# Solving Leetcode Interviews in Seconds with AI: Check If Digits Are Equal in String After Operations II


	# 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](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 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
	```python
	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())
	```