How do you solve Subarray Sums Divisible by K on LeetCode?

Subarray Sums Divisible by K (LeetCode #974) 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 without recalculating them multiple times.

What is the brute force solution for Subarray Sums Divisible by K?

The brute force approach for Subarray Sums Divisible by K is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves checking all possible subarrays and calculating their sums to see if they are divisible by k. This is straightforward but inefficient for larger arrays. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Subarray Sums Divisible by K?

Subarray Sums 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 Subarray Sums Divisible by K?

Always consider edge cases, such as when k is larger than the sum of elements. Explain your thought process clearly while coding, as it helps the interviewer understand your approach. Practice writing clean and efficient code, as readability is important in interviews.

What are common mistakes when solving Subarray Sums Divisible by K?

Forgetting to handle negative remainders when calculating modulo. Not initializing the hash map properly, which can lead to incorrect counts.

#974

Subarray Sums 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)

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Intuition

Time O(n)Space O(n)

The optimal solution uses a prefix sum and a hash map to efficiently count the number of valid subarrays. By tracking the remainders of prefix sums, we can quickly determine how many previous sums can form a valid subarray with the current sum.

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Algorithm

5 steps

1Step 1: Initialize a hash map to store the frequency of remainders and a variable for the prefix sum.
2Step 2: Iterate through the array, updating the prefix sum and calculating the remainder when divided by k.
3Step 3: If the remainder is negative, adjust it to be positive by adding k.
4Step 4: Check if this remainder has been seen before in the hash map. If so, add its frequency to the count.
5Step 5: Increment the frequency of the current remainder in the hash map.

solution.py14 lines

1# Full working Python code
2
3def subarraySumDivisibleByK(nums, k):
4    count = 0
5    prefix_sum = 0
6    remainder_count = {0: 1}
7    for num in nums:
8        prefix_sum += num
9        remainder = prefix_sum % k
10        if remainder < 0:
11            remainder += k
12        count += remainder_count.get(remainder, 0)
13        remainder_count[remainder] = remainder_count.get(remainder, 0) + 1
14    return count

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Complexity note: The optimal solution runs in linear time because we only make a single pass through the array while using a hash map to store remainders, leading to efficient lookups.

1Using prefix sums allows us to efficiently calculate subarray sums without recalculating them multiple times.
2Hash maps can be used to track the frequency of remainders, which helps in counting valid subarrays quickly.

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