Introduction

In this blog post, we will explore how to solve the LeetCode problem "2546" 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 two 0-indexed binary strings s and target of the same length n. You can do the following operation on s any number of times: Choose two different indices i and j where 0 <= i, j < n. Simultaneously, replace s[i] with (s[i] OR s[j]) and s[j] with (s[i] XOR s[j]). For example, if s = "0110", you can choose i = 0 and j = 2, then simultaneously replace s[0] with (s[0] OR s[2] = 0 OR 1 = 1), and s[2] with (s[0] XOR s[2] = 0 XOR 1 = 1), so we will have s = "1110". Return true if you can make the string s equal to target, or false otherwise. Example 1: Input: s = "1010", target = "0110" Output: true Explanation: We can do the following operations: - Choose i = 2 and j = 0. We have now s = "0010". - Choose i = 2 and j = 1. We have now s = "0110". Since we can make s equal to target, we return true. Example 2: Input: s = "11", target = "00" Output: false Explanation: It is not possible to make s equal to target with any number of operations. Constraints: n == s.length == target.length 2 <= n <= 105 s and target consist of only the digits 0 and 1.

Explanation

Here's the breakdown of the solution:

• Count Ones: The core idea is that the number of ones in the string is invariant under the given operations. This is because s[i] OR s[j] and s[i] XOR s[j] together contain the same number of ones as s[i] and s[j] combined.
• Handle All Zeros Case: If the target string contains all zeros, the source string must also contain all zeros to be transformable. Otherwise, it's impossible.

• Compare One Counts: If the target string has at least one '1', we simply need to check if the source string also has at least one '1'. If both have at least one '1', the transformation is possible.

• Time Complexity: O(n), where n is the length of the strings.

• Space Complexity: O(1)

Code

    def makeStringsEqual(s: str, target: str) -> bool:
    """
    Checks if string s can be transformed into target using the given operations.
    """

    s_ones = s.count('1')
    target_ones = target.count('1')

    if target_ones == 0:
        return s_ones == 0
    else:
        return s_ones > 0