Added tasks 1748, 1749, 1750, 1751, 1752.

Author: ThanhNIT

src/main/java/g1701_1800/s1748_sum_of_unique_elements/Solution.java

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+package g1701_1800.s1748_sum_of_unique_elements;
+
+// #Easy #Array #Hash_Table #Counting #2022_04_30_Time_2_ms_(54.08%)_Space_42_MB_(38.95%)
+
+import java.util.HashMap;
+import java.util.Map;
+
+public class Solution {
+    public int sumOfUnique(int[] nums) {
+        Map<Integer, Integer> map = new HashMap<>();
+        int sum = 0;
+        for (int num : nums) {
+            map.put(num, map.getOrDefault(num, 0) + 1);
+        }
+        for (Map.Entry<Integer, Integer> entry : map.entrySet()) {
+            if (entry.getValue() == 1) {
+                sum += entry.getKey();
+            }
+        }
+        return sum;
+    }
+}

src/main/java/g1701_1800/s1748_sum_of_unique_elements/readme.md

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+1748\. Sum of Unique Elements
+
+Easy
+
+You are given an integer array `nums`. The unique elements of an array are the elements that appear **exactly once** in the array.
+
+Return _the **sum** of all the unique elements of_ `nums`.
+
+**Example 1:**
+
+**Input:** nums = [1,2,3,2]
+
+**Output:** 4
+
+**Explanation:** The unique elements are [1,3], and the sum is 4.
+
+**Example 2:**
+
+**Input:** nums = [1,1,1,1,1]
+
+**Output:** 0
+
+**Explanation:** There are no unique elements, and the sum is 0.
+
+**Example 3:**
+
+**Input:** nums = [1,2,3,4,5]
+
+**Output:** 15
+
+**Explanation:** The unique elements are [1,2,3,4,5], and the sum is 15.
+
+**Constraints:**
+
+*   `1 <= nums.length <= 100`
+*   `1 <= nums[i] <= 100`

src/main/java/g1701_1800/s1749_maximum_absolute_sum_of_any_subarray/Solution.java

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+package g1701_1800.s1749_maximum_absolute_sum_of_any_subarray;
+
+// #Medium #Array #Dynamic_Programming #2022_04_30_Time_3_ms_(90.60%)_Space_65.8_MB_(11.86%)
+
+public class Solution {
+    public int maxAbsoluteSum(int[] nums) {
+        int min = 0;
+        int max = 0;
+        int s = 0;
+        for (int num : nums) {
+            s += num;
+            min = Math.min(min, s);
+            max = Math.max(max, s);
+        }
+        return max - min;
+    }
+}

src/main/java/g1701_1800/s1749_maximum_absolute_sum_of_any_subarray/readme.md

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+1749\. Maximum Absolute Sum of Any Subarray
+
+Medium
+
+You are given an integer array `nums`. The **absolute sum** of a subarray <code>[nums<sub>l</sub>, nums<sub>l+1</sub>, ..., nums<sub>r-1</sub>, nums<sub>r</sub>]</code> is <code>abs(nums<sub>l</sub> + nums<sub>l+1</sub> + ... + nums<sub>r-1</sub> + nums<sub>r</sub>)</code>.
+
+Return _the **maximum** absolute sum of any **(possibly empty)** subarray of_ `nums`.
+
+Note that `abs(x)` is defined as follows:
+
+*   If `x` is a negative integer, then `abs(x) = -x`.
+*   If `x` is a non-negative integer, then `abs(x) = x`.
+
+**Example 1:**
+
+**Input:** nums = [1,-3,2,3,-4]
+
+**Output:** 5
+
+**Explanation:** The subarray [2,3] has absolute sum = abs(2+3) = abs(5) = 5.
+
+**Example 2:**
+
+**Input:** nums = [2,-5,1,-4,3,-2]
+
+**Output:** 8
+
+**Explanation:** The subarray [-5,1,-4] has absolute sum = abs(-5+1-4) = abs(-8) = 8.
+
+**Constraints:**
+
+*   <code>1 <= nums.length <= 10<sup>5</sup></code>
+*   <code>-10<sup>4</sup> <= nums[i] <= 10<sup>4</sup></code>

src/main/java/g1701_1800/s1750_minimum_length_of_string_after_deleting_similar_ends/Solution.java

