Introduction

In this blog post, we will explore how to solve the LeetCode problem "1583" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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.

Explanation

Here's the solution:

• Core Idea: Iterate through each person and check if they are unhappy. A person is unhappy if their partner isn't their best preference, and someone else prefers them over their current partner and they prefer that someone over their current partner.
• Optimization: Use a preference matrix (2D array) to quickly determine if person x prefers person u over person y. This avoids repeatedly searching through the preference lists.

• Unhappiness Check: For each person, iterate through all other people and check if the conditions for unhappiness are met.

• Runtime Complexity: O(n^3), where n is the number of friends. Storage Complexity: O(n^2).

Code

    def unhappyFriends(n: int, preferences: list[list[int]], pairs: list[list[int]]) -> int:
    """
    Calculates the number of unhappy friends based on their preferences and pairings.

    Args:
        n: The number of friends.
        preferences: A list of preferences for each friend.
        pairs: A list of pairs indicating the pairings of friends.

    Returns:
        The number of unhappy friends.
    """

    # Create a preference matrix for efficient lookup
    pref_matrix = [[0] * n for _ in range(n)]
    for i in range(n):
        for j, preferred_friend in enumerate(preferences[i]):
            pref_matrix[i][preferred_friend] = j

    # Create a dictionary to store the partner of each person
    partner = {}
    for x, y in pairs:
        partner[x] = y
        partner[y] = x

    unhappy_count = 0
    unhappy = [False] * n  # To avoid double counting

    for x in range(n):
        y = partner[x]  # Current partner of x

        for u in range(n):
            if x == u:
                continue

            v = partner[u]
            # Check if x prefers u over y AND u prefers x over v
            if pref_matrix[x][u] < pref_matrix[x][y] and pref_matrix[u][x] < pref_matrix[u][v]:
                unhappy[x] = True


    return sum(unhappy)