Added tasks 3110-3123

Author: javadev

src/main/java/g3101_3200/s3110_score_of_a_string/Solution.java

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+package g3101_3200.s3110_score_of_a_string;
+
+// #Easy #String #2024_04_27_Time_1_ms_(99.93%)_Space_41.4_MB_(99.03%)
+
+public class Solution {
+    public int scoreOfString(String s) {
+        int sum = 0;
+        for (int i = 0; i < s.length() - 1; i++) {
+            sum += Math.abs((s.charAt(i) - '0') - (s.charAt(i + 1) - '0'));
+        }
+        return sum;
+    }
+}

src/main/java/g3101_3200/s3110_score_of_a_string/readme.md

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+3110\. Score of a String
+
+Easy
+
+You are given a string `s`. The **score** of a string is defined as the sum of the absolute difference between the **ASCII** values of adjacent characters.
+
+Return the **score** of `s`.
+
+**Example 1:**
+
+**Input:** s = "hello"
+
+**Output:** 13
+
+**Explanation:**
+
+The **ASCII** values of the characters in `s` are: `'h' = 104`, `'e' = 101`, `'l' = 108`, `'o' = 111`. So, the score of `s` would be `|104 - 101| + |101 - 108| + |108 - 108| + |108 - 111| = 3 + 7 + 0 + 3 = 13`.
+
+**Example 2:**
+
+**Input:** s = "zaz"
+
+**Output:** 50
+
+**Explanation:**
+
+The **ASCII** values of the characters in `s` are: `'z' = 122`, `'a' = 97`. So, the score of `s` would be `|122 - 97| + |97 - 122| = 25 + 25 = 50`.
+
+**Constraints:**
+
+*   `2 <= s.length <= 100`
+*   `s` consists only of lowercase English letters.

src/main/java/g3101_3200/s3111_minimum_rectangles_to_cover_points/Solution.java

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+package g3101_3200.s3111_minimum_rectangles_to_cover_points;
+
+// #Medium #Array #Sorting #Greedy #2024_04_27_Time_4_ms_(99.55%)_Space_97.4_MB_(47.05%)
+
+import java.util.Arrays;
+
+public class Solution {
+    public int minRectanglesToCoverPoints(int[][] points, int w) {
+        Arrays.sort(points, (a, b) -> a[0] - b[0]);
+        int res = 0;
+        int last = -1;
+        for (int[] a : points) {
+            if (a[0] > last) {
+                res++;
+                last = a[0] + w;
+            }
+        }
+        return res;
+    }
+}

