Maximum Subarray Sum With Length Divisible by K

Maximum Subarray Sum With Length Divisible by K

1. What is this problem about?

The Maximum Subarray Sum With Length Divisible by K problem adds a structural constraint to the traditional maximum sum challenge. You are given an array of integers and an integer K. You need to find the maximum sum of a contiguous subarray such that the number of elements in that subarray is a multiple of K (e.g., K, 2K, 3K, etc.). This means you cannot just pick any high-sum subarray; its length must satisfy the divisibility rule, making the search space more restricted and the logic more nuanced.

2. Why is this asked in interviews?

This "Maximum Subarray Sum With Length Divisible by K interview question" is frequently seen in Google and Bloomberg interviews. It tests a candidate's proficiency with prefix sums and their ability to optimize range sum queries. Specifically, it looks for an understanding of how to use remainders (modulo arithmetic) to group prefix sums. It's a classic example of taking a problem that seems to require O(N^2) and reducing it to O(N) using clever bookkeeping.

3. Algorithmic pattern used

The "Array, Hash Table, Prefix Sum interview pattern" is the standard approach here. By calculating a prefix sum array where P[i] is the sum of the first i elements, any subarray sum can be expressed as P[j] - P[i]. For the length j - i to be divisible by K, j and i must have the same remainder when divided by K (i.e., j % K == i % K). We can use a hash table or a small array to store the minimum prefix sum encountered so far for each possible remainder from 0 to K-1.

4. Example explanation

Suppose we have the array [1, -2, 3, 4] and K = 2.

• Prefix sums: P[0]=0, P[1]=1, P[2]=-1, P[3]=2, P[4]=6.
• Indices with remainder 0 (mod 2): 0, 2, 4. Prefix sums: {0, -1, 6}.
• Indices with remainder 1 (mod 2): 1, 3. Prefix sums: {1, 2}. To maximize the sum for remainder 0, we look at P[4] - min(P[0], P[2]) = 6 - (-1) = 7. To maximize the sum for remainder 1, we look at P[3] - P[1] = 2 - 1 = 1. The maximum sum with length divisible by 2 is 7 (which corresponds to the subarray [-2, 3, 4, -1]... wait, the example subarray is [3, 4]).

5. Common mistakes candidates make

One frequent pitfall in the "Maximum Subarray Sum With Length Divisible by K coding problem" is failing to initialize the "minimum prefix sum" storage correctly. For example, the remainder 0 should initially be associated with a prefix sum of 0 at index 0. Another mistake is not considering that the array can contain negative numbers, which means the "minimum" prefix sum for a remainder might not be the first one you find. Candidates also sometimes get confused between the remainder of the index and the remainder of the sum.

6. Interview preparation tip

Master the relationship between subarray lengths and prefix sum indices. Remember that Sum(i to j) = P[j+1] - P[i]. If the length must be a multiple of K, then (j + 1) - i must be a multiple of K. Getting this index logic right is 90% of the battle. Also, practice problems involving "subarray sum equals X" or "subarray sum divisible by X" to get comfortable with using hash tables to store previously seen prefix sum states.