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Solving Leetcode Interviews in Seconds with AI: Binary Subarrays With Sum

Updated
2 min read

Introduction

In this blog post, we will explore how to solve the LeetCode problem "930" 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 a binary array nums and an integer goal, return the number of non-empty subarrays with a sum goal. A subarray is a contiguous part of the array. Example 1: Input: nums = [1,0,1,0,1], goal = 2 Output: 4 Explanation: The 4 subarrays are bolded and underlined below: [1,0,1,0,1] [1,0,1,0,1] [1,0,1,0,1] [1,0,1,0,1] Example 2: Input: nums = [0,0,0,0,0], goal = 0 Output: 15 Constraints: 1 <= nums.length <= 3 * 104 nums[i] is either 0 or 1. 0 <= goal <= nums.length

Explanation

Here's a solution to the problem, focusing on efficiency and clarity:

  • Prefix Sum and Hash Map: Calculate the prefix sum of the array. Use a hash map to store the frequency of each prefix sum encountered.
  • Counting Subarrays: Iterate through the prefix sums. For each prefix sum prefix_sum[i], check how many times prefix_sum[i] - goal has appeared previously. This count represents the number of subarrays ending at index i with a sum equal to goal.
  • Handle Edge Cases: Correctly handle the case where the goal is 0, and initialize the hash map appropriately.

  • Runtime Complexity: O(n), where n is the length of the input array. Storage Complexity: O(n) in the worst case (when all prefix sums are distinct).

Code

    def numSubarraysWithSum(nums, goal):
    prefix_sum_count = {0: 1}  # Initialize count for prefix sum 0
    current_sum = 0
    count = 0

    for num in nums:
        current_sum += num
        diff = current_sum - goal

        if diff in prefix_sum_count:
            count += prefix_sum_count[diff]

        if current_sum in prefix_sum_count:
            prefix_sum_count[current_sum] += 1
        else:
            prefix_sum_count[current_sum] = 1

    return count

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