Added tasks 3254-3261

src/main/kotlin/g3201_3300/s3254_find_the_power_of_k_size_subarrays_i/Solution.kt

@@ -0,0 +1,27 @@
+package g3201_3300.s3254_find_the_power_of_k_size_subarrays_i
+
+// #Medium #Array #Sliding_Window #2024_08_21_Time_245_ms_(92.59%)_Space_42.2_MB_(16.67%)
+
+class Solution {
+    fun resultsArray(nums: IntArray, k: Int): IntArray {
+        val n = nums.size
+        val arr = IntArray(n - k + 1)
+        var count = 0
+        for (i in 1 until k) {
+            if (nums[i] == nums[i - 1] + 1) {
+                count++
+            }
+        }
+        arr[0] = if ((count == k - 1)) nums[k - 1] else -1
+        for (i in 1..n - k) {
+            if (nums[i] == nums[i - 1] + 1) {
+                count--
+            }
+            if (nums[i + k - 1] == nums[i + k - 2] + 1) {
+                count++
+            }
+            arr[i] = if ((count == k - 1)) nums[i + k - 1] else -1
+        }
+        return arr
+    }
+}

src/main/kotlin/g3201_3300/s3254_find_the_power_of_k_size_subarrays_i/readme.md

@@ -0,0 +1,48 @@
+3254\. Find the Power of K-Size Subarrays I
+
+Medium
+
+You are given an array of integers `nums` of length `n` and a _positive_ integer `k`.
+
+The **power** of an array is defined as:
+
+*   Its **maximum** element if _all_ of its elements are **consecutive** and **sorted** in **ascending** order.
+*   \-1 otherwise.
+
+You need to find the **power** of all subarrays of `nums` of size `k`.
+
+Return an integer array `results` of size `n - k + 1`, where `results[i]` is the _power_ of `nums[i..(i + k - 1)]`.
+
+**Example 1:**
+
+**Input:** nums = [1,2,3,4,3,2,5], k = 3
+
+**Output:** [3,4,-1,-1,-1]
+
+**Explanation:**
+
+There are 5 subarrays of `nums` of size 3:
+
+*   `[1, 2, 3]` with the maximum element 3.
+*   `[2, 3, 4]` with the maximum element 4.
+*   `[3, 4, 3]` whose elements are **not** consecutive.
+*   `[4, 3, 2]` whose elements are **not** sorted.
+*   `[3, 2, 5]` whose elements are **not** consecutive.
+
+**Example 2:**
+
+**Input:** nums = [2,2,2,2,2], k = 4
+
+**Output:** [-1,-1]
+
+**Example 3:**
+
+**Input:** nums = [3,2,3,2,3,2], k = 2
+
+**Output:** [-1,3,-1,3,-1]
+
+**Constraints:**
+
+*   `1 <= n == nums.length <= 500`
+*   <code>1 <= nums[i] <= 10<sup>5</sup></code>
+*   `1 <= k <= n`

src/main/kotlin/g3201_3300/s3255_find_the_power_of_k_size_subarrays_ii/Solution.kt

@@ -0,0 +1,28 @@
+package g3201_3300.s3255_find_the_power_of_k_size_subarrays_ii
+
+// #Medium #Array #Sliding_Window #2024_08_21_Time_892_ms_(89.36%)_Space_69.8_MB_(76.60%)
+
+class Solution {
+    fun resultsArray(nums: IntArray, k: Int): IntArray {
+        if (k == 1) {
+            return nums
+        }
+        var start = 0
+        val n = nums.size
+        val output = IntArray(n - k + 1)
+        for (i in 1 until n) {
+            if (nums[i] != nums[i - 1] + 1) {
+                start = i
+            }
+            val index = i - k + 1
+            if (index >= 0) {
+                if (start > index) {
+                    output[index] = -1
+                } else {
+                    output[index] = nums[i]
+                }
+            }
+        }
+        return output
+    }
+}

src/main/kotlin/g3201_3300/s3255_find_the_power_of_k_size_subarrays_ii/readme.md

@@ -0,0 +1,48 @@
+3255\. Find the Power of K-Size Subarrays II
+
+Medium
+
+You are given an array of integers `nums` of length `n` and a _positive_ integer `k`.
+
+The **power** of an array is defined as:
+
+*   Its **maximum** element if _all_ of its elements are **consecutive** and **sorted** in **ascending** order.
+*   \-1 otherwise.
+
+You need to find the **power** of all subarrays of `nums` of size `k`.
+
+Return an integer array `results` of size `n - k + 1`, where `results[i]` is the _power_ of `nums[i..(i + k - 1)]`.
+
+**Example 1:**
+
+**Input:** nums = [1,2,3,4,3,2,5], k = 3
+
+**Output:** [3,4,-1,-1,-1]
+
+**Explanation:**
+
+There are 5 subarrays of `nums` of size 3:
+
+*   `[1, 2, 3]` with the maximum element 3.
+*   `[2, 3, 4]` with the maximum element 4.
+*   `[3, 4, 3]` whose elements are **not** consecutive.
+*   `[4, 3, 2]` whose elements are **not** sorted.
+*   `[3, 2, 5]` whose elements are **not** consecutive.
+
+**Example 2:**
+
+**Input:** nums = [2,2,2,2,2], k = 4
+
+**Output:** [-1,-1]
+
+**Example 3:**
+
+**Input:** nums = [3,2,3,2,3,2], k = 2
+
+**Output:** [-1,3,-1,3,-1]
+
+**Constraints:**
+
+*   <code>1 <= n == nums.length <= 10<sup>5</sup></code>
+*   <code>1 <= nums[i] <= 10<sup>6</sup></code>
+*   `1 <= k <= n`