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+package g1701_1800.s1750_minimum_length_of_string_after_deleting_similar_ends;
+
+// #Medium #String #Two_Pointers #2022_04_30_Time_5_ms_(68.68%)_Space_53.9_MB_(15.47%)
+
+public class Solution {
+    public int minimumLength(String s) {
+        int i = 0;
+        int j = s.length() - 1;
+        if (s.charAt(i) == s.charAt(j)) {
+            while (i < j && s.charAt(i) == s.charAt(j)) {
+                char c = s.charAt(i);
+                i++;
+                while (c == s.charAt(i) && i < j) {
+                    i++;
+                }
+                j--;
+                while (c == s.charAt(j) && i < j) {
+                    j--;
+                }
+            }
+        }
+        return i <= j ? s.substring(i, j).length() + 1 : 0;
+    }
+}

src/main/java/g1701_1800/s1750_minimum_length_of_string_after_deleting_similar_ends/readme.md

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+1750\. Minimum Length of String After Deleting Similar Ends
+
+Medium
+
+Given a string `s` consisting only of characters `'a'`, `'b'`, and `'c'`. You are asked to apply the following algorithm on the string any number of times:
+
+1.  Pick a **non-empty** prefix from the string `s` where all the characters in the prefix are equal.
+2.  Pick a **non-empty** suffix from the string `s` where all the characters in this suffix are equal.
+3.  The prefix and the suffix should not intersect at any index.
+4.  The characters from the prefix and suffix must be the same.
+5.  Delete both the prefix and the suffix.
+
+Return _the **minimum length** of_ `s` _after performing the above operation any number of times (possibly zero times)_.
+
+**Example 1:**
+
+**Input:** s = "ca"
+
+**Output:** 2
+
+**Explanation:** You can't remove any characters, so the string stays as is.
+
+**Example 2:**
+
+**Input:** s = "cabaabac"
+
+**Output:** 0
+
+**Explanation:** An optimal sequence of operations is: 
+
+- Take prefix = "c" and suffix = "c" and remove them, s = "abaaba". 
+
+- Take prefix = "a" and suffix = "a" and remove them, s = "baab". 
+
+- Take prefix = "b" and suffix = "b" and remove them, s = "aa". 
+
+- Take prefix = "a" and suffix = "a" and remove them, s = "".
+
+**Example 3:**
+
+**Input:** s = "aabccabba"
+
+**Output:** 3
+
+**Explanation:** An optimal sequence of operations is: 
+
+- Take prefix = "aa" and suffix = "a" and remove them, s = "bccabb". 
+
+- Take prefix = "b" and suffix = "bb" and remove them, s = "cca".
+
+**Constraints:**
+
+*   <code>1 <= s.length <= 10<sup>5</sup></code>
+*   `s` only consists of characters `'a'`, `'b'`, and `'c'`.

src/main/java/g1701_1800/s1751_maximum_number_of_events_that_can_be_attended_ii/Solution.java

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+package g1701_1800.s1751_maximum_number_of_events_that_can_be_attended_ii;
+
+// #Hard #Array #Dynamic_Programming #Binary_Search
+// #2022_04_30_Time_12_ms_(98.33%)_Space_84.7_MB_(92.22%)
+
+import java.util.Arrays;
+import java.util.Comparator;
+import java.util.Optional;
+
+public class Solution {
+    public int maxValue(int[][] events, int k) {
+        if (k == 1) {
+            Optional<int[]> value = Arrays.stream(events).max(Comparator.comparingInt(e -> e[2]));
+            if (value.isPresent()) {
+                return value.get()[2];
+            } else {
+                throw new NullPointerException();
+            }
+        }
+
+        int n = events.length;
+        Arrays.sort(events, Comparator.comparingInt(a -> a[0]));
+
+        int[][] memo = new int[n][k + 1];
+
+        return dfs(events, 0, k, memo);
+    }
+
+    private int dfs(int[][] events, int i, int k, int[][] memo) {
+        if (k == 0 || i >= events.length) {
+            return 0;
+        }
+        if (memo[i][k] > 0) {
+            return memo[i][k];
+        }
+
+        int idx = binarySearch(events, events[i][1] + 1, i + 1);
+        int use = events[i][2] + dfs(events, idx, k - 1, memo);
+
+        int notUse = dfs(events, i + 1, k, memo);
+
+        int res = Math.max(use, notUse);
+        memo[i][k] = res;
+
+        return res;
+    }
+
+    private int binarySearch(int[][] events, int i, int st) {
+
+        if (st >= events.length) {
+            return st;
+        }
+
+        int end = events.length - 1;
+        while (st < end) {
+            int mid = st + (end - st) / 2;
+            if (events[mid][0] < i) {
+                st = mid + 1;
+            } else {
+                end = mid;
+            }
+        }
+
+        return events[st][0] >= i ? st : st + 1;
+    }
+}

src/main/java/g1701_1800/s1751_maximum_number_of_events_that_can_be_attended_ii/readme.md

