What is the time complexity of Subarray Sums Divisible by K?

The optimal solution has a time complexity of O(n).

What data structures are used to solve Subarray Sums Divisible by K?

This problem uses Array, Hash Table, Prefix Sum. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Subarray Sums Divisible by K?

Insight 1: The sum of a subarray from index i to j is divisible by k if (prefix_sum[j+1] - prefix_sum[i]) % k == 0. This is equivalent to prefix_sum[j+1] % k == prefix_sum[i] % k. Insight 2: Using a hash table (or in this case, an array acting as a hash table since remainders are bounded) to store the count of each remainder modulo k allows us to efficiently find subarrays with sums divisible by k.

Which companies ask LeetCode 974. Subarray Sums Divisible by K in interviews?

This problem is frequently asked at Citadel, Expedia, Tinkoff, thoughtspot.

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Subarray Sums Divisible by K - LeetCode 974 Solution

Q: How to solve Subarray Sums Divisible by K?

The solution uses the concept of prefix sums and modular arithmetic. It iterates through the array `nums`, calculates the prefix sum at each index, and then computes the remainder of the prefix sum when divided by `k`. It uses an array `remainder_counts` of size `k` to store the frequency of each remainder. For each element, it increments the count of subarrays that can be formed using the current prefix sum and the existing prefix sums with the same remainder. The key idea is based on the property that if two prefix sums have the same remainder when divided by k, then their difference (which represents the sum of the subarray between them) is divisible by k.

Subarray Sums Divisible by K - Complete Solution Guide

Subarray Sums Divisible by K is LeetCode problem 974, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

Given an integer array nums and an integer k , return the number of non-empty subarrays that have a sum divisible by k . A subarray is a contiguous part of an array. Example 1: Input: nums = [4,5,0,-2,-3,1], k = 5 Output: 7 Explanation: There are 7 subarrays with a sum divisible by k = 5: [4, 5, 0, -2, -3, 1], [5], [5, 0], [5, 0, -2, -3], [0], [0, -2, -3], [-2, -3] Example 2: Input: nums = [5], k = 9 Output: 0 Constraints: 1 <= nums.length <= 3 * 10 4 -10 4 <= nums[i] <= 10 4 2 <= k <= 10 4

Detailed Explanation

The problem asks us to find the number of non-empty subarrays in a given integer array `nums` whose sum is divisible by a given integer `k`. A subarray is a contiguous sequence of elements within the array. We need to count how many such subarrays exist.

Solution Approach

The solution uses the concept of prefix sums and modular arithmetic. It iterates through the array `nums`, calculates the prefix sum at each index, and then computes the remainder of the prefix sum when divided by `k`. It uses an array `remainder_counts` of size `k` to store the frequency of each remainder. For each element, it increments the count of subarrays that can be formed using the current prefix sum and the existing prefix sums with the same remainder. The key idea is based on the property that if two prefix sums have the same remainder when divided by k, then their difference (which represents the sum of the subarray between them) is divisible by k.

Step-by-Step Algorithm

Step 1: Initialize an array `remainder_counts` of size `k` with all elements set to 0. Set `remainder_counts[0]` to 1 because an empty subarray has a sum of 0, which is divisible by `k`.
Step 2: Initialize `count` to 0 to keep track of the number of subarrays with sums divisible by `k`. Initialize `prefix_sum` to 0.
Step 3: Iterate through the `nums` array.
Step 4: In each iteration, update `prefix_sum` by adding the current element.
Step 5: Calculate the `remainder` of the `prefix_sum` when divided by `k` using the modulo operator (`%`).
Step 6: Handle negative remainders by adding `k` to the remainder until it becomes non-negative (only needed in Java, C, and C++).
Step 7: Increment `count` by the value stored at `remainder_counts[remainder]`. This is because `remainder_counts[remainder]` represents the number of subarrays that end before the current index and have the same remainder as the current prefix sum. Therefore any of these will be the start of a valid subarray.
Step 8: Increment `remainder_counts[remainder]` by 1 to update the count of the current remainder.
Step 9: After iterating through all elements, return `count`.

Key Insights

Insight 1: The sum of a subarray from index i to j is divisible by k if (prefix_sum[j+1] - prefix_sum[i]) % k == 0. This is equivalent to prefix_sum[j+1] % k == prefix_sum[i] % k.
Insight 2: Using a hash table (or in this case, an array acting as a hash table since remainders are bounded) to store the count of each remainder modulo k allows us to efficiently find subarrays with sums divisible by k.
Insight 3: When calculating the remainder with the modulo operator, it's important to handle negative remainders correctly to ensure they are non-negative. In languages like Java and C++, negative remainders need special handling.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(k)

Topics

This problem involves: Array, Hash Table, Prefix Sum.

Companies

Asked at: Citadel, Expedia, Tinkoff, thoughtspot.

