Added tasks 3238-3245

src/main/kotlin/g3201_3300/s3238_find_the_number_of_winning_players/Solution.kt

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+package g3201_3300.s3238_find_the_number_of_winning_players
+
+// #Easy #Array #Hash_Table #Counting #2024_08_07_Time_207_ms_(90.38%)_Space_42_MB_(75.00%)
+
+@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
+    }
+}

src/main/kotlin/g3201_3300/s3238_find_the_number_of_winning_players/readme.md

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+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 <code>pick[i] = [x<sub>i</sub>, y<sub>i</sub>]</code> represents that the player <code>x<sub>i</sub></code> picked a ball of color <code>y<sub>i</sub></code>.
+
+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`
+*   <code>0 <= x<sub>i</sub> <= n - 1</code>
+*   <code>0 <= y<sub>i</sub> <= 10</code>

src/main/kotlin/g3201_3300/s3239_minimum_number_of_flips_to_make_binary_grid_palindromic_i/Solution.kt

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+package g3201_3300.s3239_minimum_number_of_flips_to_make_binary_grid_palindromic_i
+
+// #Medium #Array #Matrix #Two_Pointers #2024_08_07_Time_856_ms_(87.50%)_Space_109.2_MB_(66.67%)
+
+import kotlin.math.min
+
+class Solution {
+    fun minFlips(grid: Array<IntArray>): Int {
+        val m = grid.size
+        val n = grid[0].size
+        var rowFlips = 0
+        for (i in 0 until m / 2) {
+            for (j in 0 until n) {
+                val sum = grid[i][j] + grid[m - 1 - i][j]
+                rowFlips += min(sum, (2 - sum))
+            }
+        }
+        var columnFlips = 0
+        for (j in 0 until n / 2) {
+            for (ints in grid) {
+                val sum = ints[j] + ints[n - 1 - j]
+                columnFlips += min(sum, (2 - sum))
+            }
+        }
+        return min(rowFlips, columnFlips)
+    }
+}

src/main/kotlin/g3201_3300/s3239_minimum_number_of_flips_to_make_binary_grid_palindromic_i/readme.md

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+3239\. Minimum Number of Flips to Make Binary Grid Palindromic I
+
+Medium
+
+You are given an `m x n` binary matrix `grid`.
+
+A row or column is considered **palindromic** if its values read the same forward and backward.
+
+You can **flip** any number of cells in `grid` from `0` to `1`, or from `1` to `0`.
+
+Return the **minimum** number of cells that need to be flipped to make **either** all rows **palindromic** or all columns **palindromic**.
+
+**Example 1:**
+
+**Input:** grid = [[1,0,0],[0,0,0],[0,0,1]]
+
+**Output:** 2
+
+**Explanation:**
+
+![](https://assets.leetcode.com/uploads/2024/07/07/screenshot-from-2024-07-08-00-20-10.png)
+
+Flipping the highlighted cells makes all the rows palindromic.
+
+**Example 2:**
+
+**Input:** grid = [[0,1],[0,1],[0,0]]
+
+**Output:** 1
+
+**Explanation:**
+
+![](https://assets.leetcode.com/uploads/2024/07/07/screenshot-from-2024-07-08-00-31-23.png)
+
+Flipping the highlighted cell makes all the columns palindromic.
+
+**Example 3:**
+
+**Input:** grid = [[1],[0]]
+
+**Output:** 0
+
+**Explanation:**
+
+All rows are already palindromic.
+
+**Constraints:**
+
+*   `m == grid.length`
+*   `n == grid[i].length`
+*   <code>1 <= m * n <= 2 * 10<sup>5</sup></code>
+*   `0 <= grid[i][j] <= 1`

src/main/kotlin/g3201_3300/s3240_minimum_number_of_flips_to_make_binary_grid_palindromic_ii/Solution.kt

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+package g3201_3300.s3240_minimum_number_of_flips_to_make_binary_grid_palindromic_ii
+
+// #Medium #Array #Matrix #Two_Pointers #2024_08_07_Time_805_ms_(100.00%)_Space_99_MB_(100.00%)
+
+import kotlin.math.min
+
+class Solution {
+    fun minFlips(grid: Array<IntArray>): Int {
+        var res = 0
+        var one = 0
+        var diff = 0
+        val m = grid.size
+        val n = grid[0].size
+        // Handle quadrants
+        for (i in 0 until m / 2) {
+            for (j in 0 until n / 2) {
+                val v =
+                    (
+                        grid[i][j] +
+                            grid[i][n - 1 - j] +
+                            grid[m - 1 - i][j] +
+                            grid[m - 1 - i][n - 1 - j]
+                        )
+                res += min(v, (4 - v))
+            }
+        }
+        // Handle middle column
+        if (n % 2 > 0) {
+            for (i in 0 until m / 2) {
+                diff += grid[i][n / 2] xor grid[m - 1 - i][n / 2]
+                one += grid[i][n / 2] + grid[m - 1 - i][n / 2]
+            }
+        }
+        // Handle middle row
+        if (m % 2 > 0) {
+            for (j in 0 until n / 2) {
+                diff += grid[m / 2][j] xor grid[m / 2][n - 1 - j]
+                one += grid[m / 2][j] + grid[m / 2][n - 1 - j]
+            }
+        }
+        // Handle center point
+        if (n % 2 > 0 && m % 2 > 0) {
+            res += grid[m / 2][n / 2]
+        }
+        // Divisible by 4
+        if (diff == 0 && one % 4 > 0) {
+            res += 2
+        }
+        return res + diff
+    }
+}

src/main/kotlin/g3201_3300/s3240_minimum_number_of_flips_to_make_binary_grid_palindromic_ii/readme.md

