Subarray Sum Equals K

Subarray Sum Equals K - Problem

Given an array of integers nums and an integer k, return the total number of continuous subarrays whose sum equals to k.

A subarray is a contiguous non-empty sequence of elements within an array.

Example: In array [1,1,1] with k=2, there are 2 subarrays with sum 2: [1,1] from index 0-1 and [1,1] from index 1-2.

Input & Output

Example 1 — Basic Case

$ Input: nums = [1,1,1], k = 2

› Output: 2

💡 Note: There are 2 subarrays with sum 2: [1,1] from index 0-1 and [1,1] from index 1-2

Example 2 — Single Element Match

$ Input: nums = [1,2,3], k = 3

› Output: 2

💡 Note: Two subarrays sum to 3: [3] at index 2, and [1,2] from index 0-1

Example 3 — No Matches

$ Input: nums = [1,2,3], k = 10

› Output: 0

💡 Note: No subarray sums to 10 since maximum possible sum is 1+2+3=6

Constraints

1 ≤ nums.length ≤ 2 × 10⁴
-1000 ≤ nums[i] ≤ 1000
-10⁷ ≤ k ≤ 10⁷

Visualization

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Asked in

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

The key insight is using prefix sums with hash map: if prefixSum[j] - prefixSum[i] = k, then subarray from i+1 to j has sum k. Best approach is prefix sum with hash map in O(n) time. Time: O(n), Space: O(n)

Common Approaches

✓ Brute Force - Check All Subarrays

⏱️ Time: O(n²) Space: O(1)

Use nested loops to examine every possible subarray. For each starting position, extend the subarray one element at a time and track the running sum.

Prefix Sum with Hash Map

⏱️ Time: O(n) Space: O(n)

Calculate running prefix sum and use hash map to store frequency of each sum seen so far. For each position, check if (prefixSum - k) exists in map.

Brute Force - Check All Subarrays — Algorithm Steps

Outer loop: iterate through each starting position
Inner loop: extend subarray from current start position
Track running sum and count when it equals k

Visualization

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Step-by-Step Walkthrough

Start Position

Begin from each index

Extend Subarray

Add elements one by one

Count Matches

Increment when sum equals k

Code -

solution.c — C

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

int solution(int* nums, int numsSize, int k) {
    int count = 0;
    
    for (int i = 0; i < numsSize; i++) {
        int currentSum = 0;
        for (int j = i; j < numsSize; j++) {
            currentSum += nums[j];
            if (currentSum == k) {
                count++;
            }
        }
    }
    
    return count;
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    // Parse array
    int nums[1000];
    int numsSize = 0;
    char* token = strtok(line, "[],\n ");
    while (token != NULL) {
        nums[numsSize++] = atoi(token);
        token = strtok(NULL, "[],\n ");
    }
    
    int k;
    scanf("%d", &k);
    
    int result = solution(nums, numsSize, k);
    printf("%d\n", result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

Nested loops iterate through all possible subarray start and end positions

⚠ Quadratic Growth

Space Complexity

O(1)

Only using variables for counting and sum tracking

✓ Linear Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force - Check All Subarrays — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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