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

Solution: Degree

node with degree (in_degree – out_degree) N – 1 is the judge.

Time complexity: O(N+T)
Space complexity: O(N)

class Solution {
public:
  int findJudge(int N, vector<vector<int>>& trusts) {    
    vector<int> degrees(N + 1);    
    for (const auto& trust : trusts) {
      --degrees[trust[0]];
      ++degrees[trust[1]];
    }
    for (int i = 1; i <= N; ++i)
      if (degrees[i] == N - 1) return i;
    return -1;
  }
};

The post 花花酱 LeetCode 997. Find the Town Judge appeared first on Huahua's Tech Road.

Problem:

https://leetcode.com/problems/degree-of-an-array/description/

Given a non-empty array of non-negative integers nums, the degree of this array is defined as the maximum frequency of any one of its elements.

Your task is to find the smallest possible length of a (contiguous) subarray of nums, that has the same degree as nums.

Example 1:

Input: [1, 2, 2, 3, 1]
Output: 2
Explanation: 
The input array has a degree of 2 because both elements 1 and 2 appear twice.
Of the subarrays that have the same degree:
[1, 2, 2, 3, 1], [1, 2, 2, 3], [2, 2, 3, 1], [1, 2, 2], [2, 2, 3], [2, 2]
The shortest length is 2. So return 2.

Example 2:

Input: [1,2,2,3,1,4,2]
Output: 6

Solution 1: Hashtable

Time complexity: O(n)

Space complexity: O(n)

C++

// Author: Huahua
// Running time: 33 ms (beats 90.96%)
class Solution {
public:
  int findShortestSubArray(vector<int>& nums) {
    unordered_map<int, vector<int>> indices;
    for (int i = 0; i < nums.size(); ++i)
      indices[nums[i]].push_back(i);
    size_t degree = 0;
    for (const auto& p : indices)
      degree = max(degree, p.second.size());
    int ans = nums.size();
    for (const auto& p : indices) {
      if (p.second.size() != degree) continue;
      ans = min(ans, p.second.back() - p.second.front() + 1);
    }
    return ans;
  }
};

The post 花花酱 LeetCode 697. Degree of an Array appeared first on Huahua's Tech Road.