Added tasks 3285-3292

src/main/java/g3201_3300/s3285_find_indices_of_stable_mountains/Solution.java

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+package g3201_3300.s3285_find_indices_of_stable_mountains;
+
+// #Easy #Array #2024_09_15_Time_1_ms_(100.00%)_Space_44.5_MB_(100.00%)
+
+import java.util.ArrayList;
+import java.util.List;
+
+public class Solution {
+    public List<Integer> stableMountains(int[] height, int threshold) {
+        int n = height.length;
+        List<Integer> list = new ArrayList<>();
+        for (int i = 0; i < n - 1; i++) {
+            if (height[i] > threshold) {
+                list.add(i + 1);
+            }
+        }
+        return list;
+    }
+}

src/main/java/g3201_3300/s3285_find_indices_of_stable_mountains/readme.md

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+3285\. Find Indices of Stable Mountains
+
+Easy
+
+There are `n` mountains in a row, and each mountain has a height. You are given an integer array `height` where `height[i]` represents the height of mountain `i`, and an integer `threshold`.
+
+A mountain is called **stable** if the mountain just before it (**if it exists**) has a height **strictly greater** than `threshold`. **Note** that mountain 0 is **not** stable.
+
+Return an array containing the indices of _all_ **stable** mountains in **any** order.
+
+**Example 1:**
+
+**Input:** height = [1,2,3,4,5], threshold = 2
+
+**Output:** [3,4]
+
+**Explanation:**
+
+*   Mountain 3 is stable because `height[2] == 3` is greater than `threshold == 2`.
+*   Mountain 4 is stable because `height[3] == 4` is greater than `threshold == 2`.
+
+**Example 2:**
+
+**Input:** height = [10,1,10,1,10], threshold = 3
+
+**Output:** [1,3]
+
+**Example 3:**
+
+**Input:** height = [10,1,10,1,10], threshold = 10
+
+**Output:** []
+
+**Constraints:**
+
+*   `2 <= n == height.length <= 100`
+*   `1 <= height[i] <= 100`
+*   `1 <= threshold <= 100`

src/main/java/g3201_3300/s3286_find_a_safe_walk_through_a_grid/Solution.java

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+package g3201_3300.s3286_find_a_safe_walk_through_a_grid;
+
+// #Medium #Array #Matrix #Heap_Priority_Queue #Graph #Shortest_Path #Breadth_First_Search
+// #2024_09_15_Time_90_ms_(100.00%)_Space_46.6_MB_(100.00%)
+
+import java.util.LinkedList;
+import java.util.List;
+import java.util.Objects;
+import java.util.Queue;
+
+public class Solution {
+    public boolean findSafeWalk(List<List<Integer>> grid, int health) {
+        int n = grid.size();
+        int m = grid.get(0).size();
+        int[] dr = new int[] {0, 0, 1, -1};
+        int[] dc = new int[] {1, -1, 0, 0};
+        boolean[][][] visited = new boolean[n][m][health + 1];
+        Queue<int[]> bfs = new LinkedList<>();
+        bfs.add(new int[] {0, 0, health - grid.get(0).get(0)});
+        visited[0][0][health - grid.get(0).get(0)] = true;
+        while (!bfs.isEmpty()) {
+            int size = bfs.size();
+            while (size-- > 0) {
+                int[] currNode = bfs.poll();
+                int r = Objects.requireNonNull(currNode)[0];
+                int c = currNode[1];
+                int h = currNode[2];
+                if (r == n - 1 && c == m - 1 && h > 0) {
+                    return true;
+                }
+                for (int k = 0; k < 4; k++) {
+                    int nr = r + dr[k];
+                    int nc = c + dc[k];
+                    if (isValidMove(nr, nc, n, m)) {
+                        int nh = h - grid.get(nr).get(nc);
+                        if (nh >= 0 && !visited[nr][nc][nh]) {
+                            visited[nr][nc][nh] = true;
+                            bfs.add(new int[] {nr, nc, nh});
+                        }
+                    }
+                }
+            }
+        }
+        return false;
+    }
+
+    private boolean isValidMove(int r, int c, int n, int m) {
+        return r >= 0 && c >= 0 && r < n && c < m;
+    }
+}

