LeetCode in Kotlin

3238. Find the Number of Winning Players

Easy

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 leasti + 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 <= xi <= n - 1
• 0 <= yi <= 10

Solution

@Suppress("UNUSED_PARAMETER")
class Solution {
    fun winningPlayerCount(n: Int, pick: Array<IntArray>): Int {
        val dp = Array(11) { IntArray(11) }
        for (ints in pick) {
            val p = ints[0]
            val pi = ints[1]
            dp[p][pi] += 1
        }
        var count = 0
        for (i in 0..10) {
            var win = false
            for (j in 0..10) {
                if (dp[i][j] >= i + 1) {
                    win = true
                    break
                }
            }
            if (win) {
                count += 1
            }
        }
        return count
    }
}

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