What is the time complexity of Continuous Subarray Sum?

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

What is the space complexity of Continuous Subarray Sum?

The space complexity is O(min(n, k)).

What data structures are used to solve Continuous Subarray Sum?

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

What are the key insights for solving Continuous Subarray Sum?

Insight 1: Using prefix sums allows efficient calculation of the sum of any subarray. Instead of repeatedly summing the subarray, we can subtract prefix sums. Insight 2: If two prefix sums have the same remainder when divided by 'k', the subarray between their indices has a sum that is a multiple of 'k'. This is because (sum2 - sum1) % k == 0 if sum1 % k == sum2 % k.

Which companies ask LeetCode 523. Continuous Subarray Sum in interviews?

This problem is frequently asked at Yandex.

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Continuous Subarray Sum - LeetCode 523 Solution

Continuous Subarray Sum - Complete Solution Guide

Continuous Subarray Sum is LeetCode problem 523, 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 true if nums has a good subarray or false otherwise . A good subarray is a subarray where: its length is at least two , and the sum of the elements of the subarray is a multiple of k . Note that: A subarray is a contiguous part of the array. An integer x is a multiple of k if there exists an integer n such that x = n * k . 0 is always a multiple of k . Example 1: Input: nums = [23, 2,4 ,6,7], k = 6 Output: true Explanation: [2, 4] is a continu

Detailed Explanation

The problem asks us to determine if a given array of integers contains a contiguous subarray of length at least 2 whose sum is a multiple of a given integer 'k'. A multiple of 'k' means the sum can be expressed as n * k, where n is an integer. We need to return `true` if such a subarray exists, and `false` otherwise. Crucially, a subarray must be contiguous (elements are next to each other) and must have a length of at least 2.

Solution Approach

The solution utilizes a hash map (dictionary) to store the remainders of prefix sums modulo 'k', along with their corresponding indices. We iterate through the array, calculating the running sum and its remainder when divided by 'k'. If we encounter a remainder we've seen before, we check if the distance between the current index and the index where we first saw the remainder is at least 2. If so, we've found a valid subarray. If we haven't seen the remainder before, we store it in the hash map along with its index.

Step-by-Step Algorithm

Step 1: Initialize a hash map `remainder_map` to store remainders as keys and their first seen indices as values. Initialize the map with {0: -1} to handle the case where a subarray starting from the beginning is a multiple of k.
Step 2: Initialize a variable `running_sum` to 0. This variable will store the cumulative sum of the array up to the current index.
Step 3: Iterate through the array `nums` from left to right, with index `i` and value `num`.
Step 4: Update `running_sum` by adding the current element `num` to it: `running_sum += num`.
Step 5: Calculate the remainder of `running_sum` when divided by `k`: `remainder = running_sum % k`.
Step 6: Check if the `remainder` exists as a key in `remainder_map`.
Step 7: If the `remainder` exists in `remainder_map`, check if the difference between the current index `i` and the index stored in the map for that remainder, `remainder_map[remainder]`, is greater than or equal to 2. If it is, return `true` because we have found a valid subarray.
Step 8: If the `remainder` does not exist in `remainder_map`, add it to the map with its current index `i`: `remainder_map[remainder] = i`.
Step 9: After iterating through the entire array, if no valid subarray has been found, return `false`.

Key Insights

Insight 1: Using prefix sums allows efficient calculation of the sum of any subarray. Instead of repeatedly summing the subarray, we can subtract prefix sums.
Insight 2: If two prefix sums have the same remainder when divided by 'k', the subarray between their indices has a sum that is a multiple of 'k'. This is because (sum2 - sum1) % k == 0 if sum1 % k == sum2 % k.
Insight 3: We need to track the *first* occurrence of each remainder. If we encounter the same remainder again later, we check the length of the subarray between the first and current occurrences. If the length is >= 2, we have found a good subarray.
Insight 4: Initialize the hash map with {0: -1}. This handles the edge case where the prefix sum itself is a multiple of k from the beginning of the array. If the sum % k == 0 and the current index is i, then i - (-1) >= 2 will correctly identify a valid subarray.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(min(n, k))

Topics

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

Companies

Asked at: Yandex.

Continuous Subarray Sum | LeetCode 523

Medium

Array Hash Table Math Prefix Sum

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

Solution Code

Available in:

import java.util.HashMap;
import java.util.Map;

class Solution {
    public boolean checkSubarraySum(int[] nums, int k) {
        Map<Integer, Integer> remainderMap = new HashMap<>();
        remainderMap.put(0, -1);
        int runningSum = 0;
        
        for (int i = 0; i < nums.length; i++) {
            runningSum += nums[i];
            int remainder = runningSum % k;
            
            if (remainderMap.containsKey(remainder)) {
                if (i - remainderMap.get(remainder) >= 2) {
                    return true;
                }
            } else {
                remainderMap.put(remainder, i);
            }
        }
        
        return false;
    }
}

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Problem Info

Description

Problem 523: Continuous Subarray Sum - Detailed explanation and solution approaches.

