Added tasks 3392-3405

Author: javadev

README.md

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+[![](https://img.shields.io/github/stars/javadev/LeetCode-in-Kotlin?label=Stars&style=flat-square)](https://github.com/javadev/LeetCode-in-Kotlin)
+[![](https://img.shields.io/github/forks/javadev/LeetCode-in-Kotlin?label=Fork%20me%20on%20GitHub%20&style=flat-square)](https://github.com/javadev/LeetCode-in-Kotlin/fork)
+
+## 3392\. Count Subarrays of Length Three With a Condition
+
+Easy
+
+Given an integer array `nums`, return the number of subarrays of length 3 such that the sum of the first and third numbers equals _exactly_ half of the second number.
+
+A **subarray** is a contiguous **non-empty** sequence of elements within an array.
+
+**Example 1:**
+
+**Input:** nums = [1,2,1,4,1]
+
+**Output:** 1
+
+**Explanation:**
+
+Only the subarray `[1,4,1]` contains exactly 3 elements where the sum of the first and third numbers equals half the middle number.
+
+**Example 2:**
+
+**Input:** nums = [1,1,1]
+
+**Output:** 0
+
+**Explanation:**
+
+`[1,1,1]` is the only subarray of length 3. However, its first and third numbers do not add to half the middle number.
+
+**Constraints:**
+
+*   `3 <= nums.length <= 100`
+*   `-100 <= nums[i] <= 100`
+
+## Solution
+
+```kotlin
+class Solution {
+    fun countSubarrays(nums: IntArray): Int {
+        val window = 3
+        var cnt = 0
+        for (i in 0..nums.size - window) {
+            val first = nums[i].toFloat()
+            val second = nums[i + 1].toFloat()
+            val third = nums[i + 2].toFloat()
+            if (second / 2 == first + third) {
+                cnt++
+            }
+        }
+        return cnt
+    }
+}
+```

src/main/kotlin/g3301_3400/s3392_count_subarrays_of_length_three_with_a_condition/readme.md

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+[![](https://img.shields.io/github/stars/javadev/LeetCode-in-Kotlin?label=Stars&style=flat-square)](https://github.com/javadev/LeetCode-in-Kotlin)
+[![](https://img.shields.io/github/forks/javadev/LeetCode-in-Kotlin?label=Fork%20me%20on%20GitHub%20&style=flat-square)](https://github.com/javadev/LeetCode-in-Kotlin/fork)
+
+## 3393\. Count Paths With the Given XOR Value
+
+Medium
+
+You are given a 2D integer array `grid` with size `m x n`. You are also given an integer `k`.
+
+Your task is to calculate the number of paths you can take from the top-left cell `(0, 0)` to the bottom-right cell `(m - 1, n - 1)` satisfying the following **constraints**:
+
+*   You can either move to the right or down. Formally, from the cell `(i, j)` you may move to the cell `(i, j + 1)` or to the cell `(i + 1, j)` if the target cell _exists_.
+*   The `XOR` of all the numbers on the path must be **equal** to `k`.
+
+Return the total number of such paths.
+
+Since the answer can be very large, return the result **modulo** <code>10<sup>9</sup> + 7</code>.
+
+**Example 1:**
+
+**Input:** grid = \[\[2, 1, 5], [7, 10, 0], [12, 6, 4]], k = 11
+
+**Output:** 3
+
+**Explanation:**
+
+The 3 paths are:
+
+*   `(0, 0) → (1, 0) → (2, 0) → (2, 1) → (2, 2)`
+*   `(0, 0) → (1, 0) → (1, 1) → (1, 2) → (2, 2)`
+*   `(0, 0) → (0, 1) → (1, 1) → (2, 1) → (2, 2)`
+
+**Example 2:**
+
+**Input:** grid = \[\[1, 3, 3, 3], [0, 3, 3, 2], [3, 0, 1, 1]], k = 2
+
+**Output:** 5
+
+**Explanation:**
+
+The 5 paths are:
+
+*   `(0, 0) → (1, 0) → (2, 0) → (2, 1) → (2, 2) → (2, 3)`
+*   `(0, 0) → (1, 0) → (1, 1) → (2, 1) → (2, 2) → (2, 3)`
+*   `(0, 0) → (1, 0) → (1, 1) → (1, 2) → (1, 3) → (2, 3)`
+*   `(0, 0) → (0, 1) → (1, 1) → (1, 2) → (2, 2) → (2, 3)`
+*   `(0, 0) → (0, 1) → (0, 2) → (1, 2) → (2, 2) → (2, 3)`
+
+**Example 3:**
+
+**Input:** grid = \[\[1, 1, 1, 2], [3, 0, 3, 2], [3, 0, 2, 2]], k = 10
+
+**Output:** 0
+
+**Constraints:**
+
+*   `1 <= m == grid.length <= 300`
+*   `1 <= n == grid[r].length <= 300`
+*   `0 <= grid[r][c] < 16`
+*   `0 <= k < 16`
+
+## Solution
+
+```kotlin
+class Solution {
+    private var m = -1
+    private var n = -1
+    private lateinit var dp: Array<Array<IntArray>>
+
+    fun countPathsWithXorValue(grid: Array<IntArray>, k: Int): Int {
+        m = grid.size
+        n = grid[0].size
+        dp = Array(m) { Array(n) { IntArray(16) { -1 } } }
+        return dfs(grid, 0, k, 0, 0)
+    }
+
+    private fun dfs(grid: Array<IntArray>, xorVal: Int, k: Int, i: Int, j: Int): Int {
+        var xorVal = xorVal
+        if (i < 0 || j < 0 || j >= n || i >= m) {
+            return 0
+        }
+        xorVal = xorVal xor grid[i][j]
+        if (dp[i][j][xorVal] != -1) {
+            return dp[i][j][xorVal]
+        }
+        if (i == m - 1 && j == n - 1 && xorVal == k) {
+            return 1
+        }
+        val down = dfs(grid, xorVal, k, i + 1, j)
+        val right = dfs(grid, xorVal, k, i, j + 1)
+        dp[i][j][xorVal] = (down + right) % MOD
+        return dp[i][j][xorVal]
+    }
+
+    companion object {
+        private val MOD = (1e9 + 7).toInt()
+    }
+}
+```

