# Solving Leetcode Interviews in Seconds with AI: Find Players With Zero or One Losses


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "2225" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), 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 an integer array matches where matches[i] = [winneri, loseri] indicates that the player winneri defeated player loseri in a match. Return a list answer of size 2 where:  answer[0] is a list of all players that have not lost any matches. answer[1] is a list of all players that have lost exactly one match.  The values in the two lists should be returned in increasing order. Note:  You should only consider the players that have played at least one match. The testcases will be generated such that no two matches will have the same outcome.    Example 1:  Input: matches = [[1,3],[2,3],[3,6],[5,6],[5,7],[4,5],[4,8],[4,9],[10,4],[10,9]] Output: [[1,2,10],[4,5,7,8]] Explanation: Players 1, 2, and 10 have not lost any matches. Players 4, 5, 7, and 8 each have lost one match. Players 3, 6, and 9 each have lost two matches. Thus, answer[0] = [1,2,10] and answer[1] = [4,5,7,8].  Example 2:  Input: matches = [[2,3],[1,3],[5,4],[6,4]] Output: [[1,2,5,6],[]] Explanation: Players 1, 2, 5, and 6 have not lost any matches. Players 3 and 4 each have lost two matches. Thus, answer[0] = [1,2,5,6] and answer[1] = [].    Constraints:  1 <= matches.length <= 105 matches[i].length == 2 1 <= winneri, loseri <= 105 winneri != loseri All matches[i] are unique.  

	# Explanation
	Here's the breakdown of the approach, complexity, and Python code:

*   **High-Level Approach:**
    *   Use a dictionary (or array as a hashmap) to store the number of losses for each player.
    *   Iterate through the `matches` array and update the loss counts.
    *   Iterate through the loss counts to identify players with zero losses and exactly one loss, storing them in separate lists, and sort the lists in ascending order.

*   **Complexity:**
    *   Runtime Complexity: O(N + M), where N is the number of matches and M is the maximum player number (up to 10^5 based on the constraints).
    *   Storage Complexity: O(M) due to the `losses` dictionary/array.

*   **Python Code:**

	
	# Code
	```python
	def find_winners(mathes):    """
    Finds players with zero losses and exactly one loss in a match list.

    Args:
        matches: A list of lists, where each inner list represents a match
                 [winner, loser].

    Returns:
        A list of two lists:
        - The first list contains players with zero losses, sorted in ascending order.
        - The second list contains players with exactly one loss, sorted in ascending order.
    """

    losses = {}  # Dictionary to store the number of losses for each player

    # Count the number of losses for each player
    for winner, loser in matches:
        losses[winner] = losses.get(winner, 0)  # Initialize winner if not seen
        losses[loser] = losses.get(loser, 0) + 1

    zero_losses = []
    one_loss = []

    # Iterate through the loss counts to identify players with zero and one losses
    for player, loss_count in losses.items():
        if loss_count == 0:
            zero_losses.append(player)
        elif loss_count == 1:
            one_loss.append(player)

    # Find players who won but didn't lose, and thus might not be in the losses dict at all
    winners = set()
    for winner, _ in matches:
        winners.add(winner)
    
    for winner in winners:
        if winner not in losses:
            zero_losses.append(winner)
            losses[winner] = 0

    # Sort the lists in ascending order
    zero_losses.sort()
    one_loss.sort()

    return [zero_losses, one_loss]
	```