Added tasks 3379-3382

Author: javadev

src/main/kotlin/g3301_3400/s3379_transformed_array/Solution.kt

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+package g3301_3400.s3379_transformed_array
+
+// #Easy #Array #Simulation #2024_12_10_Time_206_ms_(84.38%)_Space_38.6_MB_(75.00%)
+
+import kotlin.math.abs
+
+class Solution {
+    fun constructTransformedArray(nums: IntArray): IntArray {
+        val n = nums.size
+        val res = IntArray(n)
+        for (i in 0..<n) {
+            if (nums[i] > 0) {
+                res[i] = nums[(i + nums[i]) % n]
+            } else if (nums[i] < 0) {
+                val r: Int = abs(nums[i]) / n
+                res[i] = nums[abs(i + nums[i] + r * n + n) % n]
+            }
+        }
+        return res
+    }
+}

src/main/kotlin/g3301_3400/s3379_transformed_array/readme.md

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+3379\. Transformed Array
+
+Easy
+
+You are given an integer array `nums` that represents a circular array. Your task is to create a new array `result` of the **same** size, following these rules:
+
+For each index `i` (where `0 <= i < nums.length`), perform the following **independent** actions:
+
+*   If `nums[i] > 0`: Start at index `i` and move `nums[i]` steps to the **right** in the circular array. Set `result[i]` to the value of the index where you land.
+*   If `nums[i] < 0`: Start at index `i` and move `abs(nums[i])` steps to the **left** in the circular array. Set `result[i]` to the value of the index where you land.
+*   If `nums[i] == 0`: Set `result[i]` to `nums[i]`.
+
+Return the new array `result`.
+
+**Note:** Since `nums` is circular, moving past the last element wraps around to the beginning, and moving before the first element wraps back to the end.
+
+**Example 1:**
+
+**Input:** nums = [3,-2,1,1]
+
+**Output:** [1,1,1,3]
+
+**Explanation:**
+
+*   For `nums[0]` that is equal to 3, If we move 3 steps to right, we reach `nums[3]`. So `result[0]` should be 1.
+*   For `nums[1]` that is equal to -2, If we move 2 steps to left, we reach `nums[3]`. So `result[1]` should be 1.
+*   For `nums[2]` that is equal to 1, If we move 1 step to right, we reach `nums[3]`. So `result[2]` should be 1.
+*   For `nums[3]` that is equal to 1, If we move 1 step to right, we reach `nums[0]`. So `result[3]` should be 3.
+
+**Example 2:**
+
+**Input:** nums = [-1,4,-1]
+
+**Output:** [-1,-1,4]
+
+**Explanation:**
+
+*   For `nums[0]` that is equal to -1, If we move 1 step to left, we reach `nums[2]`. So `result[0]` should be -1.
+*   For `nums[1]` that is equal to 4, If we move 4 steps to right, we reach `nums[2]`. So `result[1]` should be -1.
+*   For `nums[2]` that is equal to -1, If we move 1 step to left, we reach `nums[1]`. So `result[2]` should be 4.
+
+**Constraints:**
+
+*   `1 <= nums.length <= 100`
+*   `-100 <= nums[i] <= 100`

src/main/kotlin/g3301_3400/s3380_maximum_area_rectangle_with_point_constraints_i/Solution.kt

