Introduction

In this blog post, we will explore how to solve the LeetCode problem "3463" 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 string s consisting of digits. Perform the following operation repeatedly until the string has exactly two digits: For each pair of consecutive digits in s, starting from the first digit, calculate a new digit as the sum of the two digits modulo 10. Replace s with the sequence of newly calculated digits, maintaining the order in which they are computed. Return true if the final two digits in s are the same; otherwise, return false. Example 1: Input: s = "3902" Output: true Explanation: Initially, s = "3902" First operation: (s[0] + s[1]) % 10 = (3 + 9) % 10 = 2 (s[1] + s[2]) % 10 = (9 + 0) % 10 = 9 (s[2] + s[3]) % 10 = (0 + 2) % 10 = 2 s becomes "292" Second operation: (s[0] + s[1]) % 10 = (2 + 9) % 10 = 1 (s[1] + s[2]) % 10 = (9 + 2) % 10 = 1 s becomes "11" Since the digits in "11" are the same, the output is true. Example 2: Input: s = "34789" Output: false Explanation: Initially, s = "34789". After the first operation, s = "7157". After the second operation, s = "862". After the third operation, s = "48". Since '4' != '8', the output is false. Constraints: 3 <= s.length <= 105 s consists of only digits.

Explanation

Here's a breakdown of the solution:

• Iterative Reduction: The core idea is to repeatedly apply the given operation (sum of consecutive digits modulo 10) until the string's length is reduced to 2.
• In-place Update: To optimize memory usage, the string s is updated in place with the newly calculated digits in each iteration. A new string is not created in each iteration.

• Comparison: After the iterations stop when the string contains exactly two digits, the two digits are compared for equality.

• Runtime Complexity: O(n^2), where n is the initial length of the string.

• Storage Complexity: O(1), as the string is updated in-place and minimal extra space is used.

Code

    def solve():
    s = input()
    while len(s) > 2:
        new_s = ""
        for i in range(len(s) - 1):
            new_s += str((int(s[i]) + int(s[i+1])) % 10)
        s = new_s
    return s[0] == s[1]

print(solve())