Added tasks 3168-3171

Author: javadev

src/main/java/g3101_3200/s3168_minimum_number_of_chairs_in_a_waiting_room/Solution.java

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+package g3101_3200.s3168_minimum_number_of_chairs_in_a_waiting_room;
+
+// #Easy #String #Simulation #2024_06_06_Time_1_ms_(100.00%)_Space_41.9_MB_(67.53%)
+
+public class Solution {
+    public int minimumChairs(String s) {
+        int count = 0;
+        int ans = Integer.MIN_VALUE;
+        for (char ch : s.toCharArray()) {
+            if (ch == 'E') {
+                count++;
+                ans = Math.max(ans, count);
+            } else {
+                count--;
+            }
+        }
+        return ans;
+    }
+}

src/main/java/g3101_3200/s3168_minimum_number_of_chairs_in_a_waiting_room/readme.md

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+3168\. Minimum Number of Chairs in a Waiting Room
+
+Easy
+
+You are given a string `s`. Simulate events at each second `i`:
+
+*   If `s[i] == 'E'`, a person enters the waiting room and takes one of the chairs in it.
+*   If `s[i] == 'L'`, a person leaves the waiting room, freeing up a chair.
+
+Return the **minimum** number of chairs needed so that a chair is available for every person who enters the waiting room given that it is initially **empty**.
+
+**Example 1:**
+
+**Input:** s = "EEEEEEE"
+
+**Output:** 7
+
+**Explanation:**
+
+After each second, a person enters the waiting room and no person leaves it. Therefore, a minimum of 7 chairs is needed.
+
+**Example 2:**
+
+**Input:** s = "ELELEEL"
+
+**Output:** 2
+
+**Explanation:**
+
+Let's consider that there are 2 chairs in the waiting room. The table below shows the state of the waiting room at each second.
+
+| Second | Event | People in the Waiting Room | Available Chairs |
+|--------|-------|----------------------------|------------------|
+| 0      | Enter | 1                          | 1                |
+| 1      | Leave | 0                          | 2                |
+| 2      | Enter | 1                          | 1                |
+| 3      | Leave | 0                          | 2                |
+| 4      | Enter | 1                          | 1                |
+| 5      | Enter | 2                          | 0                |
+| 6      | Leave | 1                          | 1                |
+
+**Example 3:**
+
+**Input:** s = "ELEELEELLL"
+
+**Output:** 3
+
+**Explanation:**
+
+Let's consider that there are 3 chairs in the waiting room. The table below shows the state of the waiting room at each second.
+
+| Second | Event | People in the Waiting Room | Available Chairs |
+|--------|-------|----------------------------|------------------|
+| 0      | Enter | 1                          | 2                |
+| 1      | Leave | 0                          | 3                |
+| 2      | Enter | 1                          | 2                |
+| 3      | Enter | 2                          | 1                |
+| 4      | Leave | 1                          | 2                |
+| 5      | Enter | 2                          | 1                |
+| 6      | Enter | 3                          | 0                |
+| 7      | Leave | 2                          | 1                |
+| 8      | Leave | 1                          | 2                |
+| 9      | Leave | 0                          | 3                |
+
+**Constraints:**
+
+*   `1 <= s.length <= 50`
+*   `s` consists only of the letters `'E'` and `'L'`.
+*   `s` represents a valid sequence of entries and exits.

src/main/java/g3101_3200/s3169_count_days_without_meetings/Solution.java

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+package g3101_3200.s3169_count_days_without_meetings;
+
+// #Medium #Array #Sorting #2024_06_06_Time_11_ms_(99.96%)_Space_113.7_MB_(5.10%)
+
+import java.util.ArrayList;
+import java.util.List;
+
+public class Solution {
+    public int countDays(int days, int[][] meetings) {
+        List<int[]> availableDays = new ArrayList<>();
+        availableDays.add(new int[] {1, days});
+        // Iterate through each meeting
+        for (int[] meeting : meetings) {
+            int start = meeting[0];
+            int end = meeting[1];
+            List<int[]> newAvailableDays = new ArrayList<>();
+            // Iterate through available days and split the intervals
+            for (int[] interval : availableDays) {
+                if (start > interval[1] || end < interval[0]) {
+                    // No overlap, keep the interval
+                    newAvailableDays.add(interval);
+                } else {
+                    // Overlap, split the interval
+                    if (interval[0] < start) {
+                        newAvailableDays.add(new int[] {interval[0], start - 1});
+                    }
+                    if (interval[1] > end) {
+                        newAvailableDays.add(new int[] {end + 1, interval[1]});
+                    }
+                }
+            }
+            availableDays = newAvailableDays;
+        }
+        // Count the remaining available days
+        int availableDaysCount = 0;
+        for (int[] interval : availableDays) {
+            availableDaysCount += interval[1] - interval[0] + 1;
+        }
+        return availableDaysCount;
+    }
+}

