Introduction

In this blog post, we will explore how to solve the LeetCode problem "560" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

Given an array of integers nums and an integer k, return the total number of subarrays whose sum equals to k. A subarray is a contiguous non-empty sequence of elements within an array. Example 1: Input: nums = [1,1,1], k = 2 Output: 2 Example 2: Input: nums = [1,2,3], k = 3 Output: 2 Constraints: 1 <= nums.length <= 2 * 104 -1000 <= nums[i] <= 1000 -107 <= k <= 107

Explanation

Here's a solution that addresses the problem efficiently:

• Prefix Sum: Calculate the prefix sum of the array. This allows us to find the sum of any subarray in O(1) time by subtracting the prefix sums.
• Hash Map: Store the frequency of each prefix sum encountered so far in a hash map. This allows us to efficiently check if there exists a previous subarray whose sum, when added to the current element, equals k.

• Count Subarrays: Iterate through the array, updating the prefix sum and the hash map. If prefix_sum - k exists in the hash map, it means there's a subarray ending at the current index with a sum of k.

• Runtime Complexity: O(n), Storage Complexity: O(n)

Code

    def subarray_sum(nums: list[int], k: int) -> int:
    """
    Given an array of integers nums and an integer k, return the total number of subarrays whose sum equals to k.
    A subarray is a contiguous non-empty sequence of elements within an array.

    Example 1:
    Input: nums = [1,1,1], k = 2
    Output: 2

    Example 2:
    Input: nums = [1,2,3], k = 3
    Output: 2

    Constraints:
    1 <= nums.length <= 2 * 104
    -1000 <= nums[i] <= 1000
    -107 <= k <= 107
    """

    prefix_sums = {0: 1}  # Initialize with 0:1 to handle cases where a subarray starting from index 0 sums to k
    current_sum = 0
    count = 0

    for num in nums:
        current_sum += num
        if current_sum - k in prefix_sums:
            count += prefix_sums[current_sum - k]
        if current_sum in prefix_sums:
            prefix_sums[current_sum] += 1
        else:
            prefix_sums[current_sum] = 1

    return count