Added tasks 3002-3007

Author: ThanhNIT

src/main/java/g3001_3100/s3002_maximum_size_of_a_set_after_removals/Solution.java

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+package g3001_3100.s3002_maximum_size_of_a_set_after_removals;
+
+// #Medium #Array #Hash_Table #Greedy #2024_02_26_Time_26_ms_(94.01%)_Space_53.3_MB_(80.90%)
+
+import java.util.HashSet;
+
+public class Solution {
+    public int maximumSetSize(int[] nums1, int[] nums2) {
+        HashSet<Integer> uniq1 = new HashSet<>();
+        HashSet<Integer> uniq2 = new HashSet<>();
+        for (int i = 0; i < nums1.length; i++) {
+            uniq1.add(nums1[i]);
+            uniq2.add(nums2[i]);
+        }
+        int common = 0;
+        if (uniq1.size() <= uniq2.size()) {
+            for (int u : uniq1) {
+                if (uniq2.contains(u)) {
+                    common++;
+                }
+            }
+        } else {
+            for (int u : uniq2) {
+                if (uniq1.contains(u)) {
+                    common++;
+                }
+            }
+        }
+        int half = nums1.length / 2;
+        int from1 = Math.min(uniq1.size() - common, half);
+        int from2 = Math.min(uniq2.size() - common, half);
+        int takeFromCommon1 = half - from1;
+        int takeFromCommon2 = half - from2;
+        return from1 + from2 + Math.min(takeFromCommon1 + takeFromCommon2, common);
+    }
+}

src/main/java/g3001_3100/s3002_maximum_size_of_a_set_after_removals/readme.md

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+3002\. Maximum Size of a Set After Removals
+
+Medium
+
+You are given two **0-indexed** integer arrays `nums1` and `nums2` of even length `n`.
+
+You must remove `n / 2` elements from `nums1` and `n / 2` elements from `nums2`. After the removals, you insert the remaining elements of `nums1` and `nums2` into a set `s`.
+
+Return _the **maximum** possible size of the set_ `s`.
+
+**Example 1:**
+
+**Input:** nums1 = [1,2,1,2], nums2 = [1,1,1,1]
+
+**Output:** 2
+
+**Explanation:** We remove two occurences of 1 from nums1 and nums2. After the removals, the arrays become equal to nums1 = [2,2] and nums2 = [1,1]. Therefore, s = {1,2}. It can be shown that 2 is the maximum possible size of the set s after the removals.
+
+**Example 2:**
+
+**Input:** nums1 = [1,2,3,4,5,6], nums2 = [2,3,2,3,2,3]
+
+**Output:** 5
+
+**Explanation:** We remove 2, 3, and 6 from nums1, as well as 2 and two occurrences of 3 from nums2. After the removals, the arrays become equal to nums1 = [1,4,5] and nums2 = [2,3,2]. Therefore, s = {1,2,3,4,5}. It can be shown that 5 is the maximum possible size of the set s after the removals.
+
+**Example 3:**
+
+**Input:** nums1 = [1,1,2,2,3,3], nums2 = [4,4,5,5,6,6]
+
+**Output:** 6
+
+**Explanation:** We remove 1, 2, and 3 from nums1, as well as 4, 5, and 6 from nums2. After the removals, the arrays become equal to nums1 = [1,2,3] and nums2 = [4,5,6]. Therefore, s = {1,2,3,4,5,6}. It can be shown that 6 is the maximum possible size of the set s after the removals.
+
+**Constraints:**
+
+*   `n == nums1.length == nums2.length`
+*   <code>1 <= n <= 2 * 10<sup>4</sup></code>
+*   `n` is even.
+*   <code>1 <= nums1[i], nums2[i] <= 10<sup>9</sup></code>

src/main/java/g3001_3100/s3003_maximize_the_number_of_partitions_after_operations/Solution.java