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+package g3301_3400.s3380_maximum_area_rectangle_with_point_constraints_i
+
+// #Medium #Array #Math #Sorting #Enumeration #Geometry #Segment_Tree #Binary_Indexed_Tree
+// #2024_12_10_Time_10_ms_(94.74%)_Space_40.1_MB_(84.21%)
+
+import kotlin.math.max
+import kotlin.math.min
+
+class Solution {
+    fun maxRectangleArea(points: Array<IntArray>): Int {
+        val set: MutableSet<String?> = HashSet<String?>()
+        for (p in points) {
+            set.add(p.contentToString())
+        }
+        var maxArea = -1
+        for (point in points) {
+            for (j in 1..<points.size) {
+                val p2 = points[j]
+                if (point[0] == p2[0] || point[1] == p2[1] || !set.contains(
+                        intArrayOf(
+                            point[0],
+                            p2[1],
+                        ).contentToString(),
+                    ) || !set.contains(intArrayOf(p2[0], point[1]).contentToString()) ||
+                    !validate(points, point, p2)
+                ) {
+                    continue
+                }
+                maxArea = max(maxArea, (p2[1] - point[1]) * (p2[0] - point[0]))
+            }
+        }
+        return maxArea
+    }
+
+    private fun validate(points: Array<IntArray>, p1: IntArray, p2: IntArray): Boolean {
+        val top = max(p1[1], p2[1])
+        val bot = min(p1[1], p2[1])
+        val left = min(p1[0], p2[0])
+        val right = max(p1[0], p2[0])
+        for (p in points) {
+            val x = p[0]
+            val y = p[1]
+            if ((y == top || y == bot) && x > left && x < right ||
+                (x == left || x == right) && y > bot && y < top ||
+                (x > left && x < right && y > bot && y < top)
+            ) {
+                return false
+            }
+        }
+        return true
+    }
+}

src/main/kotlin/g3301_3400/s3380_maximum_area_rectangle_with_point_constraints_i/readme.md

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+3380\. Maximum Area Rectangle With Point Constraints I
+
+Medium
+
+You are given an array `points` where <code>points[i] = [x<sub>i</sub>, y<sub>i</sub>]</code> represents the coordinates of a point on an infinite plane.
+
+Your task is to find the **maximum** area of a rectangle that:
+
+*   Can be formed using **four** of these points as its corners.
+*   Does **not** contain any other point inside or on its border.
+*   Has its edges **parallel** to the axes.
+
+Return the **maximum area** that you can obtain or -1 if no such rectangle is possible.
+
+**Example 1:**
+
+**Input:** points = [[1,1],[1,3],[3,1],[3,3]]
+
+**Output:** 4
+
+**Explanation:**
+
+**![Example 1 diagram](https://assets.leetcode.com/uploads/2024/11/02/example1.png)**
+
+We can make a rectangle with these 4 points as corners and there is no other point that lies inside or on the border. Hence, the maximum possible area would be 4.
+
+**Example 2:**
+
+**Input:** points = [[1,1],[1,3],[3,1],[3,3],[2,2]]
+
+**Output:** \-1
+
+**Explanation:**
+
+**![Example 2 diagram](https://assets.leetcode.com/uploads/2024/11/02/example2.png)**
+
+There is only one rectangle possible is with points `[1,1], [1,3], [3,1]` and `[3,3]` but `[2,2]` will always lie inside it. Hence, returning -1.
+
+**Example 3:**
+
+**Input:** points = [[1,1],[1,3],[3,1],[3,3],[1,2],[3,2]]
+
+**Output:** 2
+
+**Explanation:**
+
+**![Example 3 diagram](https://assets.leetcode.com/uploads/2024/11/02/example3.png)**
+
+The maximum area rectangle is formed by the points `[1,3], [1,2], [3,2], [3,3]`, which has an area of 2. Additionally, the points `[1,1], [1,2], [3,1], [3,2]` also form a valid rectangle with the same area.
+
+**Constraints:**
+
+*   `1 <= points.length <= 10`
+*   `points[i].length == 2`
+*   <code>0 <= x<sub>i</sub>, y<sub>i</sub> <= 100</code>
+*   All the given points are **unique**.

src/main/kotlin/g3301_3400/s3381_maximum_subarray_sum_with_length_divisible_by_k/Solution.kt

