Solving Leetcode Interviews in Seconds with AI: Binary Subarrays With Sum
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 timesprefix_sum[i] - goalhas appeared previously. This count represents the number of subarrays ending at indexiwith a sum equal togoal. 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