How do you solve Find the Number of Winning Players on LeetCode?

Find the Number of Winning Players (LeetCode #3238) 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: Players win by collecting a specific number of balls of the same color.

What is the brute force solution for Find the Number of Winning Players?

The brute force approach for Find the Number of Winning Players is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks each player's picks and counts the number of balls of each color they have. Then, it compares this count to the required number of balls for each player to win. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Find the Number of Winning Players?

Find the Number of Winning Players uses the following concepts: Array, Hash Table, Counting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Find the Number of Winning Players?

Clarify the problem statement and constraints before jumping to code. Think about edge cases and how they might affect your solution. Explain your thought process clearly while coding to show your understanding.

What are common mistakes when solving Find the Number of Winning Players?

Not correctly checking the winning condition for each player. Overlooking edge cases where players have no balls picked.

#3238

Find the Number of Winning Players

Easy

ArrayHash TableCountingHash 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 solution uses a hash map to count the occurrences of each color for each player in a single pass. This reduces the time complexity significantly by avoiding nested loops.

⚙️

Algorithm

5 steps

1Step 1: Initialize a hash map to count colors for each player.
2Step 2: Iterate through the pick array and populate the hash map with counts of each color for each player.
3Step 3: For each player, check if any color count exceeds the player's index + 1.
4Step 4: Count the number of players who meet the winning condition.
5Step 5: Return the total number of winning players.

solution.py10 lines

1# Full working Python code
2
3def count_winning_players(n, pick):
4    color_count = [{} for _ in range(n)]
5    for x, y in pick:
6        color_count[x][y] = color_count[x].get(y, 0) + 1
7    return sum(1 for player in range(n) if any(count > player for count in color_count[player].values()))
8
9# Example usage
10print(count_winning_players(4, [[0,0],[1,0],[1,0],[2,1],[2,1],[2,0]]))

ℹ

Complexity note: The time complexity is O(n) because we only iterate through the picks once to build the color counts, and then we check each player's counts in linear time. The space complexity is O(n) due to the storage of color counts for each player.

1Players win by collecting a specific number of balls of the same color.
2Using hash maps allows efficient counting of occurrences.

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