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+3240\. Minimum Number of Flips to Make Binary Grid Palindromic II
+
+Medium
+
+You are given an `m x n` binary matrix `grid`.
+
+A row or column is considered **palindromic** if its values read the same forward and backward.
+
+You can **flip** any number of cells in `grid` from `0` to `1`, or from `1` to `0`.
+
+Return the **minimum** number of cells that need to be flipped to make **all** rows and columns **palindromic**, and the total number of `1`'s in `grid` **divisible** by `4`.
+
+**Example 1:**
+
+**Input:** grid = [[1,0,0],[0,1,0],[0,0,1]]
+
+**Output:** 3
+
+**Explanation:**
+
+![](https://assets.leetcode.com/uploads/2024/08/01/image.png)
+
+**Example 2:**
+
+**Input:** grid = [[0,1],[0,1],[0,0]]
+
+**Output:** 2
+
+**Explanation:**
+
+![](https://assets.leetcode.com/uploads/2024/07/08/screenshot-from-2024-07-09-01-37-48.png)
+
+**Example 3:**
+
+**Input:** grid = [[1],[1]]
+
+**Output:** 2
+
+**Explanation:**
+
+![](https://assets.leetcode.com/uploads/2024/08/01/screenshot-from-2024-08-01-23-05-26.png)
+
+**Constraints:**
+
+*   `m == grid.length`
+*   `n == grid[i].length`
+*   <code>1 <= m * n <= 2 * 10<sup>5</sup></code>
+*   `0 <= grid[i][j] <= 1`

src/main/kotlin/g3201_3300/s3241_time_taken_to_mark_all_nodes/Solution.kt

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+package g3201_3300.s3241_time_taken_to_mark_all_nodes
+
+// #Hard #Dynamic_Programming #Depth_First_Search #Tree #Graph
+// #2024_08_07_Time_1228_ms_(100.00%)_Space_108.5_MB_(100.00%)
+
+import kotlin.math.max
+
+class Solution {
+    private lateinit var head: IntArray
+    private lateinit var nxt: IntArray
+    private lateinit var to: IntArray
+    private lateinit var last: IntArray
+    private lateinit var lastNo: IntArray
+    private lateinit var second: IntArray
+    private lateinit var ans: IntArray
+
+    fun timeTaken(edges: Array<IntArray>): IntArray {
+        val n = edges.size + 1
+        head = IntArray(n)
+        nxt = IntArray(n shl 1)
+        to = IntArray(n shl 1)
+        head.fill(-1)
+        var i = 0
+        var j = 2
+        while (i < edges.size) {
+            val u = edges[i][0]
+            val v = edges[i][1]
+            nxt[j] = head[u]
+            head[u] = j
+            to[j] = v
+            j++
+            nxt[j] = head[v]
+            head[v] = j
+            to[j] = u
+            j++
+            i++
+        }
+        last = IntArray(n)
+        lastNo = IntArray(n)
+        second = IntArray(n)
+        ans = IntArray(n)
+        dfs(-1, 0)
+        System.arraycopy(last, 0, ans, 0, n)
+        dfs2(-1, 0, 0)
+        return ans
+    }
+
+    private fun dfs2(f: Int, u: Int, preLast: Int) {
+        var e = head[u]
+        var v: Int
+        while (e != -1) {
+            v = to[e]
+            if (f != v) {
+                val pl = if (v == lastNo[u]) {
+                    (
+                        max(
+                            preLast,
+                            second[u]
+                        ) + (if ((u and 1) == 0) 2 else 1)
+                        )
+                } else {
+                    (
+                        max(
+                            preLast,
+                            last[u]
+                        ) + (if ((u and 1) == 0) 2 else 1)
+                        )
+                }
+                ans[v] = max(ans[v], pl)
+                dfs2(u, v, pl)
+            }
+            e = nxt[e]
+        }
+    }
+
+    private fun dfs(f: Int, u: Int) {
+        var e = head[u]
+        var v: Int
+        while (e != -1) {
+            v = to[e]
+            if (f != v) {
+                dfs(u, v)
+                val t = last[v] + (if ((v and 1) == 0) 2 else 1)
+                if (last[u] < t) {
+                    second[u] = last[u]
+                    last[u] = t
+                    lastNo[u] = v
+                } else if (second[u] < t) {
+                    second[u] = t
+                }
+            }
+            e = nxt[e]
+        }
+    }
+}