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


	# 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](https://www.chatmagic.app), 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
	```python
	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
	```