How do you solve Count Unhappy Friends on LeetCode?

Count Unhappy Friends (LeetCode #1583) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Using a ranking matrix allows for efficient preference comparisons.

What is the brute force solution for Count Unhappy Friends?

The brute force approach for Count Unhappy Friends is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks each friend to see if they are unhappy by comparing their preferences with every other pair. This method is straightforward but inefficient as it involves nested loops. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Count Unhappy Friends?

Count Unhappy Friends uses the following concepts: Array, Simulation. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Count Unhappy Friends?

Always clarify the problem statement and constraints before diving into coding. Consider edge cases and test your solution with them. Explain your thought process clearly while coding to demonstrate your understanding.

What are common mistakes when solving Count Unhappy Friends?

Not using a ranking matrix, leading to inefficient comparisons. Overlooking edge cases where friends have the same preferences.

#1583

Count Unhappy Friends

Medium

ArraySimulationHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal approach uses a ranking matrix to quickly determine preferences, allowing us to check unhappiness in constant time. This significantly reduces the number of comparisons needed.

⚙️

Algorithm

3 steps

1Step 1: Create a rank matrix where rank[i][j] indicates the preference rank of friend j for friend i.
2Step 2: For each pair in pairs, check if either friend is unhappy by using the rank matrix.
3Step 3: Count the number of unhappy friends based on the conditions provided.

solution.py12 lines

1def countUnhappyFriends(n, preferences, pairs):
2    rank = [[0] * n for _ in range(n)]
3    for i in range(n):
4        for j, friend in enumerate(preferences[i]):
5            rank[i][friend] = j
6    unhappy_count = 0
7    for x, y in pairs:
8        for u, v in pairs:
9            if x != u and rank[x][u] < rank[x][y] and rank[u][x] < rank[u][v]:
10                unhappy_count += 1
11                break
12    return unhappy_count

ℹ

Complexity note: The optimal approach runs in linear time due to the use of a pre-computed ranking matrix, allowing for constant time checks for unhappiness.

1Using a ranking matrix allows for efficient preference comparisons.
2Understanding the conditions for unhappiness is crucial for implementing the solution.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.