src/main/java/g3101_3200/s3111_minimum_rectangles_to_cover_points/readme.md

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+3111\. Minimum Rectangles to Cover Points
+
+Medium
+
+You are given a 2D integer array `points`, where <code>points[i] = [x<sub>i</sub>, y<sub>i</sub>]</code>. You are also given an integer `w`. Your task is to **cover** **all** the given points with rectangles.
+
+Each rectangle has its lower end at some point <code>(x<sub>1</sub>, 0)</code> and its upper end at some point <code>(x<sub>2</sub>, y<sub>2</sub>)</code>, where <code>x<sub>1</sub> <= x<sub>2</sub></code>, <code>y<sub>2</sub> >= 0</code>, and the condition <code>x<sub>2</sub> - x<sub>1</sub> <= w</code> **must** be satisfied for each rectangle.
+
+A point is considered covered by a rectangle if it lies within or on the boundary of the rectangle.
+
+Return an integer denoting the **minimum** number of rectangles needed so that each point is covered by **at least one** rectangle_._
+
+**Note:** A point may be covered by more than one rectangle.
+
+**Example 1:**
+
+![](https://assets.leetcode.com/uploads/2024/03/04/screenshot-from-2024-03-04-20-33-05.png)
+
+**Input:** points = [[2,1],[1,0],[1,4],[1,8],[3,5],[4,6]], w = 1
+
+**Output:** 2
+
+**Explanation:**
+
+The image above shows one possible placement of rectangles to cover the points:
+
+*   A rectangle with a lower end at `(1, 0)` and its upper end at `(2, 8)`
+*   A rectangle with a lower end at `(3, 0)` and its upper end at `(4, 8)`
+
+**Example 2:**
+
+![](https://assets.leetcode.com/uploads/2024/03/04/screenshot-from-2024-03-04-18-59-12.png)
+
+**Input:** points = [[0,0],[1,1],[2,2],[3,3],[4,4],[5,5],[6,6]], w = 2
+
+**Output:** 3
+
+**Explanation:**
+
+The image above shows one possible placement of rectangles to cover the points:
+
+*   A rectangle with a lower end at `(0, 0)` and its upper end at `(2, 2)`
+*   A rectangle with a lower end at `(3, 0)` and its upper end at `(5, 5)`
+*   A rectangle with a lower end at `(6, 0)` and its upper end at `(6, 6)`
+
+**Example 3:**
+
+![](https://assets.leetcode.com/uploads/2024/03/04/screenshot-from-2024-03-04-20-24-03.png)
+
+**Input:** points = [[2,3],[1,2]], w = 0
+
+**Output:** 2
+
+**Explanation:**
+
+The image above shows one possible placement of rectangles to cover the points:
+
+*   A rectangle with a lower end at `(1, 0)` and its upper end at `(1, 2)`
+*   A rectangle with a lower end at `(2, 0)` and its upper end at `(2, 3)`
+
+**Constraints:**
+
+*   <code>1 <= points.length <= 10<sup>5</sup></code>
+*   `points[i].length == 2`
+*   <code>0 <= x<sub>i</sub> == points[i][0] <= 10<sup>9</sup></code>
+*   <code>0 <= y<sub>i</sub> == points[i][1] <= 10<sup>9</sup></code>
+*   <code>0 <= w <= 10<sup>9</sup></code>
+*   All pairs <code>(x<sub>i</sub>, y<sub>i</sub>)</code> are distinct.

src/main/java/g3101_3200/s3112_minimum_time_to_visit_disappearing_nodes/Solution.java

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+package g3101_3200.s3112_minimum_time_to_visit_disappearing_nodes;
+
+// #Medium #Array #Heap_Priority_Queue #Graph #Shortest_Path
+// #2024_04_27_Time_10_ms_(100.00%)_Space_85.4_MB_(99.80%)
+
+import java.util.Arrays;
+
+public class Solution {
+    public int[] minimumTime(int n, int[][] edges, int[] disappear) {
+        int[] dist = new int[n];
+        Arrays.fill(dist, Integer.MAX_VALUE);
+        boolean exit = false;
+        int i, src, dest, cost;
+        dist[0] = 0;
+        for (i = 0; i < n && !exit; ++i) {
+            exit = true;
+            for (int[] edge : edges) {
+                src = edge[0];
+                dest = edge[1];
+                cost = edge[2];
+                if (dist[src] != -1
+                        && dist[src] != Integer.MAX_VALUE
+                        && dist[src] < disappear[src]) {
+                    if (dist[src] + cost < dist[dest]) {
+                        exit = false;
+                        dist[dest] = dist[src] + cost;
+                    }
+                }
+                if (dist[dest] != -1
+                        && dist[dest] != Integer.MAX_VALUE
+                        && dist[dest] < disappear[dest]) {
+                    if (dist[dest] + cost < dist[src]) {
+                        exit = false;
+                        dist[src] = dist[dest] + cost;
+                    }
+                }
+            }
+        }
+        for (i = 0; i < dist.length; ++i) {
+            if (dist[i] == Integer.MAX_VALUE || dist[i] >= disappear[i]) {
+                dist[i] = -1;
+            }
+        }
+        return dist;
+    }
+}

src/main/java/g3101_3200/s3112_minimum_time_to_visit_disappearing_nodes/readme.md