src/main/java/g3201_3300/s3286_find_a_safe_walk_through_a_grid/readme.md

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+3286\. Find a Safe Walk Through a Grid
+
+Medium
+
+You are given an `m x n` binary matrix `grid` and an integer `health`.
+
+You start on the upper-left corner `(0, 0)` and would like to get to the lower-right corner `(m - 1, n - 1)`.
+
+You can move up, down, left, or right from one cell to another adjacent cell as long as your health _remains_ **positive**.
+
+Cells `(i, j)` with `grid[i][j] = 1` are considered **unsafe** and reduce your health by 1.
+
+Return `true` if you can reach the final cell with a health value of 1 or more, and `false` otherwise.
+
+**Example 1:**
+
+**Input:** grid = [[0,1,0,0,0],[0,1,0,1,0],[0,0,0,1,0]], health = 1
+
+**Output:** true
+
+**Explanation:**
+
+The final cell can be reached safely by walking along the gray cells below.
+
+![](https://assets.leetcode.com/uploads/2024/08/04/3868_examples_1drawio.png)
+
+**Example 2:**
+
+**Input:** grid = [[0,1,1,0,0,0],[1,0,1,0,0,0],[0,1,1,1,0,1],[0,0,1,0,1,0]], health = 3
+
+**Output:** false
+
+**Explanation:**
+
+A minimum of 4 health points is needed to reach the final cell safely.
+
+![](https://assets.leetcode.com/uploads/2024/08/04/3868_examples_2drawio.png)
+
+**Example 3:**
+
+**Input:** grid = [[1,1,1],[1,0,1],[1,1,1]], health = 5
+
+**Output:** true
+
+**Explanation:**
+
+The final cell can be reached safely by walking along the gray cells below.
+
+![](https://assets.leetcode.com/uploads/2024/08/04/3868_examples_3drawio.png)
+
+Any path that does not go through the cell `(1, 1)` is unsafe since your health will drop to 0 when reaching the final cell.
+
+**Constraints:**
+
+*   `m == grid.length`
+*   `n == grid[i].length`
+*   `1 <= m, n <= 50`
+*   `2 <= m * n`
+*   `1 <= health <= m + n`
+*   `grid[i][j]` is either 0 or 1.

src/main/java/g3201_3300/s3287_find_the_maximum_sequence_value_of_array/Solution.java

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+package g3201_3300.s3287_find_the_maximum_sequence_value_of_array;
+
+// #Hard #Array #Dynamic_Programming #Bit_Manipulation
+// #2024_09_15_Time_1140_ms_(100.00%)_Space_285.4_MB_(100.00%)
+
+import java.util.HashSet;
+import java.util.Set;
+
+@SuppressWarnings("unchecked")
+public class Solution {
+    public int maxValue(int[] nums, int k) {
+        int n = nums.length;
+        Set<Integer>[][] left = new Set[n][k + 1];
+        Set<Integer>[][] right = new Set[n][k + 1];
+        for (int i = 0; i < n; i++) {
+            for (int j = 0; j <= k; j++) {
+                left[i][j] = new HashSet<>();
+                right[i][j] = new HashSet<>();
+            }
+        }
+        left[0][0].add(0);
+        left[0][1].add(nums[0]);
+        for (int i = 1; i < n - k; i++) {
+            left[i][0].add(0);
+            for (int j = 1; j <= k; j++) {
+                left[i][j].addAll(left[i - 1][j]);
+                for (int v : left[i - 1][j - 1]) {
+                    left[i][j].add(v | nums[i]);
+                }
+            }
+        }
+        right[n - 1][0].add(0);
+        right[n - 1][1].add(nums[n - 1]);
+        int result = 0;
+        if (k == 1) {
+            for (int l : left[n - 2][k]) {
+                result = Math.max(result, l ^ nums[n - 1]);
+            }
+        }
+        for (int i = n - 2; i >= k; i--) {
+            right[i][0].add(0);
+            for (int j = 1; j <= k; j++) {
+                right[i][j].addAll(right[i + 1][j]);
+                for (int v : right[i + 1][j - 1]) {
+                    right[i][j].add(v | nums[i]);
+                }
+            }
+            for (int l : left[i - 1][k]) {
+                for (int r : right[i][k]) {
+                    result = Math.max(result, l ^ r);
+                }
+            }
+        }
+        return result;
+    }
+}

src/main/java/g3201_3300/s3287_find_the_maximum_sequence_value_of_array/readme.md

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+3287\. Find the Maximum Sequence Value of Array
+
+Hard
+
+You are given an integer array `nums` and a **positive** integer `k`.
+
+The **value** of a sequence `seq` of size `2 * x` is defined as:
+
+*   `(seq[0] OR seq[1] OR ... OR seq[x - 1]) XOR (seq[x] OR seq[x + 1] OR ... OR seq[2 * x - 1])`.
+
+Return the **maximum** **value** of any subsequence of `nums` having size `2 * k`.
+
+**Example 1:**
+
+**Input:** nums = [2,6,7], k = 1
+
+**Output:** 5
+
+**Explanation:**
+
+The subsequence `[2, 7]` has the maximum value of `2 XOR 7 = 5`.
+
+**Example 2:**
+
+**Input:** nums = [4,2,5,6,7], k = 2
+
+**Output:** 2
+
+**Explanation:**
+
+The subsequence `[4, 5, 6, 7]` has the maximum value of `(4 OR 5) XOR (6 OR 7) = 2`.
+
+**Constraints:**
+
+*   `2 <= nums.length <= 400`
+*   <code>1 <= nums[i] < 2<sup>7</sup></code>
+*   `1 <= k <= nums.length / 2`

src/main/java/g3201_3300/s3288_length_of_the_longest_increasing_path/Solution.java