src/main/java/g3101_3200/s3169_count_days_without_meetings/readme.md

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+3169\. Count Days Without Meetings
+
+Medium
+
+You are given a positive integer `days` representing the total number of days an employee is available for work (starting from day 1). You are also given a 2D array `meetings` of size `n` where, `meetings[i] = [start_i, end_i]` represents the starting and ending days of meeting `i` (inclusive).
+
+Return the count of days when the employee is available for work but no meetings are scheduled.
+
+**Note:** The meetings may overlap.
+
+**Example 1:**
+
+**Input:** days = 10, meetings = [[5,7],[1,3],[9,10]]
+
+**Output:** 2
+
+**Explanation:**
+
+There is no meeting scheduled on the 4<sup>th</sup> and 8<sup>th</sup> days.
+
+**Example 2:**
+
+**Input:** days = 5, meetings = [[2,4],[1,3]]
+
+**Output:** 1
+
+**Explanation:**
+
+There is no meeting scheduled on the 5<sup>th</sup> day.
+
+**Example 3:**
+
+**Input:** days = 6, meetings = [[1,6]]
+
+**Output:** 0
+
+**Explanation:**
+
+Meetings are scheduled for all working days.
+
+**Constraints:**
+
+*   <code>1 <= days <= 10<sup>9</sup></code>
+*   <code>1 <= meetings.length <= 10<sup>5</sup></code>
+*   `meetings[i].length == 2`
+*   `1 <= meetings[i][0] <= meetings[i][1] <= days`

src/main/java/g3101_3200/s3170_lexicographically_minimum_string_after_removing_stars/Solution.java

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+package g3101_3200.s3170_lexicographically_minimum_string_after_removing_stars;
+
+// #Medium #String #Hash_Table #Greedy #Stack #Heap_Priority_Queue
+// #2024_06_06_Time_29_ms_(99.93%)_Space_45.6_MB_(92.80%)
+
+import java.util.Arrays;
+
+public class Solution {
+    public String clearStars(String s) {
+        char[] arr = s.toCharArray();
+        int[] idxChain = new int[arr.length];
+        int[] lastIdx = new int[26];
+        Arrays.fill(idxChain, -1);
+        Arrays.fill(lastIdx, -1);
+        for (int i = 0; i < arr.length; i++) {
+            if (arr[i] == '*') {
+                for (int j = 0; j < 26; j++) {
+                    if (lastIdx[j] != -1) {
+                        arr[lastIdx[j]] = '#';
+                        lastIdx[j] = idxChain[lastIdx[j]];
+                        break;
+                    }
+                }
+                arr[i] = '#';
+            } else {
+                idxChain[i] = lastIdx[arr[i] - 'a'];
+                lastIdx[arr[i] - 'a'] = i;
+            }
+        }
+        StringBuilder sb = new StringBuilder();
+        for (char c : arr) {
+            if (c != '#') {
+                sb.append(c);
+            }
+        }
+        return sb.toString();
+    }
+}

src/main/java/g3101_3200/s3170_lexicographically_minimum_string_after_removing_stars/readme.md

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+3170\. Lexicographically Minimum String After Removing Stars
+
+Medium
+
+You are given a string `s`. It may contain any number of `'*'` characters. Your task is to remove all `'*'` characters.
+
+While there is a `'*'`, do the following operation:
+
+*   Delete the leftmost `'*'` and the **smallest** non-`'*'` character to its _left_. If there are several smallest characters, you can delete any of them.
+
+Return the lexicographically smallest resulting string after removing all `'*'` characters.
+
+**Example 1:**
+
+**Input:** s = "aaba\*"
+
+**Output:** "aab"
+
+**Explanation:**
+
+We should delete one of the `'a'` characters with `'*'`. If we choose `s[3]`, `s` becomes the lexicographically smallest.
+
+**Example 2:**
+
+**Input:** s = "abc"
+
+**Output:** "abc"
+
+**Explanation:**
+
+There is no `'*'` in the string.
+
+**Constraints:**
+
+*   <code>1 <= s.length <= 10<sup>5</sup></code>
+*   `s` consists only of lowercase English letters and `'*'`.
+*   The input is generated such that it is possible to delete all `'*'` characters.

src/main/java/g3101_3200/s3171_find_subarray_with_bitwise_and_closest_to_k/Solution.java