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+3112\. Minimum Time to Visit Disappearing Nodes
+
+Medium
+
+There is an undirected graph of `n` nodes. You are given a 2D array `edges`, where <code>edges[i] = [u<sub>i</sub>, v<sub>i</sub>, length<sub>i</sub>]</code> describes an edge between node <code>u<sub>i</sub></code> and node <code>v<sub>i</sub></code> with a traversal time of <code>length<sub>i</sub></code> units.
+
+Additionally, you are given an array `disappear`, where `disappear[i]` denotes the time when the node `i` disappears from the graph and you won't be able to visit it.
+
+**Notice** that the graph might be disconnected and might contain multiple edges.
+
+Return the array `answer`, with `answer[i]` denoting the **minimum** units of time required to reach node `i` from node 0. If node `i` is **unreachable** from node 0 then `answer[i]` is `-1`.
+
+**Example 1:**
+
+![](https://assets.leetcode.com/uploads/2024/03/09/example1.png)
+
+**Input:** n = 3, edges = [[0,1,2],[1,2,1],[0,2,4]], disappear = [1,1,5]
+
+**Output:** [0,-1,4]
+
+**Explanation:**
+
+We are starting our journey from node 0, and our goal is to find the minimum time required to reach each node before it disappears.
+
+*   For node 0, we don't need any time as it is our starting point.
+*   For node 1, we need at least 2 units of time to traverse `edges[0]`. Unfortunately, it disappears at that moment, so we won't be able to visit it.
+*   For node 2, we need at least 4 units of time to traverse `edges[2]`.
+
+**Example 2:**
+
+![](https://assets.leetcode.com/uploads/2024/03/09/example2.png)
+
+**Input:** n = 3, edges = [[0,1,2],[1,2,1],[0,2,4]], disappear = [1,3,5]
+
+**Output:** [0,2,3]
+
+**Explanation:**
+
+We are starting our journey from node 0, and our goal is to find the minimum time required to reach each node before it disappears.
+
+*   For node 0, we don't need any time as it is the starting point.
+*   For node 1, we need at least 2 units of time to traverse `edges[0]`.
+*   For node 2, we need at least 3 units of time to traverse `edges[0]` and `edges[1]`.
+
+**Example 3:**
+
+**Input:** n = 2, edges = [[0,1,1]], disappear = [1,1]
+
+**Output:** [0,-1]
+
+**Explanation:**
+
+Exactly when we reach node 1, it disappears.
+
+**Constraints:**
+
+*   <code>1 <= n <= 5 * 10<sup>4</sup></code>
+*   <code>0 <= edges.length <= 10<sup>5</sup></code>
+*   <code>edges[i] == [u<sub>i</sub>, v<sub>i</sub>, length<sub>i</sub>]</code>
+*   <code>0 <= u<sub>i</sub>, v<sub>i</sub> <= n - 1</code>
+*   <code>1 <= length<sub>i</sub> <= 10<sup>5</sup></code>
+*   `disappear.length == n`
+*   <code>1 <= disappear[i] <= 10<sup>5</sup></code>

src/main/java/g3101_3200/s3113_find_the_number_of_subarrays_where_boundary_elements_are_maximum/Solution.java

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+package g3101_3200.s3113_find_the_number_of_subarrays_where_boundary_elements_are_maximum;
+
+// #Hard #Array #Binary_Search #Stack #Monotonic_Stack
+// #2024_04_27_Time_13_ms_(98.83%)_Space_60.4_MB_(67.66%)
+
+import java.util.ArrayDeque;
+
+public class Solution {
+    public long numberOfSubarrays(int[] nums) {
+        ArrayDeque<int[]> stack = new ArrayDeque<>();
+        long res = 0;
+        for (int a : nums) {
+            while (!stack.isEmpty() && stack.peek()[0] < a) stack.pop();
+            if (stack.isEmpty() || stack.peek()[0] != a) stack.push(new int[] {a, 0});
+            res += ++stack.peek()[1];
+        }
+        return res;
+    }
+}

src/main/java/g3101_3200/s3113_find_the_number_of_subarrays_where_boundary_elements_are_maximum/readme.md