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+package g3301_3400.s3381_maximum_subarray_sum_with_length_divisible_by_k
+
+// #Medium #Array #Hash_Table #Prefix_Sum #2024_12_10_Time_6_ms_(100.00%)_Space_83.5_MB_(69.23%)
+
+import kotlin.math.max
+
+class Solution {
+    fun maxSubarraySum(nums: IntArray, k: Int): Long {
+        val n = nums.size
+        val maxSum = LongArray(n)
+        var minSum: Long = 0
+        for (i in n - 1 downTo n - k + 1) {
+            maxSum[i] = Int.Companion.MIN_VALUE.toLong()
+            minSum += nums[i]
+        }
+        minSum += nums[n - k]
+        maxSum[n - k] = minSum
+        var ans = minSum
+        for (i in n - k - 1 downTo 0) {
+            minSum = minSum + nums[i] - nums[i + k]
+            maxSum[i] = max(minSum, (minSum + maxSum[i + k]))
+            ans = max(maxSum[i], ans)
+        }
+        return ans
+    }
+}

src/main/kotlin/g3301_3400/s3381_maximum_subarray_sum_with_length_divisible_by_k/readme.md

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+3381\. Maximum Subarray Sum With Length Divisible by K
+
+Medium
+
+You are given an array of integers `nums` and an integer `k`.
+
+Return the **maximum** sum of a **non-empty subarray** of `nums`, such that the size of the subarray is **divisible** by `k`.
+
+A **subarray** is a contiguous **non-empty** sequence of elements within an array.
+
+**Example 1:**
+
+**Input:** nums = [1,2], k = 1
+
+**Output:** 3
+
+**Explanation:**
+
+The subarray `[1, 2]` with sum 3 has length equal to 2 which is divisible by 1.
+
+**Example 2:**
+
+**Input:** nums = [-1,-2,-3,-4,-5], k = 4
+
+**Output:** \-10
+
+**Explanation:**
+
+The maximum sum subarray is `[-1, -2, -3, -4]` which has length equal to 4 which is divisible by 4.
+
+**Example 3:**
+
+**Input:** nums = [-5,1,2,-3,4], k = 2
+
+**Output:** 4
+
+**Explanation:**
+
+The maximum sum subarray is `[1, 2, -3, 4]` which has length equal to 4 which is divisible by 2.
+
+**Constraints:**
+
+*   <code>1 <= k <= nums.length <= 2 * 10<sup>5</sup></code>
+*   <code>-10<sup>9</sup> <= nums[i] <= 10<sup>9</sup></code>

src/main/kotlin/g3301_3400/s3382_maximum_area_rectangle_with_point_constraints_ii/Solution.kt

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+package g3301_3400.s3382_maximum_area_rectangle_with_point_constraints_ii
+
+// #Hard #Array #Math #Sorting #Geometry #Segment_Tree #Binary_Indexed_Tree
+// #2024_12_10_Time_518_ms_(100.00%)_Space_103.7_MB_(100.00%)
+
+import java.util.TreeSet
+import kotlin.math.max
+
+class Solution {
+    fun maxRectangleArea(xCoord: IntArray, yCoord: IntArray): Long {
+        if (xCoord.size < 4) {
+            return -1
+        }
+        val pair = Array<Pair>(xCoord.size) { i -> Pair(0, 0) }
+        for (i in xCoord.indices) {
+            val x0 = xCoord[i]
+            val y0 = yCoord[i]
+            pair[i] = Pair(x0, y0)
+        }
+        pair.sort()
+        val map = HashMap<Int?, Pair>()
+        val yVals = TreeSet<Int?>()
+        var best: Long = -1
+        for (i in 0..<pair.size - 1) {
+            if (!yVals.isEmpty()) {
+                val y0 = pair[i].y
+                var y1 = yVals.floor(y0)
+                while (y1 != null) {
+                    val p1: Pair = map.get(y1)!!
+                    if (p1.y < y0) {
+                        break
+                    }
+                    if (y1 == y0 && pair[i + 1].x == pair[i].x && pair[i + 1].y == p1.y) {
+                        val dY = p1.y - y0.toLong()
+                        val dX = pair[i].x - p1.x.toLong()
+                        best = max((dY * dX).toDouble(), best.toDouble()).toLong()
+                    }
+                    if (p1.x != pair[i].x) {
+                        yVals.remove(y1)
+                    }
+                    y1 = yVals.lower(y1)
+                }
+            }
+            if (pair[i].x == pair[i + 1].x) {
+                yVals.add(pair[i].y)
+                map.put(pair[i].y, pair[i + 1])
+            }
+        }
+        return best
+    }
+
+    private class Pair(val x: Int, val y: Int) : Comparable<Pair> {
+        override fun compareTo(p: Pair): Int {
+            return if (x == p.x) y - p.y else x - p.x
+        }
+    }
+}

