LeetCode #3238 — EASY

Find the Number of Winning Players

Build confidence with an intuition-first walkthrough focused on array fundamentals.

The Problem

Problem Statement

You are given an integer n representing the number of players in a game and a 2D array pick where pick[i] = [x_i, y_i] represents that the player x_i picked a ball of color y_i.

Player i wins the game if they pick strictly more than i balls of the same color. In other words,

Player 0 wins if they pick any ball.
Player 1 wins if they pick at least two balls of the same color.
...
Player i wins if they pick at least i + 1 balls of the same color.

Return the number of players who win the game.

Note that multiple players can win the game.

Example 1:

Input: n = 4, pick = [[0,0],[1,0],[1,0],[2,1],[2,1],[2,0]]

Output: 2

Explanation:

Player 0 and player 1 win the game, while players 2 and 3 do not win.

Example 2:

Input: n = 5, pick = [[1,1],[1,2],[1,3],[1,4]]

Output: 0

Explanation:

No player wins the game.

Example 3:

Input: n = 5, pick = [[1,1],[2,4],[2,4],[2,4]]

Output: 1

Explanation:

Player 2 wins the game by picking 3 balls with color 4.

Constraints:

2 <= n <= 10
1 <= pick.length <= 100
pick[i].length == 2
0 <= x_i <= n - 1
0 <= y_i <= 10

Step 01

Brute Force Baseline

Problem summary: You are given an integer n representing the number of players in a game and a 2D array pick where pick[i] = [xi, yi] represents that the player xi picked a ball of color yi. Player i wins the game if they pick strictly more than i balls of the same color. In other words, Player 0 wins if they pick any ball. Player 1 wins if they pick at least two balls of the same color. ... Player i wins if they pick at least i + 1 balls of the same color. Return the number of players who win the game. Note that multiple players can win the game.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Hash Map

Example 1

4
[[0,0],[1,0],[1,0],[2,1],[2,1],[2,0]]

Example 2

5
[[1,1],[1,2],[1,3],[1,4]]

Example 3

5
[[1,1],[2,4],[2,4],[2,4]]

Core Insight

What unlocks the optimal approach

Keep track of the number of balls of each color for each user using hashing.
Find the maximum color that occurred for each player.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3238: Find the Number of Winning Players
class Solution {
    public int winningPlayerCount(int n, int[][] pick) {
        int[][] cnt = new int[n][11];
        Set<Integer> s = new HashSet<>();
        for (var p : pick) {
            int x = p[0], y = p[1];
            if (++cnt[x][y] > x) {
                s.add(x);
            }
        }
        return s.size();
    }
}

// Accepted solution for LeetCode #3238: Find the Number of Winning Players
func winningPlayerCount(n int, pick [][]int) int {
	cnt := make([][11]int, n)
	s := map[int]struct{}{}
	for _, p := range pick {
		x, y := p[0], p[1]
		cnt[x][y]++
		if cnt[x][y] > x {
			s[x] = struct{}{}
		}
	}
	return len(s)
}

# Accepted solution for LeetCode #3238: Find the Number of Winning Players
class Solution:
    def winningPlayerCount(self, n: int, pick: List[List[int]]) -> int:
        cnt = [[0] * 11 for _ in range(n)]
        s = set()
        for x, y in pick:
            cnt[x][y] += 1
            if cnt[x][y] > x:
                s.add(x)
        return len(s)

// Accepted solution for LeetCode #3238: Find the Number of Winning Players
/**
 * [3238] Find the Number of Winning Players
 */
pub struct Solution {}

// submission codes start here

impl Solution {
    pub fn winning_player_count(n: i32, pick: Vec<Vec<i32>>) -> i32 {
        use std::collections::HashMap;
        let n = n as usize;

        let mut player = vec![HashMap::new(); n];
        for p in pick {
            let (p, b) = (p[0] as usize, p[1]);

            let mut entry = player[p].entry(b).or_insert(0);
            *entry += 1;
        }

        player
            .iter()
            .enumerate()
            .filter_map(|(i, map)| {
                if map.iter().any(|(_, v)| *v > i) {
                    Some(())
                } else {
                    None
                }
            })
            .count() as i32
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    fn test_3238() {
        assert_eq!(
            2,
            Solution::winning_player_count(
                4,
                vec![
                    vec![0, 0],
                    vec![1, 0],
                    vec![1, 0],
                    vec![2, 1],
                    vec![2, 1],
                    vec![2, 0]
                ]
            )
        );
    }
}

// Accepted solution for LeetCode #3238: Find the Number of Winning Players
function winningPlayerCount(n: number, pick: number[][]): number {
    const cnt: number[][] = Array.from({ length: n }, () => Array(11).fill(0));
    const s = new Set<number>();
    for (const [x, y] of pick) {
        if (++cnt[x][y] > x) {
            s.add(x);
        }
    }
    return s.size;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(m + n × M)

Space

O(n × M)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

Find the Number of Winning Players

Problem Statement

Roadmap

Brute Force Baseline

Baseline thinking

Example 1

Example 2

Example 3

Related Problems

Core Insight

What unlocks the optimal approach

Algorithm Walkthrough

Iteration Checklist

Edge Cases

Full Annotated Code

Interactive Study Demo

Complexity Analysis

Approach Breakdown

Common Mistakes

Off-by-one on range boundaries

Mutating counts without cleanup