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Count Unhappy Friends

You are given a list of preferences for n friends, where n is always even.

For each person i, preferences[i] contains a list of friends sorted in the order of preference. In other words, a friend earlier in the list is more preferred than a friend later in the list. Friends in each list are denoted by integers from 0 to n-1.

All the friends are divided into pairs. The pairings are given in a list pairs, where pairs[i] = [xi, yi] denotes xi is paired with yi and yi is paired with xi.

However, this pairing may cause some of the friends to be unhappy. A friend x is unhappy if x is paired with y and there exists a friend u who is paired with v but:

  • x prefers u over y, and
  • u prefers x over v.

Return the number of unhappy friends.

 

Example 1:

Input: n = 4, preferences = [[1, 2, 3], [3, 2, 0], [3, 1, 0], [1, 2, 0]], pairs = [[0, 1], [2, 3]]
Output: 2
Explanation:
Friend 1 is unhappy because:
- 1 is paired with 0 but prefers 3 over 0, and
- 3 prefers 1 over 2.
Friend 3 is unhappy because:
- 3 is paired with 2 but prefers 1 over 2, and
- 1 prefers 3 over 0.
Friends 0 and 2 are happy.

Example 2:

Input: n = 2, preferences = [[1], [0]], pairs = [[1, 0]]
Output: 0
Explanation: Both friends 0 and 1 are happy.

Example 3:

Input: n = 4, preferences = [[1, 3, 2], [2, 3, 0], [1, 3, 0], [0, 2, 1]], pairs = [[1, 3], [0, 2]]
Output: 4

 

Constraints:

  • 2 <= n <= 500
  • n is even.
  • preferences.length == n
  • preferences[i].length == n - 1
  • 0 <= preferences[i][j] <= n - 1
  • preferences[i] does not contain i.
  • All values in preferences[i] are unique.
  • pairs.length == n/2
  • pairs[i].length == 2
  • xi != yi
  • 0 <= xi, yi <= n - 1
  • Each person is contained in exactly one pair.

Solution Explanation: Counting Unhappy Friends

This problem involves determining the number of unhappy friends given a list of friend preferences and pairings. A friend is unhappy if they prefer another friend over their assigned partner, and that other friend also prefers them over their assigned partner.

Approach:

The solution utilizes a clever approach leveraging the inherent structure of the problem to efficiently count unhappy friends. It avoids brute-force comparisons of all possible friend combinations.

  1. Preprocessing:

    • A 2D array d is created to store the "distance" of a friend in another friend's preference list. d[i][j] represents the index (position) of friend j in friend i's preference list. A smaller value indicates a higher preference.
    • A 1D array p stores the assigned partner for each friend. p[i] is the friend paired with friend i.

  2. Iterating through Friends:

    • The code iterates through each friend x.
    • For each friend x, it retrieves their assigned partner y from p[x].
    • It then iterates through x's preference list (preferences[x]) up to the index of their partner y (meaning only the friends preferred to y). This optimized iteration is key to efficiency.
    • For each friend u preferred by x over y, the code checks if u prefers x over their partner v (p[u]). This check is done using the d array. If both conditions are true, x is unhappy, and the counter ans is incremented. The inner loop breaks because once a friend is found who makes x unhappy, there's no need to check the rest.

  3. Returning the Result:

    • Finally, the function returns ans, representing the total number of unhappy friends.

Time Complexity Analysis:

  • The preprocessing step (creating d and p) takes O(n²) time due to nested loops iterating through the preferences and pairs.
  • The main loop iterates through each friend (n times). The inner loop iterates only up to the index of the paired friend in the preference list, which is at most n-1. Therefore, the main loop's complexity is roughly O(n²).
  • Overall, the dominant factor is O(n²), so the time complexity is O(n²).

Space Complexity Analysis:

  • The d array uses O(n²) space.
  • The p array uses O(n) space.
  • The overall space complexity is O(n²).

Code Examples (Python):

def unhappyFriends(n, preferences, pairs):
    d = [[preferences[i].index(j) for j in range(n)] for i in range(n)]  # Preprocessing
    p = {}
    for x, y in pairs:
        p[x] = y
        p[y] = x
 
    ans = 0
    for x in range(n):
        y = p[x]
        for i in range(d[x][y]): # Optimized Iteration
            u = preferences[x][i]
            v = p[u]
            if d[u][x] < d[u][v]:
                ans += 1
                break  # Optimization: Break inner loop once unhappy friend is found
 
    return ans

This detailed explanation, including the optimized code and complexity analysis, provides a comprehensive understanding of the solution to the "Count Unhappy Friends" problem. The key to efficient solving is the pre-processing step and the clever use of the d array to quickly determine preferences without exhaustive comparisons.