Added tasks 3423-3430

Author: javadev

src/main/java/g3401_3500/s3423_maximum_difference_between_adjacent_elements_in_a_circular_array/Solution.java

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+package g3401_3500.s3423_maximum_difference_between_adjacent_elements_in_a_circular_array;
+
+// #Easy #Array #2025_01_22_Time_1_(99.86%)_Space_43.72_(36.06%)
+
+public class Solution {
+    public int maxAdjacentDistance(int[] nums) {
+        int maxDiff = 0;
+        for (int i = 0; i < nums.length; i++) {
+            int nextIndex = (i + 1) % nums.length;
+            int diff = Math.abs(nums[i] - nums[nextIndex]);
+            maxDiff = Math.max(maxDiff, diff);
+        }
+        return maxDiff;
+    }
+}

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

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+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/java/g3401_3500/s3424_minimum_cost_to_make_arrays_identical/Solution.java

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+package g3401_3500.s3424_minimum_cost_to_make_arrays_identical;
+
+// #Medium #Array #Sorting #Greedy #2025_01_22_Time_60_(98.08%)_Space_57.68_(39.04%)
+
+import java.util.Arrays;
+
+public class Solution {
+    public long minCost(int[] arr, int[] brr, long k) {
+        long res1 = 0;
+        long res2 = 0;
+        for (int i = 0; i < arr.length; ++i) {
+            res1 += Math.abs(arr[i] - brr[i]);
+        }
+        Arrays.sort(arr);
+        Arrays.sort(brr);
+        for (int i = 0; i < arr.length; ++i) {
+            res2 += Math.abs(arr[i] - brr[i]);
+        }
+        return Math.min(res1, res2 + k);
+    }
+}

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

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+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/java/g3401_3500/s3425_longest_special_path/Solution.java

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+package g3401_3500.s3425_longest_special_path;
+
+// #Hard #Array #Hash_Table #Depth_First_Search #Tree #Sliding_Window
+// #2025_01_22_Time_49_(74.66%)_Space_98.04_(44.26%)
+
+import java.util.ArrayList;
+import java.util.Arrays;
+
+@SuppressWarnings("unchecked")
+public class Solution {
+    private ArrayList<int[]>[] adj;
+    private int[] nums;
+    private int[] dist;
+    private int[] lastOccur;
+    private ArrayList<Integer> pathStack;
+    private int minIndex;
+    private int maxLen;
+    private int minNodesForMaxLen;
+
+    public int[] longestSpecialPath(int[][] edges, int[] nums) {
+        int n = nums.length;
+        this.nums = nums;
+        adj = new ArrayList[n];
+        for (int i = 0; i < n; i++) {
+            adj[i] = new ArrayList<>();
+        }
+        for (int[] e : edges) {
+            int u = e[0];
+            int v = e[1];
+            int w = e[2];
+            adj[u].add(new int[] {v, w});
+            adj[v].add(new int[] {u, w});
+        }
+        dist = new int[n];
+        buildDist(0, -1, 0);
+        int maxVal = 0;
+        for (int val : nums) {
+            if (val > maxVal) {
+                maxVal = val;
+            }
+        }
+        lastOccur = new int[maxVal + 1];
+        Arrays.fill(lastOccur, -1);
+        pathStack = new ArrayList<>();
+        minIndex = 0;
+        maxLen = 0;
+        minNodesForMaxLen = Integer.MAX_VALUE;
+        dfs(0, -1);
+        return new int[] {maxLen, minNodesForMaxLen};
+    }
+
+    private void buildDist(int u, int parent, int currDist) {
+        dist[u] = currDist;
+        for (int[] edge : adj[u]) {
+            int v = edge[0];
+            int w = edge[1];
+            if (v == parent) {
+                continue;
+            }
+            buildDist(v, u, currDist + w);
+        }
+    }
+
+    private void dfs(int u, int parent) {
+        int stackPos = pathStack.size();
+        pathStack.add(u);
+        int val = nums[u];
+        int oldPos = lastOccur[val];
+        int oldMinIndex = minIndex;
+        lastOccur[val] = stackPos;
+        if (oldPos >= minIndex) {
+            minIndex = oldPos + 1;
+        }
+        if (minIndex <= stackPos) {
+            int ancestor = pathStack.get(minIndex);
+            int pathLength = dist[u] - dist[ancestor];
+            int pathNodes = stackPos - minIndex + 1;
+            if (pathLength > maxLen) {
+                maxLen = pathLength;
+                minNodesForMaxLen = pathNodes;
+            } else if (pathLength == maxLen && pathNodes < minNodesForMaxLen) {
+                minNodesForMaxLen = pathNodes;
+            }
+        }
+        for (int[] edge : adj[u]) {
+            int v = edge[0];
+            if (v == parent) {
+                continue;
+            }
+            dfs(v, u);
+        }
+        pathStack.remove(pathStack.size() - 1);
+        lastOccur[val] = oldPos;
+        minIndex = oldMinIndex;
+    }
+}

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

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+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/java/g3401_3500/s3426_manhattan_distances_of_all_arrangements_of_pieces/Solution.java

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+package g3401_3500.s3426_manhattan_distances_of_all_arrangements_of_pieces;
+
+// #Hard #Math #Combinatorics #2025_01_22_Time_20_(87.92%)_Space_40.82_(98.07%)
+
+public class Solution {
+    private long comb(long a, long b, long mod) {
+        if (b > a) {
+            return 0;
+        }
+        long numer = 1;
+        long denom = 1;
+        for (long i = 0; i < b; ++i) {
+            numer = numer * (a - i) % mod;
+            denom = denom * (i + 1) % mod;
+        }
+        long denomInv = 1;
+        long exp = mod - 2;
+        while (exp > 0) {
+            if (exp % 2 > 0) {
+                denomInv = denomInv * denom % mod;
+            }
+            denom = denom * denom % mod;
+            exp /= 2;
+        }
+        return numer * denomInv % mod;
+    }
+
+    public int distanceSum(int m, int n, int k) {
+        long res = 0;
+        long mod = 1000000007;
+        long base = comb((long) m * n - 2, k - 2L, mod);
+        for (int d = 1; d < n; ++d) {
+            res = (res + (long) d * (n - d) % mod * m % mod * m % mod) % mod;
+        }
+        for (int d = 1; d < m; ++d) {
+            res = (res + (long) d * (m - d) % mod * n % mod * n % mod) % mod;
+        }
+        return (int) (res * base % mod);
+    }
+}

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

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+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/java/g3401_3500/s3427_sum_of_variable_length_subarrays/Solution.java

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+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%)
+
+public class Solution {
+    public int subarraySum(int[] nums) {
+        int res = nums[0];
+        for (int i = 1; i < nums.length; i++) {
+            int j = i - nums[i] - 1;
+            nums[i] += nums[i - 1];
+            res += nums[i] - (j < 0 ? 0 : nums[j]);
+        }
+        return res;
+    }
+}