Subarray Sums Divisible by K | LeetCode 974

Medium

Array Hash Table Prefix Sum

Time: O(n)Space: O(k)

Solution Code

Available in:

class Solution {
    public int subarraysDivByK(int[] nums, int k) {
        int[] remainder_counts = new int[k];
        remainder_counts[0] = 1;
        
        int count = 0;
        int prefix_sum = 0;
        
        for (int num : nums) {
            prefix_sum += num;
            int remainder = prefix_sum % k;
            if (remainder < 0) {
                remainder += k; // Ensure remainder is non-negative
            }

            count += remainder_counts[remainder];
            remainder_counts[remainder] += 1;
        }
        
        return count;
    }
}

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Problem Info

Description

Problem 974: Subarray Sums Divisible by K - Detailed explanation and solution approaches.

Topics

Array Hash Table Prefix Sum

Companies

CitadelExpediaTinkoffthoughtspot

Stats

Difficulty:Medium

Languages:4

Companies:4

Problem #:974

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

9 steps

1
Step 1: Initialize an array `remainder_counts` of size `k` with all elements set to 0. Set `remainder_counts[0]` to 1 because an empty subarray has a sum of 0, which is divisible by `k`.
2
Step 2: Initialize `count` to 0 to keep track of the number of subarrays with sums divisible by `k`. Initialize `prefix_sum` to 0.
3
Step 3: Iterate through the `nums` array.
4
Step 4: In each iteration, update `prefix_sum` by adding the current element.
5
Step 5: Calculate the `remainder` of the `prefix_sum` when divided by `k` using the modulo operator (`%`).
6
Step 6: Handle negative remainders by adding `k` to the remainder until it becomes non-negative (only needed in Java, C, and C++).
7
Step 7: Increment `count` by the value stored at `remainder_counts[remainder]`. This is because `remainder_counts[remainder]` represents the number of subarrays that end before the current index and have the same remainder as the current prefix sum. Therefore any of these will be the start of a valid subarray.
8
Step 8: Increment `remainder_counts[remainder]` by 1 to update the count of the current remainder.
9
Step 9: After iterating through all elements, return `count`.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(k)

📝

Why?

The time complexity is determined by the single loop iterating through the `nums` array. All other operations inside the loop take constant time. The space complexity is determined by the size of the `remainder_counts` array, which is proportional to `k`.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Brute Force

Generate all possible subarrays and check if the sum of each subarray is divisible by `k`. This involves iterating through all possible start and end indices of subarrays.

Time: O(n^2), where n is the length of the `nums` array. Generating all subarrays takes O(n^2) time, and calculating the sum of each subarray takes O(n) in the worst case (although it can be optimized using prefix sums). Thus it is overall O(n^2).Space: O(1), as it uses only a constant amount of extra space.

Trade-offs: The brute force approach is simpler to understand but significantly less efficient than the prefix sum approach for larger arrays. It's generally not recommended for problems with larger input sizes due to its quadratic time complexity. The primary solution trades space O(k) for time O(n), versus this O(n^2) solution. The main solution also avoids nested loops.

✨

Key Insights

1Insight 1: The sum of a subarray from index i to j is divisible by k if (prefix_sum[j+1] - prefix_sum[i]) % k == 0. This is equivalent to prefix_sum[j+1] % k == prefix_sum[i] % k.
2Insight 2: Using a hash table (or in this case, an array acting as a hash table since remainders are bounded) to store the count of each remainder modulo k allows us to efficiently find subarrays with sums divisible by k.
3Insight 3: When calculating the remainder with the modulo operator, it's important to handle negative remainders correctly to ensure they are non-negative. In languages like Java and C++, negative remainders need special handling.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

Edge Cases

Edge case 1: When the input array is empty. The code will correctly return 0 since the loop will not execute.
Edge case 2: When all numbers are 0. The prefix sum will always be 0, so the remainder will always be 0. The `remainder_counts[0]` will increment with each number, correctly counting all subarrays.
Edge case 3: When `k` is 1. All sums are divisible by 1, so every subarray is a valid subarray. The algorithm handles this correctly.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to handle negative remainders correctly. In languages like Java and C++, the modulo operator can return negative values when the dividend is negative. Adding `k` to the remainder ensures it becomes non-negative.

✗Common mistake 2: Not initializing `remainder_counts[0]` to 1. This is crucial because it accounts for the empty subarray (or rather, a prefix sum of 0) which has a sum divisible by `k`.

✗Common mistake 3: Incorrectly calculating the prefix sum, leading to wrong remainder values. Double-checking the calculation of the prefix sum is essential.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Financial analysis, where you need to find periods of profitability that are divisible by a certain number of months or quarters.

🏦Real-world scenario 2: Signal processing, where you want to identify patterns in a signal that repeat every `k` samples, indicating divisibility.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: August 14, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 13

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 6

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

What are the trade-offs of using a hash table?

Hash tables provide O(1) average-case lookup, insertion, and deletion, making them ideal for frequency counting and duplicate detection. However, they use O(n) extra space and may degrade to O(n) time in worst-case scenarios with hash collisions. Always consider if the space trade-off is acceptable.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Subarray Sums Divisible by K"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether hash table techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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