Understand what the interviewer is asking for by using test cases and questions about the problem.
- Established a set (2-3) of test cases to verify their own solution later.
- Established a set (1-2) of edge cases to verify their solution handles complexities.
- Have fully understood the problem and have no clarifying questions.
- Have you verified any Time/Space Constraints for this problem?
O(V + E)
time and O(V)
space. HAPPY CASE
Input: n = 2, trust = [[1,2]]
Output: 2
Input: n = 3, trust = [[1,3],[2,3]]
Output: 3
EDGE CASE
Input: n = 1, trust = []
Output: 1
Match what this problem looks like to known categories of problems, e.g. Linked List or Dynamic Programming, and strategies or patterns in those categories.
For Graph Problems, common solution patterns include:
Plan the solution with appropriate visualizations and pseudocode.
General Idea: Since all trust pairs are unique, we can ensure that any person with n-1 trust is the town judge, by reducing trust for the truster and increase trust for the trustee. This way a person who trust another person cannot achieve n-1 trust.
1. Create a list of of person and their trust value
2. Increase trust value of trustee and decrease trust value of truster for each pair
3. Check if anyone achieves n-1 trust. This person is the town judge
4. If no one achieves n-1 trust, then there is no town judge.
⚠️ Common Mistakes
Implement the code to solve the algorithm.
class Solution:
def findJudge(self, n: int, trust: List[List[int]]) -> int:
# Create a list of of person and their trust value
numTrust = [0] * (n + 1)
# Increase trust value of trustee and decrease trust value of truster for each pair
for person1, person2 in trust:
numTrust[person1] -= 1
numTrust[person2] += 1
# Check if anyone achieves n-1 trust. This person is the town judge
for i in range(1, len(numTrust)):
if numTrust[i] == n - 1:
return i
# If no one achieves n-1 trust, then there is no town judge
return -1
class Solution {
public int findJudge(int n, int[][] trust) {
if (trust.length == 0 && n == 1)
return 1;
// Create a list of of person and their trust value
int[] count = new int[n + 1];
// Increase trust value of trustee and decrease trust value of truster for each pair
for (int[] person : trust) {
count[person[0]]--;
count[person[1]]++;
}
// Check if anyone achieves n-1 trust. This person is the town judge
for (int person = 0; person < count.length; person++) {
if (count[person] == n - 1) return person;
}
// If no one achieves n-1 trust, then there is no town judge
return -1;
}
}
Review the code by running specific example(s) and recording values (watchlist) of your code's variables along the way.
Evaluate the performance of your algorithm and state any strong/weak or future potential work.
Assume V
represents the number of vertices/nodes.
Assume E
represents the number of edges