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Solving Leetcode Interviews in Seconds with AI: Binary Prefix Divisible By 5

Updated
2 min read

Introduction

In this blog post, we will explore how to solve the LeetCode problem "1018" 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

You are given a binary array nums (0-indexed). We define xi as the number whose binary representation is the subarray nums[0..i] (from most-significant-bit to least-significant-bit). For example, if nums = [1,0,1], then x0 = 1, x1 = 2, and x2 = 5. Return an array of booleans answer where answer[i] is true if xi is divisible by 5. Example 1: Input: nums = [0,1,1] Output: [true,false,false] Explanation: The input numbers in binary are 0, 01, 011; which are 0, 1, and 3 in base-10. Only the first number is divisible by 5, so answer[0] is true. Example 2: Input: nums = [1,1,1] Output: [false,false,false] Constraints: 1 <= nums.length <= 105 nums[i] is either 0 or 1.

Explanation

Here's a solution to the problem:

  • Key Idea: We don't need to compute the full decimal value of each binary prefix. Instead, we can efficiently calculate the remainder when divided by 5 at each step. If x_i is the decimal value of nums[0...i], then x_{i+1} = 2 * x_i + nums[i+1]. We can apply the modulo operator at each step to keep the numbers small and avoid potential overflow issues.
  • Modulo Arithmetic: Using the property (a * b) % m = ((a % m) * (b % m)) % m and (a + b) % m = ((a % m) + (b % m)) % m, we can efficiently compute the remainders.
  • Iterative Calculation: We initialize the remainder to 0 and iterate through the nums array, updating the remainder at each step using the formula remainder = (2 * remainder + nums[i]) % 5.

  • Complexity: O(n) runtime, O(n) storage.

Code

    def prefixesDivBy5(nums):
    """
    Given a binary array nums, return an array of booleans answer where answer[i] is true if xi is divisible by 5.

    Args:
        nums (list[int]): A list of integers representing a binary array.

    Returns:
        list[bool]: A list of booleans where the i-th element is True if the number represented by nums[0..i] is divisible by 5.
    """
    result = []
    remainder = 0
    for num in nums:
        remainder = (2 * remainder + num) % 5
        result.append(remainder == 0)
    return result

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