Continuous Subarray Sum - Practice Coding Problems

Continuous Subarray Sum - Problem

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
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

Input & Output

Example 1 — Valid Subarray Found

$ Input: nums = [23,2,4,6,7], k = 6

› Output: true

💡 Note: The subarray [2,4] has sum 6, which is divisible by 6. Also [4,6] has sum 10... wait, let me recalculate: [2,4] = 6, [4,6] = 10, [6,7] = 13. Actually [2,4] with sum 6 is divisible by 6, so return true.

Example 2 — No Valid Subarray

$ Input: nums = [23,2,4,6,6], k = 7

› Output: false

💡 Note: No subarray of length ≥ 2 has sum divisible by 7. Checking: [23,2]=25, [2,4]=6, [4,6]=10, [6,6]=12, [23,2,4]=29, etc. None are divisible by 7.

Example 3 — Zero Sum Edge Case

$ Input: nums = [5,0,0], k = 3

› Output: true

💡 Note: The subarray [0,0] has sum 0, and 0 is always a multiple of any k. Length is 2, so this is valid.

Constraints

1 ≤ nums.length ≤ 10⁵
0 ≤ nums[i] ≤ 10⁹
1 ≤ k ≤ 2³¹ - 1

Visualization

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Understanding the Visualization

Input

Array [23,2,4,6,7] and divisor k=6

Process

Check subarrays for sum divisibility

Output

Return true if valid subarray exists

Key Takeaway

🎯 Key Insight: Use modular arithmetic with hash map to detect when prefix sum remainders repeat

Asked in

f Facebook 45 G Google 38 a Amazon 32 M Microsoft 28

The key insight is using modular arithmetic: if two prefix sums have the same remainder when divided by k, their difference is divisible by k. The optimal approach uses a hash map to track remainders in O(n) time with O(k) space.

Common Approaches

Approach	Time	Space	Notes
✓ Optimized Brute Force - Prefix Sum	O(n²)	O(n)	Use prefix sums to calculate subarray sums in O(1) time
Hash Map with Modular Arithmetic	O(n)	O(min(n,k))	Use hash map to store remainder patterns of prefix sums
Brute Force - Check All Subarrays	O(n³)	O(1)	Check every possible subarray of length ≥ 2 and test if sum is divisible by k

Optimized Brute Force - Prefix Sum — Algorithm Steps

Build prefix sum array
For each pair (i,j), calculate sum using prefix[j+1] - prefix[i]
Check if sum is divisible by k

Visualization

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Step-by-Step Walkthrough

Build Prefix

Calculate cumulative sums

Range Query

Get subarray sum in O(1)

Check

Test divisibility by k

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int solution(int* nums, int numsSize, int k) {
    long long* prefix = (long long*)malloc((numsSize + 1) * sizeof(long long));
    prefix[0] = 0;
    for (int i = 0; i < numsSize; i++) {
        prefix[i + 1] = prefix[i] + nums[i];
    }
    
    for (int i = 0; i < numsSize - 1; i++) {
        for (int j = i + 1; j < numsSize; j++) {
            long long subarraySum = prefix[j + 1] - prefix[i];
            if (subarraySum % k == 0) {
                free(prefix);
                return 1;
            }
        }
    }
    free(prefix);
    return 0;
}

int main() {
    char line[100000];
    fgets(line, sizeof(line), stdin);
    
    int nums[10000];
    int numsSize = 0;
    char* token = strtok(line + 1, ",]");
    while (token != NULL) {
        nums[numsSize++] = atoi(token);
        token = strtok(NULL, ",]");
    }
    
    int k;
    scanf("%d", &k);
    
    int result = solution(nums, numsSize, k);
    printf(result ? "true\n" : "false\n");
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

Two nested loops to check all pairs, but sum calculation is O(1) with prefix sums

⚠ Quadratic Growth

Space Complexity

O(n)

Prefix sum array stores n+1 cumulative sum values

⚡ Linearithmic Space

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High Frequency

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Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Optimized Brute Force - Prefix Sum — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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