Added tasks 3423-3430

Author: javadev

src/main/kotlin/g3401_3500/s3423_maximum_difference_between_adjacent_elements_in_a_circular_array/Solution.kt

@@ -0,0 +1,18 @@
+package g3401_3500.s3423_maximum_difference_between_adjacent_elements_in_a_circular_array
+
+// #Easy #2025_01_19_Time_2_(100.00%)_Space_38.80_(100.00%)
+
+import kotlin.math.abs
+import kotlin.math.max
+
+class Solution {
+    fun maxAdjacentDistance(nums: IntArray): Int {
+        var maxDiff = 0
+        for (i in nums.indices) {
+            val nextIndex = (i + 1) % nums.size
+            val diff = abs((nums[i] - nums[nextIndex]))
+            maxDiff = max(maxDiff, diff)
+        }
+        return maxDiff
+    }
+}

src/main/kotlin/g3401_3500/s3423_maximum_difference_between_adjacent_elements_in_a_circular_array/readme.md

@@ -0,0 +1,32 @@
+3423\. Maximum Difference Between Adjacent Elements in a Circular Array
+
+Easy
+
+Given a **circular** array `nums`, find the **maximum** absolute difference between adjacent elements.
+
+**Note**: In a circular array, the first and last elements are adjacent.
+
+**Example 1:**
+
+**Input:** nums = [1,2,4]
+
+**Output:** 3
+
+**Explanation:**
+
+Because `nums` is circular, `nums[0]` and `nums[2]` are adjacent. They have the maximum absolute difference of `|4 - 1| = 3`.
+
+**Example 2:**
+
+**Input:** nums = [-5,-10,-5]
+
+**Output:** 5
+
+**Explanation:**
+
+The adjacent elements `nums[0]` and `nums[1]` have the maximum absolute difference of `|-5 - (-10)| = 5`.
+
+**Constraints:**
+
+*   `2 <= nums.length <= 100`
+*   `-100 <= nums[i] <= 100`

src/main/kotlin/g3401_3500/s3424_minimum_cost_to_make_arrays_identical/Solution.kt

@@ -0,0 +1,28 @@
+package g3401_3500.s3424_minimum_cost_to_make_arrays_identical
+
+// #Medium #Array #Sorting #Greedy #2025_01_23_Time_38_(100.00%)_Space_64.36_(97.14%)
+
+import kotlin.math.abs
+import kotlin.math.min
+
+class Solution {
+    fun minCost(arr: IntArray, brr: IntArray, k: Long): Long {
+        val n = arr.size
+        var sum1: Long = 0
+        var sum2: Long
+        for (i in 0..<n) {
+            sum1 = sum1 + abs((arr[i] - brr[i]))
+        }
+        if (k < sum1) {
+            arr.sort()
+            brr.sort()
+            sum2 = k
+            for (i in 0..<n) {
+                sum2 = sum2 + abs((arr[i] - brr[i]))
+            }
+        } else {
+            return sum1
+        }
+        return min(sum1, sum2)
+    }
+}

