find-the-town-judge

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Description

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:

1. The town judge trusts nobody.
2. Everybody (except for the town judge) trusts the town judge.
3. 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.

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

Solution(Python)

class Solution:
    # The key idea is to visulaize this problem as grpah of n people with trust as
    # graph problem
    # According to the problem Town judge
    # 1.trusts nobody (i.e zero out going connections)
    # 2.everybody trusts him (i.e n incoiming connections)
    #
    # Instead of creating and traversing graph for finding a node with n-1             # connections
    #
    #  A counter array of size n indicating n people can be used to count
    #  number of outgoing connections by incrementing incoming connetion and
    #  decrementing the out going connection
    #
    # Time Complexity : O(T) ,T = length of trust array since 1<= n<=1000 nis               negligible
    # Space Complexity: O(T)
    def findJudge(self, n: int, trust: List[List[int]]) -> int:
        cntr = [0] * (n)

        for (a, b) in trust:
            cntr[a - 1] -= 1
            cntr[b - 1] += 1

        for i, cnt in enumerate(cntr):
            if cnt == n - 1:
                return i + 1

        return -1