Subarray Sum Equals K


1. Clarify the problem: Before diving into solving the problem, let's clarify the requirements:

  • We are given an array of integers and a target integer k.
  • We need to find the total number of continuous subarrays whose sum equals k.

2. Analyze the problem: To solve this problem, we can use a cumulative sum approach combined with a hashmap to keep track of the count of subarrays.

3. Design an algorithm: Here is the algorithm to solve the problem:

  1. Initialize a hashmap prefixSum to store the cumulative sum frequencies.
  2. Initialize variables count and sum to keep track of the count of subarrays and the current sum, respectively.
  3. Iterate through the array of integers.
    • Update the sum by adding the current element.
    • Check if sum equals k. If so, increment count by 1.
    • If the difference sum - k exists in the prefixSum hashmap, increment count by the frequency of sum - k stored in the hashmap.
    • Update the frequency of sum in the prefixSum hashmap by 1.
  4. Return the final count.

4. Explain your approach: The approach involves using a cumulative sum approach to calculate the sum of subarrays. We iterate through the array and keep track of the cumulative sum. At each step, we check if the current sum is equal to the target k or if there exists a previous cumulative sum sum - k in the hashmap. If either condition is satisfied, we update the count of subarrays accordingly.

5. Write clean and readable code:

def subarraySum(nums, k): prefixSum = {0: 1} count = 0 sum = 0 for num in nums: sum += num if sum - k in prefixSum: count += prefixSum[sum - k] prefixSum[sum] = prefixSum.get(sum, 0) + 1 return count # Test case nums = [1, 1, 1] k = 2 result = subarraySum(nums, k) print(result)

6. Test your code: Let's test the code with the given test case:

nums = [1, 1, 1] k = 2 result = subarraySum(nums, k) print(result)

The expected output is 2 since there are two subarrays with a sum equal to k = 2: [1, 1] and [2].

7. Optimize if necessary: The current solution has a time complexity of O(n) since we iterate through the array once. There are no further optimizations needed for this problem.

8. Handle error cases: The code assumes that the input array nums is not None and contains integers. If the array is empty or None, the code will return 0.

9. Discuss complexity analysis:

  • The time complexity of the solution is O(n), where n is the number of elements in the input array.
  • The space complexity is O(n) since the hashmap can store up to n cumulative sums in the worst case.
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