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+3113\. Find the Number of Subarrays Where Boundary Elements Are Maximum
+
+Hard
+
+You are given an array of **positive** integers `nums`.
+
+Return the number of subarrays of `nums`, where the **first** and the **last** elements of the subarray are _equal_ to the **largest** element in the subarray.
+
+**Example 1:**
+
+**Input:** nums = [1,4,3,3,2]
+
+**Output:** 6
+
+**Explanation:**
+
+There are 6 subarrays which have the first and the last elements equal to the largest element of the subarray:
+
+*   subarray <code>[**<ins>1</ins>**,4,3,3,2]</code>, with its largest element 1. The first element is 1 and the last element is also 1.
+*   subarray <code>[1,<ins>**4**</ins>,3,3,2]</code>, with its largest element 4. The first element is 4 and the last element is also 4.
+*   subarray <code>[1,4,<ins>**3**</ins>,3,2]</code>, with its largest element 3. The first element is 3 and the last element is also 3.
+*   subarray <code>[1,4,3,<ins>**3**</ins>,2]</code>, with its largest element 3. The first element is 3 and the last element is also 3.
+*   subarray <code>[1,4,3,3,<ins>**2**</ins>]</code>, with its largest element 2. The first element is 2 and the last element is also 2.
+*   subarray <code>[1,4,<ins>**3,3**</ins>,2]</code>, with its largest element 3. The first element is 3 and the last element is also 3.
+
+Hence, we return 6.
+
+**Example 2:**
+
+**Input:** nums = [3,3,3]
+
+**Output:** 6
+
+**Explanation:**
+
+There are 6 subarrays which have the first and the last elements equal to the largest element of the subarray:
+
+*   subarray <code>[<ins>**3**</ins>,3,3]</code>, with its largest element 3. The first element is 3 and the last element is also 3.
+*   subarray <code>[3,**<ins>3</ins>**,3]</code>, with its largest element 3. The first element is 3 and the last element is also 3.
+*   subarray <code>[3,3,<ins>**3**</ins>]</code>, with its largest element 3. The first element is 3 and the last element is also 3.
+*   subarray <code>[**<ins>3,3</ins>**,3]</code>, with its largest element 3. The first element is 3 and the last element is also 3.
+*   subarray <code>[3,<ins>**3,3**</ins>]</code>, with its largest element 3. The first element is 3 and the last element is also 3.
+*   subarray <code>[<ins>**3,3,3**</ins>]</code>, with its largest element 3. The first element is 3 and the last element is also 3.
+
+Hence, we return 6.
+
+**Example 3:**
+
+**Input:** nums = [1]
+
+**Output:** 1
+
+**Explanation:**
+
+There is a single subarray of `nums` which is <code>[**<ins>1</ins>**]</code>, with its largest element 1. The first element is 1 and the last element is also 1.
+
+Hence, we return 1.
+
+**Constraints:**
+
+*   <code>1 <= nums.length <= 10<sup>5</sup></code>
+*   <code>1 <= nums[i] <= 10<sup>9</sup></code>

src/main/java/g3101_3200/s3114_latest_time_you_can_obtain_after_replacing_characters/Solution.java

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+package g3101_3200.s3114_latest_time_you_can_obtain_after_replacing_characters;
+
+// #Easy #String #Enumeration #2024_04_27_Time_1_ms_(100.00%)_Space_42.5_MB_(85.42%)
+
+public class Solution {
+    public String findLatestTime(String s) {
+        StringBuilder nm = new StringBuilder();
+        if (s.charAt(0) == '?' && s.charAt(1) == '?') {
+            nm.append("11");
+        } else if (s.charAt(0) != '?' && s.charAt(1) == '?') {
+            nm.append(s.charAt(0));
+            if (s.charAt(0) == '1') {
+                nm.append("1");
+            } else {
+                nm.append("9");
+            }
+        } else if (s.charAt(0) == '?' && s.charAt(1) != '?') {
+            if (s.charAt(1) >= '2' && s.charAt(1) <= '9') {
+                nm.append("0");
+            } else {
+                nm.append("1");
+            }
+            nm.append(s.charAt(1));
+        } else {
+            nm.append(s.charAt(0));
+            nm.append(s.charAt(1));
+        }
+        nm.append(":");
+        if (s.charAt(3) == '?' && s.charAt(4) == '?') {
+            nm.append("59");
+        } else if (s.charAt(3) != '?' && s.charAt(4) == '?') {
+            nm.append(s.charAt(3));
+            nm.append("9");
+        } else if (s.charAt(3) == '?' && s.charAt(4) != '?') {
+            nm.append("5");
+            nm.append(s.charAt(4));
+        } else {
+            nm.append(s.charAt(3));
+            nm.append(s.charAt(4));
+        }
+        return nm.toString();
+    }
+}