May Leetcode Challenge - Day 10
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 differenttrust[i][0] != trust[i][1]
1 <= trust[i][0], trust[i][1] <= N
Solution
Just a straight loop to count number of trust for each person and how many person they trust
class Solution {
func findJudge(_ N: Int, _ trust: [[Int]]) -> Int {
var trustMe = Array(repeating: 0, count: N + 1)
var iTrustSomeone = Array(repeating: 0, count: N + 1)
for t in trust {
trustMe[t[1]] += 1
iTrustSomeone[t[0]] += 1
}
for i in 1...N {
if (trustMe[i] == N - 1) && (iTrustSomeone[i] == 0) {
return i
}
}
return -1
}
}
We could optimize memory by using only trustMe array