Introduction

In this blog post, we will explore how to solve the LeetCode problem "2840" 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 strings s1 and s2, both of length n, consisting of lowercase English letters. You can apply the following operation on any of the two strings any number of times: Choose any two indices i and j such that i < j and the difference j - i is even, then swap the two characters at those indices in the string. Return true if you can make the strings s1 and s2 equal, and false otherwise. Example 1: Input: s1 = "abcdba", s2 = "cabdab" Output: true Explanation: We can apply the following operations on s1: - Choose the indices i = 0, j = 2. The resulting string is s1 = "cbadba". - Choose the indices i = 2, j = 4. The resulting string is s1 = "cbbdaa". - Choose the indices i = 1, j = 5. The resulting string is s1 = "cabdab" = s2. Example 2: Input: s1 = "abe", s2 = "bea" Output: false Explanation: It is not possible to make the two strings equal. Constraints: n == s1.length == s2.length 1 <= n <= 105 s1 and s2 consist only of lowercase English letters.

Explanation

Here's the breakdown of the solution:

• Key Idea: The operation allows swapping characters at even distances. This means we can rearrange characters at even indices among themselves and characters at odd indices among themselves. Thus, we just need to check if the counts of characters at even indices are the same in both strings, and similarly for odd indices.
• Implementation: Separate even and odd indexed characters into separate lists. Sort the lists, then check if they are equal.
• Complexity: Time: O(n log n), Space: O(n) due to sorting and list creation.

Code

    def solve():
    s1 = input()
    s2 = input()
    n = len(s1)

    s1_even = sorted([s1[i] for i in range(n) if i % 2 == 0])
    s1_odd = sorted([s1[i] for i in range(n) if i % 2 != 0])
    s2_even = sorted([s2[i] for i in range(n) if i % 2 == 0])
    s2_odd = sorted([s2[i] for i in range(n) if i % 2 != 0])

    if s1_even == s2_even and s1_odd == s2_odd:
        return True
    else:
        return False

s1 = input()
s2 = input()
n = len(s1)

s1_even = sorted([s1[i] for i in range(n) if i % 2 == 0])
s1_odd = sorted([s1[i] for i in range(n) if i % 2 != 0])
s2_even = sorted([s2[i] for i in range(n) if i % 2 == 0])
s2_odd = sorted([s2[i] for i in range(n) if i % 2 != 0])

if s1_even == s2_even and s1_odd == s2_odd:
    print(True)
else:
    print(False)