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, 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

    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