# Solving Leetcode Interviews in Seconds with AI: Lexicographically Smallest String After a Swap


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "3216" 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
	> Given a string s containing only digits, return the lexicographically smallest string that can be obtained after swapping adjacent digits in s with the same parity at most once. Digits have the same parity if both are odd or both are even. For example, 5 and 9, as well as 2 and 4, have the same parity, while 6 and 9 do not.   Example 1:  Input: s = "45320" Output: "43520" Explanation:  s[1] == '5' and s[2] == '3' both have the same parity, and swapping them results in the lexicographically smallest string.  Example 2:  Input: s = "001" Output: "001" Explanation: There is no need to perform a swap because s is already the lexicographically smallest.    Constraints:  2 <= s.length <= 100 s consists only of digits.  

	# Explanation
	Here's the solution:

*   **Iterate and Find:** Iterate through the string. For each digit, find the lexicographically smallest digit with the same parity to its right.
*   **Swap if Smaller:** If a smaller digit with the same parity is found, swap them to minimize the string lexicographically. Only the *first* such swap from left to right matters.
*   **Early Exit:** After performing a swap, proceed to the next digit.

*   **Runtime Complexity:** O(N^2), where N is the length of the string. **Storage Complexity:** O(N)

	
	# Code
	```python
	def smallest_string(s: str) -> str:
    """
    Given a string s containing only digits, return the lexicographically smallest string that can be obtained
    after swapping adjacent digits in s with the same parity at most once. Digits have the same parity if both
    are odd or both are even.

    For example:
    smallest_string("45320") == "43520"
    smallest_string("001") == "001"
    """
    s_list = list(s)
    n = len(s)

    for i in range(n):
        best_idx = -1
        for j in range(i + 1, n):
            if (int(s_list[i]) % 2 == int(s_list[j]) % 2):
                if best_idx == -1 or s_list[j] < s_list[best_idx]:
                    best_idx = j

        if best_idx != -1 and s_list[best_idx] < s_list[i]:
            s_list[i], s_list[best_idx] = s_list[best_idx], s_list[i]
            break  # Only one swap is allowed

    return "".join(s_list)
	```