src/main/kotlin/g3201_3300/s3256_maximum_value_sum_by_placing_three_rooks_i/Solution.kt

@@ -0,0 +1,63 @@
+package g3201_3300.s3256_maximum_value_sum_by_placing_three_rooks_i
+
+// #Hard #Array #Dynamic_Programming #Matrix #Enumeration
+// #2024_08_21_Time_279_ms_(100.00%)_Space_41.6_MB_(93.33%)
+
+import kotlin.math.max
+
+class Solution {
+    fun maximumValueSum(board: Array<IntArray>): Long {
+        val n = board.size
+        val m = board[0].size
+        val tb = Array(n) { IntArray(m) }
+        tb[0] = board[0].copyOf(m)
+        for (i in 1 until n) {
+            for (j in 0 until m) {
+                tb[i][j] = max(tb[i - 1][j], board[i][j])
+            }
+        }
+        val bt = Array(n) { IntArray(m) }
+        bt[n - 1] = board[n - 1].copyOf(m)
+        for (i in n - 2 downTo 0) {
+            for (j in 0 until m) {
+                bt[i][j] = max(bt[i + 1][j], board[i][j])
+            }
+        }
+        var ans = Long.MIN_VALUE
+        for (i in 1 until n - 1) {
+            val max3Top = getMax3(tb[i - 1])
+            val max3Cur = getMax3(board[i])
+            val max3Bottom = getMax3(bt[i + 1])
+            for (topCand in max3Top) {
+                for (curCand in max3Cur) {
+                    for (bottomCand in max3Bottom) {
+                        if (topCand[1] != curCand[1] && topCand[1] != bottomCand[1] && curCand[1] != bottomCand[1]) {
+                            val cand = topCand[0].toLong() + curCand[0] + bottomCand[0]
+                            ans = max(ans, cand)
+                        }
+                    }
+                }
+            }
+        }
+        return ans
+    }
+
+    private fun getMax3(row: IntArray): Array<IntArray> {
+        val m = row.size
+        val ans = Array(3) { IntArray(2) }
+        ans.fill(intArrayOf(Int.MIN_VALUE, -1))
+        for (j in 0 until m) {
+            if (row[j] >= ans[0][0]) {
+                ans[2] = ans[1]
+                ans[1] = ans[0]
+                ans[0] = intArrayOf(row[j], j)
+            } else if (row[j] >= ans[1][0]) {
+                ans[2] = ans[1]
+                ans[1] = intArrayOf(row[j], j)
+            } else if (row[j] > ans[2][0]) {
+                ans[2] = intArrayOf(row[j], j)
+            }
+        }
+        return ans
+    }
+}

src/main/kotlin/g3201_3300/s3256_maximum_value_sum_by_placing_three_rooks_i/readme.md

@@ -0,0 +1,47 @@
+3256\. Maximum Value Sum by Placing Three Rooks I
+
+Hard
+
+You are given a `m x n` 2D array `board` representing a chessboard, where `board[i][j]` represents the **value** of the cell `(i, j)`.
+
+Rooks in the **same** row or column **attack** each other. You need to place _three_ rooks on the chessboard such that the rooks **do not** **attack** each other.
+
+Return the **maximum** sum of the cell **values** on which the rooks are placed.
+
+**Example 1:**
+
+**Input:** board = [[-3,1,1,1],[-3,1,-3,1],[-3,2,1,1]]
+
+**Output:** 4
+
+**Explanation:**
+
+![](https://assets.leetcode.com/uploads/2024/08/08/rooks2.png)
+
+We can place the rooks in the cells `(0, 2)`, `(1, 3)`, and `(2, 1)` for a sum of `1 + 1 + 2 = 4`.
+
+**Example 2:**
+
+**Input:** board = [[1,2,3],[4,5,6],[7,8,9]]
+
+**Output:** 15
+
+**Explanation:**
+
+We can place the rooks in the cells `(0, 0)`, `(1, 1)`, and `(2, 2)` for a sum of `1 + 5 + 9 = 15`.
+
+**Example 3:**
+
+**Input:** board = [[1,1,1],[1,1,1],[1,1,1]]
+
+**Output:** 3
+
+**Explanation:**
+
+We can place the rooks in the cells `(0, 2)`, `(1, 1)`, and `(2, 0)` for a sum of `1 + 1 + 1 = 3`.
+
+**Constraints:**
+
+*   `3 <= m == board.length <= 100`
+*   `3 <= n == board[i].length <= 100`
+*   <code>-10<sup>9</sup> <= board[i][j] <= 10<sup>9</sup></code>

src/main/kotlin/g3201_3300/s3257_maximum_value_sum_by_placing_three_rooks_ii/Solution.kt

