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May-10 2020 Challenge


10. Find the Town Judge 
In a town, there are N people labelled 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 trust, an array of pairs trust[i] = [a, b] representing that the person labelled a trusts the person labelled b.
If the town judge exists and can be identified, return the label of the town judge.  Otherwise, return -1.
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
Example 4:
Input: N = 3, trust = [[1,2],[2,3]]
Output: -1
Example 5:
Input: N = 4, trust = [[1,3],[1,4],[2,3],[2,4],[4,3]]
Output: 3
Note:
  1. 1 <= N <= 1000
  2. trust.length <= 10000
  3. trust[i] are all different
  4. trust[i][0] != trust[i][1]
  5. 1 <= trust[i][0], trust[i][1] <=N
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Solution in C++:

class Solution {
public:
    int findJudge(int N, vector<vector<int>>& trust) {
        vector<vector<int>> knows(N + 1, vector<int>(N + 1));
  for (auto &t : trust) knows[t[0]][t[1]] = 1;
  return findCelebrity(N, knows);
}
int findCelebrity(int n, vector<vector<int>>& knows, int i = 1) {
  for (auto j = i + 1; j <= n; ++j) if (knows[i][j]) i = j;
  for (auto j = 1; j < i; ++j) if (knows[i][j]) return -1;
  for (auto j = 1; j <= n; ++j) if (i != j && !knows[j][i]) return -1;
  return i;
    }
};

   

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