src/main/kotlin/g3401_3500/s3424_minimum_cost_to_make_arrays_identical/readme.md

@@ -0,0 +1,43 @@
+3424\. Minimum Cost to Make Arrays Identical
+
+Medium
+
+You are given two integer arrays `arr` and `brr` of length `n`, and an integer `k`. You can perform the following operations on `arr` _any_ number of times:
+
+*   Split `arr` into _any_ number of **contiguous** **non-empty subarrays** and rearrange these subarrays in _any order_. This operation has a fixed cost of `k`.
+*   Choose any element in `arr` and add or subtract a positive integer `x` to it. The cost of this operation is `x`.
+    
+
+Return the **minimum** total cost to make `arr` **equal** to `brr`.
+
+**Example 1:**
+
+**Input:** arr = [-7,9,5], brr = [7,-2,-5], k = 2
+
+**Output:** 13
+
+**Explanation:**
+
+*   Split `arr` into two contiguous subarrays: `[-7]` and `[9, 5]` and rearrange them as `[9, 5, -7]`, with a cost of 2.
+*   Subtract 2 from element `arr[0]`. The array becomes `[7, 5, -7]`. The cost of this operation is 2.
+*   Subtract 7 from element `arr[1]`. The array becomes `[7, -2, -7]`. The cost of this operation is 7.
+*   Add 2 to element `arr[2]`. The array becomes `[7, -2, -5]`. The cost of this operation is 2.
+
+The total cost to make the arrays equal is `2 + 2 + 7 + 2 = 13`.
+
+**Example 2:**
+
+**Input:** arr = [2,1], brr = [2,1], k = 0
+
+**Output:** 0
+
+**Explanation:**
+
+Since the arrays are already equal, no operations are needed, and the total cost is 0.
+
+**Constraints:**
+
+*   <code>1 <= arr.length == brr.length <= 10<sup>5</sup></code>
+*   <code>0 <= k <= 2 * 10<sup>10</sup></code>
+*   <code>-10<sup>5</sup> <= arr[i] <= 10<sup>5</sup></code>
+*   <code>-10<sup>5</sup> <= brr[i] <= 10<sup>5</sup></code>

src/main/kotlin/g3401_3500/s3425_longest_special_path/Solution.kt

@@ -0,0 +1,91 @@
+package g3401_3500.s3425_longest_special_path
+
+// #Hard #2025_01_19_Time_106_(100.00%)_Space_187.68_(100.00%)
+
+class Solution {
+    private lateinit var adj: Array<ArrayList<IntArray>>
+    private lateinit var nums: IntArray
+    private lateinit var dist: IntArray
+    private lateinit var lastOccur: IntArray
+    private lateinit var pathStack: ArrayList<Int>
+    private var minIndex = 0
+    private var maxLen = 0
+    private var minNodesForMaxLen = 0
+
+    fun longestSpecialPath(edges: Array<IntArray>, nums: IntArray): IntArray {
+        val n = nums.size
+        this.nums = nums
+        adj = Array<ArrayList<IntArray>>(n) { ArrayList<IntArray>() }
+        for (i in 0..<n) {
+            adj[i] = ArrayList<IntArray>()
+        }
+        for (e in edges) {
+            val u = e[0]
+            val v = e[1]
+            val w = e[2]
+            adj[u].add(intArrayOf(v, w))
+            adj[v].add(intArrayOf(u, w))
+        }
+        dist = IntArray(n)
+        buildDist(0, -1, 0)
+        var maxVal = 0
+        for (`val` in nums) {
+            if (`val` > maxVal) {
+                maxVal = `val`
+            }
+        }
+        lastOccur = IntArray(maxVal + 1)
+        lastOccur.fill(-1)
+        pathStack = ArrayList<Int>()
+        minIndex = 0
+        maxLen = 0
+        minNodesForMaxLen = Int.Companion.MAX_VALUE
+        dfs(0, -1)
+        return intArrayOf(maxLen, minNodesForMaxLen)
+    }
+
+    private fun buildDist(u: Int, parent: Int, currDist: Int) {
+        dist[u] = currDist
+        for (edge in adj[u]) {
+            val v = edge[0]
+            val w = edge[1]
+            if (v == parent) {
+                continue
+            }
+            buildDist(v, u, currDist + w)
+        }
+    }
+
+    private fun dfs(u: Int, parent: Int) {
+        val stackPos = pathStack.size
+        pathStack.add(u)
+        val `val` = nums[u]
+        val oldPos = lastOccur[`val`]
+        val oldMinIndex = minIndex
+        lastOccur[`val`] = stackPos
+        if (oldPos >= minIndex) {
+            minIndex = oldPos + 1
+        }
+        if (minIndex <= stackPos) {
+            val ancestor = pathStack[minIndex]
+            val pathLength = dist[u] - dist[ancestor]
+            val pathNodes = stackPos - minIndex + 1
+            if (pathLength > maxLen) {
+                maxLen = pathLength
+                minNodesForMaxLen = pathNodes
+            } else if (pathLength == maxLen && pathNodes < minNodesForMaxLen) {
+                minNodesForMaxLen = pathNodes
+            }
+        }
+        for (edge in adj[u]) {
+            val v = edge[0]
+            if (v == parent) {
+                continue
+            }
+            dfs(v, u)
+        }
+        pathStack.removeAt(pathStack.size - 1)
+        lastOccur[`val`] = oldPos
+        minIndex = oldMinIndex
+    }
+}