src/main/kotlin/g3301_3400/s3382_maximum_area_rectangle_with_point_constraints_ii/readme.md

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+3382\. Maximum Area Rectangle With Point Constraints II
+
+Hard
+
+There are n points on an infinite plane. You are given two integer arrays `xCoord` and `yCoord` where `(xCoord[i], yCoord[i])` represents the coordinates of the <code>i<sup>th</sup></code> point.
+
+Your task is to find the **maximum** area of a rectangle that:
+
+*   Can be formed using **four** of these points as its corners.
+*   Does **not** contain any other point inside or on its border.
+*   Has its edges **parallel** to the axes.
+
+Return the **maximum area** that you can obtain or -1 if no such rectangle is possible.
+
+**Example 1:**
+
+**Input:** xCoord = [1,1,3,3], yCoord = [1,3,1,3]
+
+**Output:** 4
+
+**Explanation:**
+
+**![Example 1 diagram](https://assets.leetcode.com/uploads/2024/11/02/example1.png)**
+
+We can make a rectangle with these 4 points as corners and there is no other point that lies inside or on the border. Hence, the maximum possible area would be 4.
+
+**Example 2:**
+
+**Input:** xCoord = [1,1,3,3,2], yCoord = [1,3,1,3,2]
+
+**Output:** \-1
+
+**Explanation:**
+
+**![Example 2 diagram](https://assets.leetcode.com/uploads/2024/11/02/example2.png)**
+
+There is only one rectangle possible is with points `[1,1], [1,3], [3,1]` and `[3,3]` but `[2,2]` will always lie inside it. Hence, returning -1.
+
+**Example 3:**
+
+**Input:** xCoord = [1,1,3,3,1,3], yCoord = [1,3,1,3,2,2]
+
+**Output:** 2
+
+**Explanation:**
+
+**![Example 3 diagram](https://assets.leetcode.com/uploads/2024/11/02/example3.png)**
+
+The maximum area rectangle is formed by the points `[1,3], [1,2], [3,2], [3,3]`, which has an area of 2. Additionally, the points `[1,1], [1,2], [3,1], [3,2]` also form a valid rectangle with the same area.
+
+**Constraints:**
+
+*   <code>1 <= xCoord.length == yCoord.length <= 2 * 10<sup>5</sup></code>
+*   <code>0 <= xCoord[i], yCoord[i] <= 8 * 10<sup>7</sup></code>
+*   All the given points are **unique**.

src/test/kotlin/g3301_3400/s3379_transformed_array/SolutionTest.kt

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+package g3301_3400.s3379_transformed_array
+
+import org.hamcrest.CoreMatchers.equalTo
+import org.hamcrest.MatcherAssert.assertThat
+import org.junit.jupiter.api.Test
+
+internal class SolutionTest {
+    @Test
+    fun constructTransformedArray() {
+        assertThat<IntArray>(
+            Solution().constructTransformedArray(intArrayOf(3, -2, 1, 1)),
+            equalTo<IntArray>(intArrayOf(1, 1, 1, 3)),
+        )
+    }
+
+    @Test
+    fun constructTransformedArray2() {
+        assertThat<IntArray>(
+            Solution().constructTransformedArray(intArrayOf(-1, 4, -1)),
+            equalTo<IntArray>(intArrayOf(-1, -1, 4)),
+        )
+    }
+}