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+package g3201_3300.s3288_length_of_the_longest_increasing_path;
+
+// #Hard #Array #Sorting #Binary_Search #2024_09_15_Time_34_ms_(100.00%)_Space_106.2_MB_(50.00%)
+
+import java.util.ArrayList;
+import java.util.List;
+
+public class Solution {
+    public int maxPathLength(int[][] coordinates, int k) {
+        List<int[]> upper = new ArrayList<>();
+        List<int[]> lower = new ArrayList<>();
+        for (int[] pair : coordinates) {
+            if (pair[0] > coordinates[k][0] && pair[1] > coordinates[k][1]) {
+                upper.add(pair);
+            }
+            if (pair[0] < coordinates[k][0] && pair[1] < coordinates[k][1]) {
+                lower.add(pair);
+            }
+        }
+        upper.sort(
+                (a, b) -> {
+                    if (a[0] == b[0]) {
+                        return b[1] - a[1];
+                    } else {
+                        return a[0] - b[0];
+                    }
+                });
+        lower.sort(
+                (a, b) -> {
+                    if (a[0] == b[0]) {
+                        return b[1] - a[1];
+                    } else {
+                        return a[0] - b[0];
+                    }
+                });
+        return longestIncreasingLength(upper) + longestIncreasingLength(lower) + 1;
+    }
+
+    private int longestIncreasingLength(List<int[]> array) {
+        List<Integer> list = new ArrayList<>();
+        for (int[] pair : array) {
+            int m = list.size();
+            if (m == 0 || list.get(m - 1) < pair[1]) {
+                list.add(pair[1]);
+            } else {
+                int idx = binarySearch(list, pair[1]);
+                list.set(idx, pair[1]);
+            }
+        }
+        return list.size();
+    }
+
+    private int binarySearch(List<Integer> list, int target) {
+        int n = list.size();
+        int left = 0;
+        int right = n - 1;
+        while (left < right) {
+            int mid = (left + right) / 2;
+            if (list.get(mid) == target) {
+                return mid;
+            } else if (list.get(mid) > target) {
+                right = mid;
+            } else {
+                left = mid + 1;
+            }
+        }
+        return left;
+    }
+}

src/main/java/g3201_3300/s3288_length_of_the_longest_increasing_path/readme.md

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+3288\. Length of the Longest Increasing Path
+
+Hard
+
+You are given a 2D array of integers `coordinates` of length `n` and an integer `k`, where `0 <= k < n`.
+
+<code>coordinates[i] = [x<sub>i</sub>, y<sub>i</sub>]</code> indicates the point <code>(x<sub>i</sub>, y<sub>i</sub>)</code> in a 2D plane.
+
+An **increasing path** of length `m` is defined as a list of points <code>(x<sub>1</sub>, y<sub>1</sub>)</code>, <code>(x<sub>2</sub>, y<sub>2</sub>)</code>, <code>(x<sub>3</sub>, y<sub>3</sub>)</code>, ..., <code>(x<sub>m</sub>, y<sub>m</sub>)</code> such that:
+
+*   <code>x<sub>i</sub> < x<sub>i + 1</sub></code> and <code>y<sub>i</sub> < y<sub>i + 1</sub></code> for all `i` where `1 <= i < m`.
+*   <code>(x<sub>i</sub>, y<sub>i</sub>)</code> is in the given coordinates for all `i` where `1 <= i <= m`.
+
+Return the **maximum** length of an **increasing path** that contains `coordinates[k]`.
+
+**Example 1:**
+
+**Input:** coordinates = [[3,1],[2,2],[4,1],[0,0],[5,3]], k = 1
+
+**Output:** 3
+
+**Explanation:**
+
+`(0, 0)`, `(2, 2)`, `(5, 3)` is the longest increasing path that contains `(2, 2)`.
+
+**Example 2:**
+
+**Input:** coordinates = [[2,1],[7,0],[5,6]], k = 2
+
+**Output:** 2
+
+**Explanation:**
+
+`(2, 1)`, `(5, 6)` is the longest increasing path that contains `(5, 6)`.
+
+**Constraints:**
+
+*   <code>1 <= n == coordinates.length <= 10<sup>5</sup></code>
+*   `coordinates[i].length == 2`
+*   <code>0 <= coordinates[i][0], coordinates[i][1] <= 10<sup>9</sup></code>
+*   All elements in `coordinates` are **distinct**.
+*   `0 <= k <= n - 1`