@@ -0,0 +1,63 @@
+package g3201_3300.s3257_maximum_value_sum_by_placing_three_rooks_ii
+
+// #Hard #Array #Dynamic_Programming #Matrix #Enumeration
+// #2024_08_21_Time_770_ms_(100.00%)_Space_87.5_MB_(33.33%)
+
+import kotlin.math.max
+
+class Solution {
+    fun maximumValueSum(board: Array<IntArray>): Long {
+        val n = board.size
+        val m = board[0].size
+        val tb = Array(n) { IntArray(m) }
+        tb[0] = board[0].copyOf(m)
+        for (i in 1 until n) {
+            for (j in 0 until m) {
+                tb[i][j] = max(tb[i - 1][j], board[i][j])
+            }
+        }
+        val bt = Array(n) { IntArray(m) }
+        bt[n - 1] = board[n - 1].copyOf(m)
+        for (i in n - 2 downTo 0) {
+            for (j in 0 until m) {
+                bt[i][j] = max(bt[i + 1][j], board[i][j])
+            }
+        }
+        var ans = Long.MIN_VALUE
+        for (i in 1 until n - 1) {
+            val max3Top = getMax3(tb[i - 1])
+            val max3Cur = getMax3(board[i])
+            val max3Bottom = getMax3(bt[i + 1])
+            for (topCand in max3Top) {
+                for (curCand in max3Cur) {
+                    for (bottomCand in max3Bottom) {
+                        if (topCand[1] != curCand[1] && topCand[1] != bottomCand[1] && curCand[1] != bottomCand[1]) {
+                            val cand = topCand[0].toLong() + curCand[0] + bottomCand[0]
+                            ans = max(ans, cand)
+                        }
+                    }
+                }
+            }
+        }
+        return ans
+    }
+
+    private fun getMax3(row: IntArray): Array<IntArray> {
+        val m = row.size
+        val ans = Array(3) { IntArray(2) }
+        ans.fill(intArrayOf(Int.MIN_VALUE, -1))
+        for (j in 0 until m) {
+            if (row[j] >= ans[0][0]) {
+                ans[2] = ans[1]
+                ans[1] = ans[0]
+                ans[0] = intArrayOf(row[j], j)
+            } else if (row[j] >= ans[1][0]) {
+                ans[2] = ans[1]
+                ans[1] = intArrayOf(row[j], j)
+            } else if (row[j] > ans[2][0]) {
+                ans[2] = intArrayOf(row[j], j)
+            }
+        }
+        return ans
+    }
+}

src/main/kotlin/g3201_3300/s3257_maximum_value_sum_by_placing_three_rooks_ii/readme.md

@@ -0,0 +1,47 @@
+3257\. Maximum Value Sum by Placing Three Rooks II
+
+Hard
+
+You are given a `m x n` 2D array `board` representing a chessboard, where `board[i][j]` represents the **value** of the cell `(i, j)`.
+
+Rooks in the **same** row or column **attack** each other. You need to place _three_ rooks on the chessboard such that the rooks **do not** **attack** each other.
+
+Return the **maximum** sum of the cell **values** on which the rooks are placed.
+
+**Example 1:**
+
+**Input:** board = [[-3,1,1,1],[-3,1,-3,1],[-3,2,1,1]]
+
+**Output:** 4
+
+**Explanation:**
+
+![](https://assets.leetcode.com/uploads/2024/08/08/rooks2.png)
+
+We can place the rooks in the cells `(0, 2)`, `(1, 3)`, and `(2, 1)` for a sum of `1 + 1 + 2 = 4`.
+
+**Example 2:**
+
+**Input:** board = [[1,2,3],[4,5,6],[7,8,9]]
+
+**Output:** 15
+
+**Explanation:**
+
+We can place the rooks in the cells `(0, 0)`, `(1, 1)`, and `(2, 2)` for a sum of `1 + 5 + 9 = 15`.
+
+**Example 3:**
+
+**Input:** board = [[1,1,1],[1,1,1],[1,1,1]]
+
+**Output:** 3
+
+**Explanation:**
+
+We can place the rooks in the cells `(0, 2)`, `(1, 1)`, and `(2, 0)` for a sum of `1 + 1 + 1 = 3`.
+
+**Constraints:**
+
+*   `3 <= m == board.length <= 500`
+*   `3 <= n == board[i].length <= 500`
+*   <code>-10<sup>9</sup> <= board[i][j] <= 10<sup>9</sup></code>