculate subarray sums.

What is the time complexity of Maximum Subarray Sum With Length Divisible by K?

The optimal solution for Maximum Subarray Sum With Length Divisible by K runs in O(n) time and uses O(n) space. This complexity is linear because we only traverse the array once and use a HashMap to store prefix sums, which allows for efficient lookups.

What is the brute force solution for Maximum Subarray Sum With Length Divisible by K?

The brute force approach for Maximum Subarray Sum With Length Divisible by K is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking every possible subarray of the given array. For each subarray, we check if its length is divisible by k and calculate its sum, keeping track of the maximum sum found. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Maximum Subarray Sum With Length Divisible by K?

Maximum Subarray Sum With Length Divisible by K uses the following concepts: Array, Hash Table, Prefix Sum. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Maximum Subarray Sum With Length Divisible by K?

Always consider edge cases, such as arrays with negative numbers or very small sizes. Explain your thought process clearly while coding; it shows your understanding. Practice recognizing patterns like prefix sums and HashMaps in problems.

What are common mistakes when solving Maximum Subarray Sum With Length Divisible by K?

Failing to handle negative numbers correctly when calculating modulo. Not initializing the HashMap properly, which can lead to incorrect results.

#3381

Maximum Subarray Sum With Length Divisible by K

Medium

ArrayHash TablePrefix SumHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal solution uses prefix sums and a HashMap to efficiently track the minimum prefix sums for each modulo k value. This allows us to calculate the maximum sum of subarrays with lengths divisible by k in linear time.

⚙️

Algorithm

5 steps

1Step 1: Initialize a prefix sum variable and a HashMap to store minimum prefix sums for each modulo k value.
2Step 2: Iterate through the array, updating the prefix sum.
3Step 3: For each prefix sum, calculate the current modulo k value.
4Step 4: If this modulo value has been seen before, calculate the potential maximum sum using the difference between the current prefix sum and the minimum prefix sum for that modulo value.
5Step 5: Update the HashMap with the minimum prefix sum for the current modulo value.

solution.py14 lines

1# Full working Python code
2
3def maxSubarraySumDivK(nums, k):
4    prefix_sum = 0
5    min_prefix = {0: 0}
6    max_sum = float('-inf')
7    for i, num in enumerate(nums):
8        prefix_sum += num
9        mod = prefix_sum % k
10        if mod in min_prefix:
11            max_sum = max(max_sum, prefix_sum - min_prefix[mod])
12        else:
13            min_prefix[mod] = prefix_sum
14    return max_sum

ℹ

Complexity note: This complexity is linear because we only traverse the array once and use a HashMap to store prefix sums, which allows for efficient lookups.

1Using prefix sums allows us to efficiently calculate subarray sums.
2HashMaps can be used to track minimum prefix sums for each modulo value, enabling quick lookups.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.