Day9 : 30 Day LeetCode may challenges

find_the_judge.cpp

class Solution {
public:
    int findJudge(int N, vector<vector<int>>& trust) {
        unordered_map<int,pair<int,int>>mp;
        //pair<int,int> for  outgoing(trust[i][0]) and incoming(trust[i][1]) 
        //judge has incoming = N-1 and outgoing = 0
        int n = trust.size();
        for(int i=0; i<n; i++){
            mp[trust[i][0]].first++;//first for outgoing from a number 
            mp[trust[i][1]].second++;//second for incoming to a number
        }
        for(int i=1; i<=N; i++){
            if (mp[i].first == 0 && mp[i].second == N-1)
                return i;
        }
        return -1;
    }
};
/*
        vector<int> in(N + 1), out(N + 1);
		for (auto a : trust) {
			++out[a[0]];
			++in[a[1]];
		}
		for (int i = 1; i <= N; ++i) {
			if (in[i] == N - 1 && out[i] == 0) return i;
		}
		return -1;
*/

Problem_statement9.txt

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:

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 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 <= N <= 1000
trust.length <= 10000
trust[i] are all different
trust[i][0] != trust[i][1]
1 <= trust[i][0], trust[i][1] <= N