src/main/kotlin/g3301_3400/s3393_count_paths_with_the_given_xor_value/readme.md

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+[![](https://img.shields.io/github/stars/javadev/LeetCode-in-Kotlin?label=Stars&style=flat-square)](https://github.com/javadev/LeetCode-in-Kotlin)
+[![](https://img.shields.io/github/forks/javadev/LeetCode-in-Kotlin?label=Fork%20me%20on%20GitHub%20&style=flat-square)](https://github.com/javadev/LeetCode-in-Kotlin/fork)
+
+## 3394\. Check if Grid can be Cut into Sections
+
+Medium
+
+You are given an integer `n` representing the dimensions of an `n x n` grid, with the origin at the bottom-left corner of the grid. You are also given a 2D array of coordinates `rectangles`, where `rectangles[i]` is in the form <code>[start<sub>x</sub>, start<sub>y</sub>, end<sub>x</sub>, end<sub>y</sub>]</code>, representing a rectangle on the grid. Each rectangle is defined as follows:
+
+*   <code>(start<sub>x</sub>, start<sub>y</sub>)</code>: The bottom-left corner of the rectangle.
+*   <code>(end<sub>x</sub>, end<sub>y</sub>)</code>: The top-right corner of the rectangle.
+
+**Note** that the rectangles do not overlap. Your task is to determine if it is possible to make **either two horizontal or two vertical cuts** on the grid such that:
+
+*   Each of the three resulting sections formed by the cuts contains **at least** one rectangle.
+*   Every rectangle belongs to **exactly** one section.
+
+Return `true` if such cuts can be made; otherwise, return `false`.
+
+**Example 1:**
+
+**Input:** n = 5, rectangles = \[\[1,0,5,2],[0,2,2,4],[3,2,5,3],[0,4,4,5]]
+
+**Output:** true
+
+**Explanation:**
+
+![](https://assets.leetcode.com/uploads/2024/10/23/tt1drawio.png)
+
+The grid is shown in the diagram. We can make horizontal cuts at `y = 2` and `y = 4`. Hence, output is true.
+
+**Example 2:**
+
+**Input:** n = 4, rectangles = \[\[0,0,1,1],[2,0,3,4],[0,2,2,3],[3,0,4,3]]
+
+**Output:** true
+
+**Explanation:**
+
+![](https://assets.leetcode.com/uploads/2024/10/23/tc2drawio.png)
+
+We can make vertical cuts at `x = 2` and `x = 3`. Hence, output is true.
+
+**Example 3:**
+
+**Input:** n = 4, rectangles = \[\[0,2,2,4],[1,0,3,2],[2,2,3,4],[3,0,4,2],[3,2,4,4]]
+
+**Output:** false
+
+**Explanation:**
+
+We cannot make two horizontal or two vertical cuts that satisfy the conditions. Hence, output is false.
+
+**Constraints:**
+
+*   <code>3 <= n <= 10<sup>9</sup></code>
+*   <code>3 <= rectangles.length <= 10<sup>5</sup></code>
+*   `0 <= rectangles[i][0] < rectangles[i][2] <= n`
+*   `0 <= rectangles[i][1] < rectangles[i][3] <= n`
+*   No two rectangles overlap.
+
+## Solution
+
+```kotlin
+import kotlin.math.max
+
+@Suppress("unused")
+class Solution {
+    fun checkValidCuts(n: Int, rectangles: Array<IntArray>): Boolean {
+        val m = rectangles.size
+        val xAxis = Array<IntArray>(m) { IntArray(2) }
+        val yAxis = Array<IntArray>(m) { IntArray(2) }
+        var ind = 0
+        for (axis in rectangles) {
+            val startX = axis[0]
+            val startY = axis[1]
+            val endX = axis[2]
+            val endY = axis[3]
+            xAxis[ind] = intArrayOf(startX, endX)
+            yAxis[ind] = intArrayOf(startY, endY)
+            ind++
+        }
+
+        xAxis.sortWith<IntArray>(
+            Comparator { a: IntArray, b: IntArray -> if (a[0] == b[0]) a[1] - b[1] else a[0] - b[0] },
+        )
+
+        yAxis.sortWith<IntArray>(
+            Comparator { a: IntArray, b: IntArray -> if (a[0] == b[0]) a[1] - b[1] else a[0] - b[0] },
+        )
+        val verticalCuts = findSections(xAxis)
+        if (verticalCuts > 2) {
+            return true
+        }
+        val horizontalCuts = findSections(yAxis)
+        return horizontalCuts > 2
+    }
+
+    private fun findSections(axis: Array<IntArray>): Int {
+        var end = axis[0][1]
+        var sections = 1
+        for (i in 1..<axis.size) {
+            if (end > axis[i][0]) {
+                end = max(end, axis[i][1])
+            } else {
+                sections++
+                end = axis[i][1]
+            }
+            if (sections > 2) {
+                return sections
+            }
+        }
+        return sections
+    }
+}
+```