Topics

Array Hash Table Math Prefix Sum

Companies

Yandex

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:523

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 a hash map `remainder_map` to store remainders as keys and their first seen indices as values. Initialize the map with {0: -1} to handle the case where a subarray starting from the beginning is a multiple of k.
2
Step 2: Initialize a variable `running_sum` to 0. This variable will store the cumulative sum of the array up to the current index.
3
Step 3: Iterate through the array `nums` from left to right, with index `i` and value `num`.
4
Step 4: Update `running_sum` by adding the current element `num` to it: `running_sum += num`.
5
Step 5: Calculate the remainder of `running_sum` when divided by `k`: `remainder = running_sum % k`.
6
Step 6: Check if the `remainder` exists as a key in `remainder_map`.
7
Step 7: If the `remainder` exists in `remainder_map`, check if the difference between the current index `i` and the index stored in the map for that remainder, `remainder_map[remainder]`, is greater than or equal to 2. If it is, return `true` because we have found a valid subarray.
8
Step 8: If the `remainder` does not exist in `remainder_map`, add it to the map with its current index `i`: `remainder_map[remainder] = i`.
9
Step 9: After iterating through the entire array, if no valid subarray has been found, return `false`.

💡

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(min(n, k))

📝

Why?

The time complexity is linear because we iterate through the array once. The space complexity is determined by the number of distinct remainders we store in the hash map. Since the remainder is always less than k, we can store at most k distinct remainders. However, we can also have at most n remainders if all prefix sums have different remainders mod 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

Iterate through all possible subarrays and check if their sum is a multiple of k. Calculate subarray sums using nested loops.

Time: O(n^2), where n is the length of the input array `nums`.Space: O(1).

Trade-offs: The brute force approach is simpler to understand but significantly less efficient than the hash map approach. It's only suitable for very small input sizes due to its quadratic time complexity.

✨

Key Insights

1Insight 1: Using prefix sums allows efficient calculation of the sum of any subarray. Instead of repeatedly summing the subarray, we can subtract prefix sums.
2Insight 2: If two prefix sums have the same remainder when divided by 'k', the subarray between their indices has a sum that is a multiple of 'k'. This is because (sum2 - sum1) % k == 0 if sum1 % k == sum2 % k.
3Insight 3: We need to track the *first* occurrence of each remainder. If we encounter the same remainder again later, we check the length of the subarray between the first and current occurrences. If the length is >= 2, we have found a good subarray.
4Insight 4: Initialize the hash map with {0: -1}. This handles the edge case where the prefix sum itself is a multiple of k from the beginning of the array. If the sum % k == 0 and the current index is i, then i - (-1) >= 2 will correctly identify a valid subarray.

📚

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: `k` is 0. The provided code implicitly handles k = 0 correctly as the modulo operator (%) is only used if k is not 0 in the C code. Java, Python, and C++ will throw an error with `k=0`, although the constraint specifies that k >=1.
Edge case 2: A valid subarray starts from the beginning of the array. The initialization of the `remainder_map` with `{0: -1}` handles this. If the sum from the beginning is a multiple of `k`, the remainder will be 0, and we'll find `0` in the map. Since the index for 0 is -1, the length will be i - (-1) = i + 1, where `i` is the current index. The condition `i+1 >=2` thus will evaluate to `i>=1`, so the solution is valid.
Edge case 3: All numbers in the array are zero. In this case, the `running_sum` will always be zero, and thus the remainder will always be zero. The hash map will contain {0: -1}, and on the second iteration, the length condition `i - remainder_map[0] >= 2` will evaluate to `i + 1 >= 2` which resolves to `i >= 1`. Since the loop begins at `i=0`, this edge case is correctly handled.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to check the length of the subarray. If you find a repeated remainder, you must verify that the subarray between the two indices has a length of at least 2.

✗Common mistake 2: Not handling the case where a valid subarray starts from the beginning of the array. Initializing the hash map with {0: -1} is crucial for this.

✗Common mistake 3: Incorrectly calculating the remainder when k is zero or negative (although the constraints specify k >= 1). While not directly applicable here due to problem constraints, it's a good practice to validate the input if it comes from external sources to prevent exceptions.

✅

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: Finding patterns in financial data. For example, identifying periods where stock prices exhibit cyclical behavior (multiples of a certain pattern length).

🏦Real-world scenario 2: Network traffic analysis. Detecting periodic congestion patterns by analyzing network traffic data and looking for recurring intervals related to certain events/attacks

💡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

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ 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: November 11, 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

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 14

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

@TechLeadJul 7

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

@DevExpert92Jun 30

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

How do I choose between a hash map and a hash set?

Use a hash set when you only need to track presence/absence of elements (like detecting duplicates). Use a hash map when you need to associate values with keys (like counting frequencies or storing indices). Hash sets use less memory since they only store keys.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Continuous Subarray Sum"?

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.

🤖 Need help with this problem?