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+1751\. Maximum Number of Events That Can Be Attended II
+
+Hard
+
+You are given an array of `events` where <code>events[i] = [startDay<sub>i</sub>, endDay<sub>i</sub>, value<sub>i</sub>]</code>. The <code>i<sup>th</sup></code> event starts at <code>startDay<sub>i</sub></code> and ends at <code>endDay<sub>i</sub></code>, and if you attend this event, you will receive a value of <code>value<sub>i</sub></code>. You are also given an integer `k` which represents the maximum number of events you can attend.
+
+You can only attend one event at a time. If you choose to attend an event, you must attend the **entire** event. Note that the end day is **inclusive**: that is, you cannot attend two events where one of them starts and the other ends on the same day.
+
+Return _the **maximum sum** of values that you can receive by attending events._
+
+**Example 1:**
+
+![](https://assets.leetcode.com/uploads/2021/01/10/screenshot-2021-01-11-at-60048-pm.png)
+
+**Input:** events = [[1,2,4],[3,4,3],[2,3,1]], k = 2
+
+**Output:** 7
+
+**Explanation:** Choose the green events, 0 and 1 (0-indexed) for a total value of 4 + 3 = 7.
+
+**Example 2:**
+
+![](https://assets.leetcode.com/uploads/2021/01/10/screenshot-2021-01-11-at-60150-pm.png)
+
+**Input:** events = [[1,2,4],[3,4,3],[2,3,10]], k = 2
+
+**Output:** 10
+
+**Explanation:** Choose event 2 for a total value of 10. Notice that you cannot attend any other event as they overlap, and that you do **not** have to attend k events.
+
+**Example 3:**
+
+**![](https://assets.leetcode.com/uploads/2021/01/10/screenshot-2021-01-11-at-60703-pm.png)**
+
+**Input:** events = [[1,1,1],[2,2,2],[3,3,3],[4,4,4]], k = 3
+
+**Output:** 9
+
+**Explanation:** Although the events do not overlap, you can only attend 3 events. Pick the highest valued three.
+
+**Constraints:**
+
+*   `1 <= k <= events.length`
+*   <code>1 <= k * events.length <= 10<sup>6</sup></code>
+*   <code>1 <= startDay<sub>i</sub> <= endDay<sub>i</sub> <= 10<sup>9</sup></code>
+*   <code>1 <= value<sub>i</sub> <= 10<sup>6</sup></code>

src/main/java/g1701_1800/s1752_check_if_array_is_sorted_and_rotated/Solution.java

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+package g1701_1800.s1752_check_if_array_is_sorted_and_rotated;
+
+// #Easy #Array #2022_04_30_Time_3_ms_(5.38%)_Space_42.8_MB_(6.77%)
+
+import java.util.Arrays;
+
+public class Solution {
+    public boolean check(int[] nums) {
+        int[] copy = Arrays.copyOf(nums, nums.length);
+        Arrays.sort(copy);
+        for (int i = 1; i <= nums.length; i++) {
+            int[] rotated = rotate(nums, i);
+            if (Arrays.equals(rotated, copy)) {
+                return true;
+            }
+        }
+        return false;
+    }
+
+    private int[] rotate(int[] nums, int start) {
+        int[] rotated = new int[nums.length];
+        int j = 0;
+        for (int i = start; i < nums.length; i++, j++) {
+            rotated[j] = nums[i];
+        }
+        for (int i = 0; i < start; i++) {
+            rotated[j++] = nums[i];
+        }
+        return rotated;
+    }
+}

src/main/java/g1701_1800/s1752_check_if_array_is_sorted_and_rotated/readme.md

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+1752\. Check if Array Is Sorted and Rotated
+
+Easy
+
+Given an array `nums`, return `true` _if the array was originally sorted in non-decreasing order, then rotated **some** number of positions (including zero)_. Otherwise, return `false`.
+
+There may be **duplicates** in the original array.
+
+**Note:** An array `A` rotated by `x` positions results in an array `B` of the same length such that `A[i] == B[(i+x) % A.length]`, where `%` is the modulo operation.
+
+**Example 1:**
+
+**Input:** nums = [3,4,5,1,2]
+
+**Output:** true
+
+**Explanation:** [1,2,3,4,5] is the original sorted array. You can rotate the array by x = 3 positions to begin on the the element of value 3: [3,4,5,1,2].
+
+**Example 2:**
+
+**Input:** nums = [2,1,3,4]
+
+**Output:** false
+
+**Explanation:** There is no sorted array once rotated that can make nums.
+
+**Example 3:**
+
+**Input:** nums = [1,2,3]
+
+**Output:** true
+
+**Explanation:** [1,2,3] is the original sorted array. You can rotate the array by x = 0 positions (i.e. no rotation) to make nums.
+
+**Constraints:**
+
+*   `1 <= nums.length <= 100`
+*   `1 <= nums[i] <= 100`