src/main/kotlin/g3401_3500/s3425_longest_special_path/readme.md

@@ -0,0 +1,48 @@
+3425\. Longest Special Path
+
+Hard
+
+You are given an undirected tree rooted at node `0` with `n` nodes numbered from `0` to `n - 1`, represented by a 2D array `edges` of length `n - 1`, where <code>edges[i] = [u<sub>i</sub>, v<sub>i</sub>, length<sub>i</sub>]</code> indicates an edge between nodes <code>u<sub>i</sub></code> and <code>v<sub>i</sub></code> with length <code>length<sub>i</sub></code>. You are also given an integer array `nums`, where `nums[i]` represents the value at node `i`.
+
+A **special path** is defined as a **downward** path from an ancestor node to a descendant node such that all the values of the nodes in that path are **unique**.
+
+**Note** that a path may start and end at the same node.
+
+Return an array `result` of size 2, where `result[0]` is the **length** of the **longest** special path, and `result[1]` is the **minimum** number of nodes in all _possible_ **longest** special paths.
+
+**Example 1:**
+
+**Input:** edges = [[0,1,2],[1,2,3],[1,3,5],[1,4,4],[2,5,6]], nums = [2,1,2,1,3,1]
+
+**Output:** [6,2]
+
+**Explanation:**
+
+#### In the image below, nodes are colored by their corresponding values in `nums`
+
+![](https://assets.leetcode.com/uploads/2024/11/02/tree3.jpeg)
+
+The longest special paths are `2 -> 5` and `0 -> 1 -> 4`, both having a length of 6. The minimum number of nodes across all longest special paths is 2.
+
+**Example 2:**
+
+**Input:** edges = [[1,0,8]], nums = [2,2]
+
+**Output:** [0,1]
+
+**Explanation:**
+
+![](https://assets.leetcode.com/uploads/2024/11/02/tree4.jpeg)
+
+The longest special paths are `0` and `1`, both having a length of 0. The minimum number of nodes across all longest special paths is 1.
+
+**Constraints:**
+
+*   <code>2 <= n <= 5 * 10<sup>4</sup></code>
+*   `edges.length == n - 1`
+*   `edges[i].length == 3`
+*   <code>0 <= u<sub>i</sub>, v<sub>i</sub> < n</code>
+*   <code>1 <= length<sub>i</sub> <= 10<sup>3</sup></code>
+*   `nums.length == n`
+*   <code>0 <= nums[i] <= 5 * 10<sup>4</sup></code>
+*   The input is generated such that `edges` represents a valid tree.

src/main/kotlin/g3401_3500/s3426_manhattan_distances_of_all_arrangements_of_pieces/Solution.kt

