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Copy path277. Find the Celebrity.cpp
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277. Find the Celebrity.cpp
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/*
Link: https://leetcode.com/problems/find-the-celebrity
Suppose you are at a party with n people (labeled from 0 to n - 1) and among them, there may exist one celebrity. The definition of a celebrity is that all the other n - 1 people know him/her but he/she does not know any of them.
Now you want to find out who the celebrity is or verify that there is not one. The only thing you are allowed to do is to ask questions like: "Hi, A. Do you know B?" to get information of whether A knows B. You need to find out the celebrity (or verify there is not one) by asking as few questions as possible (in the asymptotic sense).
You are given a helper function bool knows(a, b) which tells you whether A knows B. Implement a function int findCelebrity(n), your function should minimize the number of calls to knows.
There will be exactly one celebrity if he/she is in the party. Return the celebrity's label if there is a celebrity in the party. If there is no celebrity, return -1.
Example1
Input:
2 // next n * (n - 1) lines
0 knows 1
1 does not know 0
Output: 1
Explanation:
Everyone knows 1,and 1 knows no one.
Example2
Input:
3 // next n * (n - 1) lines
0 does not know 1
0 does not know 2
1 knows 0
1 does not know 2
2 knows 0
2 knows 1
Output: 0
Explanation:
Everyone knows 0,and 0 knows no one.
0 does not know 1,and 1 knows 0.
2 knows everyone,but 1 does not know 2.
*/
// Solution
// We eliminate 1 person by asking any two persons,
// and then, eliminate one by asking one more and
// the one from above result. Hence at the end n - 1
// persons would have been eliminated. Now we check
// with other peson if they know him or not.
/*
TC = O(n)
SC = O(1)
*/
// Forward declaration of the knows API.
bool knows(int a, int b);
class Solution {
public:
/**
* @param n a party with n people
* @return the celebrity's label or -1
*/
int findCelebrity(int n) {
std::vector<int> v;
int i, candidate = 0;
for (i = 1; i < n; i++) {
if (knows(candidate, i))
candidate = i;
}
for (i = 0; i < n; i++) {
if (candidate != i && (!knows(i, candidate) || knows(candidate, i)))
return -1;
}
return candidate;
}
};