# Solving Leetcode Interviews in Seconds with AI: Find the Town Judge


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "997" 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
	> In a town, there are n people labeled from 1 to n. There is a rumor that one of these people is secretly the town judge. If the town judge exists, then:  The town judge trusts nobody. Everybody (except for the town judge) trusts the town judge. There is exactly one person that satisfies properties 1 and 2.  You are given an array trust where trust[i] = [ai, bi] representing that the person labeled ai trusts the person labeled bi. If a trust relationship does not exist in trust array, then such a trust relationship does not exist. Return the label of the town judge if the town judge exists and can be identified, or return -1 otherwise.   Example 1:  Input: n = 2, trust = [[1,2]] Output: 2  Example 2:  Input: n = 3, trust = [[1,3],[2,3]] Output: 3  Example 3:  Input: n = 3, trust = [[1,3],[2,3],[3,1]] Output: -1    Constraints:  1 <= n <= 1000 0 <= trust.length <= 104 trust[i].length == 2 All the pairs of trust are unique. ai != bi 1 <= ai, bi <= n  

	# Explanation
	Here's the breakdown of the solution and the Python code:

*   **Key Idea:** The judge trusts no one, meaning they won't appear in the first position of any `trust` pair. Everyone else trusts the judge, meaning the judge will be the second element in `n-1` `trust` pairs. We can use an in-degree and out-degree array to keep track of trust relationships.
*   **Optimization:** We can solve this in a single pass through the `trust` array and then a final linear scan to find the judge, avoiding nested loops.

*   **Complexity:** O(T + N) time, where T is the length of the `trust` array, and O(N) space, where N is the number of people in the town.

	
	# Code
	```python
	def findJudge(n: int, trust: list[list[int]]) -> int:
    """
    Finds the town judge if it exists.

    Args:
        n: The number of people in the town.
        trust: A list of trust relationships, where trust[i] = [ai, bi] means ai trusts bi.

    Returns:
        The label of the town judge if it exists and can be identified, or -1 otherwise.
    """

    if n == 1 and not trust:
        return 1

    in_degree = [0] * (n + 1)
    out_degree = [0] * (n + 1)

    for a, b in trust:
        out_degree[a] += 1
        in_degree[b] += 1

    for i in range(1, n + 1):
        if in_degree[i] == n - 1 and out_degree[i] == 0:
            return i

    return -1
	```