@@ -0,0 +1,40 @@
+package g3401_3500.s3426_manhattan_distances_of_all_arrangements_of_pieces
+
+// #Hard #2025_01_19_Time_21_(100.00%)_Space_34.61_(100.00%)
+
+class Solution {
+    private fun comb(a: Long, b: Long, mod: Long): Long {
+        if (b > a) {
+            return 0
+        }
+        var numer: Long = 1
+        var denom: Long = 1
+        for (i in 0..<b) {
+            numer = numer * (a - i) % mod
+            denom = denom * (i + 1) % mod
+        }
+        var denomInv: Long = 1
+        var exp = mod - 2
+        while (exp > 0) {
+            if (exp % 2 > 0) {
+                denomInv = denomInv * denom % mod
+            }
+            denom = denom * denom % mod
+            exp /= 2
+        }
+        return numer * denomInv % mod
+    }
+
+    fun distanceSum(m: Int, n: Int, k: Int): Int {
+        var res: Long = 0
+        val mod: Long = 1000000007
+        val base = comb(m.toLong() * n - 2, k - 2L, mod)
+        for (d in 1..<n) {
+            res = (res + d.toLong() * (n - d) % mod * m % mod * m % mod) % mod
+        }
+        for (d in 1..<m) {
+            res = (res + d.toLong() * (m - d) % mod * n % mod * n % mod) % mod
+        }
+        return (res * base % mod).toInt()
+    }
+}

src/main/kotlin/g3401_3500/s3426_manhattan_distances_of_all_arrangements_of_pieces/readme.md

@@ -0,0 +1,53 @@
+3426\. Manhattan Distances of All Arrangements of Pieces
+
+Hard
+
+You are given three integers `m`, `n`, and `k`.
+
+There is a rectangular grid of size `m × n` containing `k` identical pieces. Return the sum of Manhattan distances between every pair of pieces over all **valid arrangements** of pieces.
+
+A **valid arrangement** is a placement of all `k` pieces on the grid with **at most** one piece per cell.
+
+Since the answer may be very large, return it **modulo** <code>10<sup>9</sup> + 7</code>.
+
+The Manhattan Distance between two cells <code>(x<sub>i</sub>, y<sub>i</sub>)</code> and <code>(x<sub>j</sub>, y<sub>j</sub>)</code> is <code>|x<sub>i</sub> - x<sub>j</sub>| + |y<sub>i</sub> - y<sub>j</sub>|</code>.
+
+**Example 1:**
+
+**Input:** m = 2, n = 2, k = 2
+
+**Output:** 8
+
+**Explanation:**
+
+The valid arrangements of pieces on the board are:
+
+![](https://assets.leetcode.com/uploads/2024/12/25/4040example1.drawio)![](https://assets.leetcode.com/uploads/2024/12/25/untitled-diagramdrawio.png)
+
+*   In the first 4 arrangements, the Manhattan distance between the two pieces is 1.
+*   In the last 2 arrangements, the Manhattan distance between the two pieces is 2.
+
+Thus, the total Manhattan distance across all valid arrangements is `1 + 1 + 1 + 1 + 2 + 2 = 8`.
+
+**Example 2:**
+
+**Input:** m = 1, n = 4, k = 3
+
+**Output:** 20
+
+**Explanation:**
+
+The valid arrangements of pieces on the board are:
+
+![](https://assets.leetcode.com/uploads/2024/12/25/4040example2drawio.png)
+
+*   The first and last arrangements have a total Manhattan distance of `1 + 1 + 2 = 4`.
+*   The middle two arrangements have a total Manhattan distance of `1 + 2 + 3 = 6`.
+
+The total Manhattan distance between all pairs of pieces across all arrangements is `4 + 6 + 6 + 4 = 20`.
+
+**Constraints:**
+
+*   <code>1 <= m, n <= 10<sup>5</sup></code>
+*   <code>2 <= m * n <= 10<sup>5</sup></code>
+*   `2 <= k <= m * n`

src/main/kotlin/g3401_3500/s3427_sum_of_variable_length_subarrays/Solution.kt

@@ -0,0 +1,15 @@
+package g3401_3500.s3427_sum_of_variable_length_subarrays
+
+// #Easy #Array #Prefix_Sum #2025_01_22_Time_0_(100.00%)_Space_43.77_(58.41%)
+
+class Solution {
+    fun subarraySum(nums: IntArray): Int {
+        var res = nums[0]
+        for (i in 1..<nums.size) {
+            val j = i - nums[i] - 1
+            nums[i] += nums[i - 1]
+            res += nums[i] - (if (j < 0) 0 else nums[j])
+        }
+        return res
+    }
+}