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+package g3101_3200.s3171_find_subarray_with_bitwise_and_closest_to_k;
+
+// #Hard #Array #Binary_Search #Bit_Manipulation #Segment_Tree
+// #2024_06_06_Time_10_ms_(98.04%)_Space_56.3_MB_(79.06%)
+
+public class Solution {
+    public int minimumDifference(int[] nums, int k) {
+        int res = Integer.MAX_VALUE;
+        for (int i = 0; i < nums.length; i++) {
+            res = Math.min(res, Math.abs(nums[i] - k));
+            for (int j = i - 1; j >= 0 && (nums[j] & nums[i]) != nums[j]; j--) {
+                nums[j] &= nums[i];
+                res = Math.min(res, Math.abs(nums[j] - k));
+            }
+        }
+        return res;
+    }
+}

src/main/java/g3101_3200/s3171_find_subarray_with_bitwise_and_closest_to_k/readme.md

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+3171\. Find Subarray With Bitwise AND Closest to K
+
+Hard
+
+You are given an array `nums` and an integer `k`. You need to find a subarray of `nums` such that the **absolute difference** between `k` and the bitwise `AND` of the subarray elements is as **small** as possible. In other words, select a subarray `nums[l..r]` such that `|k - (nums[l] AND nums[l + 1] ... AND nums[r])|` is minimum.
+
+Return the **minimum** possible value of the absolute difference.
+
+A **subarray** is a contiguous **non-empty** sequence of elements within an array.
+
+**Example 1:**
+
+**Input:** nums = [1,2,4,5], k = 3
+
+**Output:** 1
+
+**Explanation:**
+
+The subarray `nums[2..3]` has `AND` value 4, which gives the minimum absolute difference `|3 - 4| = 1`.
+
+**Example 2:**
+
+**Input:** nums = [1,2,1,2], k = 2
+
+**Output:** 0
+
+**Explanation:**
+
+The subarray `nums[1..1]` has `AND` value 2, which gives the minimum absolute difference `|2 - 2| = 0`.
+
+**Example 3:**
+
+**Input:** nums = [1], k = 10
+
+**Output:** 9
+
+**Explanation:**
+
+There is a single subarray with `AND` value 1, which gives the minimum absolute difference `|10 - 1| = 9`.
+
+**Constraints:**
+
+*   <code>1 <= nums.length <= 10<sup>5</sup></code>
+*   <code>1 <= nums[i] <= 10<sup>9</sup></code>
+*   <code>1 <= k <= 10<sup>9</sup></code>

src/test/java/g3101_3200/s3168_minimum_number_of_chairs_in_a_waiting_room/SolutionTest.java

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+package g3101_3200.s3168_minimum_number_of_chairs_in_a_waiting_room;
+
+import static org.hamcrest.CoreMatchers.equalTo;
+import static org.hamcrest.MatcherAssert.assertThat;
+
+import org.junit.jupiter.api.Test;
+
+class SolutionTest {
+    @Test
+    void minimumChairs() {
+        assertThat(new Solution().minimumChairs("EEEEEEE"), equalTo(7));
+    }
+
+    @Test
+    void minimumChairs2() {
+        assertThat(new Solution().minimumChairs("ELELEEL"), equalTo(2));
+    }
+
+    @Test
+    void minimumChairs3() {
+        assertThat(new Solution().minimumChairs("ELEELEELLL"), equalTo(3));
+    }
+}

src/test/java/g3101_3200/s3169_count_days_without_meetings/SolutionTest.java

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+package g3101_3200.s3169_count_days_without_meetings;
+
+import static org.hamcrest.CoreMatchers.equalTo;
+import static org.hamcrest.MatcherAssert.assertThat;
+
+import org.junit.jupiter.api.Test;
+
+class SolutionTest {
+    @Test
+    void countDays() {
+        assertThat(new Solution().countDays(10, new int[][] {{5, 7}, {1, 3}, {9, 10}}), equalTo(2));
+    }
+
+    @Test
+    void countDays2() {
+        assertThat(new Solution().countDays(5, new int[][] {{2, 4}, {1, 3}}), equalTo(1));
+    }
+
+    @Test
+    void countDays3() {
+        assertThat(new Solution().countDays(6, new int[][] {{1, 6}}), equalTo(0));
+    }
+}

src/test/java/g3101_3200/s3170_lexicographically_minimum_string_after_removing_stars/SolutionTest.java

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+package g3101_3200.s3170_lexicographically_minimum_string_after_removing_stars;
+
+import static org.hamcrest.CoreMatchers.equalTo;
+import static org.hamcrest.MatcherAssert.assertThat;
+
+import org.junit.jupiter.api.Test;
+
+class SolutionTest {
+    @Test
+    void clearStars() {
+        assertThat(new Solution().clearStars("aaba*"), equalTo("aab"));
+    }
+
+    @Test
+    void clearStars2() {
+        assertThat(new Solution().clearStars("abc"), equalTo("abc"));
+    }
+}