Day 8 - Subarrays with sum K.md

Difficulty	Source	Tags
Medium	160 Days of Problem Solving	Hash

🚀 Day 8. Subarrays with Sum K 🧠

The problem can be found at the following link: Question Link

💡 Problem Description:

Given an unsorted array of integers, find the number of continuous subarrays having a sum exactly equal to a given number k.

🔍 Example Walkthrough:

Input:

arr = [10, 2, -2, -20, 10], k = -10

Output:

3

Explanation: Subarrays: arr[0...3], arr[1...4], and arr[3...4] have a sum exactly equal to -10.

Input:

arr = [9, 4, 20, 3, 10, 5], k = 33

Output:

2

Explanation: Subarrays: arr[0...2] and arr[2...4] have a sum exactly equal to 33.

Input:

arr = [1, 3, 5], k = 0

Output:

0

Explanation: No subarray has a sum of 0.

Constraints:

• $(1 \leq \text{arr.size()} \leq 10^5)$
• $(-10^3 \leq \text{arr[i]} \leq 10^3)$
• $(-10^7 \leq k \leq 10^7)$

🎯 My Approach:

To solve this problem efficiently:

1. Prefix Sum with Hash Map:

   • Maintain a running sum (sum) as you traverse the array.
   • Use a hash map (prefixSumCount) to store the count of prefix sums encountered so far.
   • For each element, check if the difference sum - k exists in the hash map. If it does, it means there is a subarray ending at the current index with a sum equal to k.

2. Increment Hash Map:

   • Add the current running sum to the hash map or increment its count if it already exists.

3. Count Matching Subarrays:

   • Increment the count by the number of times sum - k has been encountered so far.

4. Final Answer:

   • The variable count will hold the number of subarrays with a sum equal to k.

🕒 Time and Auxiliary Space Complexity

• Expected Time Complexity: (O(n)), as we traverse the array once and perform (O(1)) operations for each element.
• Expected Auxiliary Space Complexity: (O(n)), to store the hash map containing prefix sums.

📝 Solution Code

Code (C++)

class Solution {
public:
    int countSubarrays(vector<int>& arr, int k) {
        unordered_map<int, int> prefixSumCount = {{0, 1}};
        int sum = 0, count = 0;

        for (int num : arr) {
            sum += num;
            count += prefixSumCount[sum - k];
            prefixSumCount[sum]++;
        }

        return count;
    }
};

Code (Java)

class Solution {
    public int countSubarrays(int[] arr, int k) {
        Map<Integer, Integer> prefixSumCount = new HashMap<>();
        prefixSumCount.put(0, 1);
        int sum = 0, count = 0;

        for (int num : arr) {
            sum += num;
            count += prefixSumCount.getOrDefault(sum - k, 0);
            prefixSumCount.put(sum, prefixSumCount.getOrDefault(sum, 0) + 1);
        }

        return count;
    }
}

Code (Python)

class Solution:
    def countSubarrays(self, arr, k):
        prefix_sum_count = {0: 1}
        sum_, count = 0, 0

        for num in arr:
            sum_ += num
            count += prefix_sum_count.get(sum_ - k, 0)
            prefix_sum_count[sum_] = prefix_sum_count.get(sum_, 0) + 1

        return count

🎯 Contribution and Support:

For discussions, questions, or doubts related to this solution, please visit my LinkedIn: Any Questions. Thank you for your input; together, we strive to create a space where learning is a collaborative endeavor.

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