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+package g3001_3100.s3003_maximize_the_number_of_partitions_after_operations;
+
+// #Hard #String #Dynamic_Programming #Bit_Manipulation #Bitmask
+// #2024_02_26_Time_1_ms_(100.00%)_Space_42.1_MB_(99.44%)
+
+public class Solution {
+    private static final int ALPHABET_SIZE = 'z' - 'a' + 1;
+
+    public int maxPartitionsAfterOperations(String s, int k) {
+        if (k == ALPHABET_SIZE) {
+            return 1;
+        }
+        int n = s.length();
+        int[] ansr = new int[n];
+        int[] usedr = new int[n];
+        int used = 0;
+        int cntUsed = 0;
+        int ans = 1;
+        for (int i = n - 1; i >= 0; --i) {
+            int ch = s.charAt(i) - 'a';
+            if ((used & (1 << ch)) == 0) {
+                if (cntUsed == k) {
+                    cntUsed = 0;
+                    used = 0;
+                    ans++;
+                }
+                used |= (1 << ch);
+                cntUsed++;
+            }
+            ansr[i] = ans;
+            usedr[i] = used;
+        }
+
+        int ansl = 0;
+        ans = ansr[0];
+        for (int l = 0; l < n; ) {
+            used = 0;
+            cntUsed = 0;
+            int usedBeforeLast = 0;
+            int usedTwiceBeforeLast = 0;
+            int last = -1;
+            int r = l;
+            while (r < n) {
+                int ch = s.charAt(r) - 'a';
+                if ((used & (1 << ch)) == 0) {
+                    if (cntUsed == k) {
+                        break;
+                    }
+                    usedBeforeLast = used;
+                    last = r;
+                    used |= (1 << ch);
+                    cntUsed++;
+                } else if (cntUsed < k) {
+                    usedTwiceBeforeLast |= (1 << ch);
+                }
+                r++;
+            }
+
+            if (cntUsed == k) {
+                if (last - l > Integer.bitCount(usedBeforeLast)) {
+                    ans = Math.max(ans, ansl + 1 + ansr[last]);
+                }
+                if (last + 1 < r) {
+                    if (last + 2 >= n) {
+                        ans = Math.max(ans, ansl + 1 + 1);
+                    } else {
+                        if (Integer.bitCount(usedr[last + 2]) == k) {
+                            int canUse = ((1 << ALPHABET_SIZE) - 1) & ~used & ~usedr[last + 2];
+                            if (canUse > 0) {
+                                ans = Math.max(ans, ansl + 1 + 1 + ansr[last + 2]);
+                            } else {
+                                ans = Math.max(ans, ansl + 1 + ansr[last + 2]);
+                            }
+                            int l1 = s.charAt(last + 1) - 'a';
+                            if ((usedTwiceBeforeLast & (1 << l1)) == 0) {
+                                ans = Math.max(ans, ansl + 1 + ansr[last + 1]);
+                            }
+                        } else {
+                            ans = Math.max(ans, ansl + 1 + ansr[last + 2]);
+                        }
+                    }
+                }
+            }
+            l = r;
+            ansl++;
+        }
+
+        return ans;
+    }
+}

src/main/java/g3001_3100/s3003_maximize_the_number_of_partitions_after_operations/readme.md

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+3003\. Maximize the Number of Partitions After Operations
+
+Hard
+
+You are given a **0-indexed** string `s` and an integer `k`.
+
+You are to perform the following partitioning operations until `s` is **empty**:
+
+*   Choose the **longest** **prefix** of `s` containing at most `k` **distinct** characters.
+*   **Delete** the prefix from `s` and increase the number of partitions by one. The remaining characters (if any) in `s` maintain their initial order.
+
+**Before** the operations, you are allowed to change **at most** **one** index in `s` to another lowercase English letter.
+
+Return _an integer denoting the **maximum** number of resulting partitions after the operations by optimally choosing at most one index to change._
+
+**Example 1:**
+
+**Input:** s = "accca", k = 2
+
+**Output:** 3
+
+**Explanation:** In this example, to maximize the number of resulting partitions, s[2] can be changed to 'b'. s becomes "acbca". The operations can now be performed as follows until s becomes empty: 
+- Choose the longest prefix containing at most 2 distinct characters, "<ins>ac</ins>bca". 
+- Delete the prefix, and s becomes "bca". The number of partitions is now 1.
+- Choose the longest prefix containing at most 2 distinct characters, "<ins>bc</ins>a". 
+- Delete the prefix, and s becomes "a". The number of partitions is now 2. 
+- Choose the longest prefix containing at most 2 distinct characters, "<ins>a</ins>". 
+- Delete the prefix, and s becomes empty. The number of partitions is now 3. 
+
+Hence, the answer is 3. It can be shown that it is not possible to obtain more than 3 partitions.
+
+**Example 2:**
+
+**Input:** s = "aabaab", k = 3
+
+**Output:** 1
+
+**Explanation:** In this example, to maximize the number of resulting partitions we can leave s as it is. The operations can now be performed as follows until s becomes empty: 
+- Choose the longest prefix containing at most 3 distinct characters, "<ins>aabaab</ins>". 
+- Delete the prefix, and s becomes empty. The number of partitions becomes 1.
+
+Hence, the answer is 1. It can be shown that it is not possible to obtain more than 1 partition.
+
+**Example 3:**
+
+**Input:** s = "xxyz", k = 1
+
+**Output:** 4
+
+**Explanation:** In this example, to maximize the number of resulting partitions, s[1] can be changed to 'a'. s becomes "xayz". The operations can now be performed as follows until s becomes empty: 
+- Choose the longest prefix containing at most 1 distinct character, "<ins>x</ins>ayz". 
+- Delete the prefix, and s becomes "ayz". The number of partitions is now 1. 
+- Choose the longest prefix containing at most 1 distinct character, "<ins>a</ins>yz". 
+- Delete the prefix, and s becomes "yz". The number of partitions is now 2. 
+- Choose the longest prefix containing at most 1 distinct character, "<ins>y</ins>z". 
+- Delete the prefix, and s becomes "z". The number of partitions is now 3. 
+- Choose the longest prefix containing at most 1 distinct character, "<ins>z</ins>". 
+- Delete the prefix, and s becomes empty. The number of partitions is now 4. 
+
+Hence, the answer is 4. It can be shown that it is not possible to obtain more than 4 partitions.
+
+**Constraints:**
+
+*   <code>1 <= s.length <= 10<sup>4</sup></code>
+*   `s` consists only of lowercase English letters.
+*   `1 <= k <= 26`

src/main/java/g3001_3100/s3005_count_elements_with_maximum_frequency/Solution.java

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+package g3001_3100.s3005_count_elements_with_maximum_frequency;
+
+// #Easy #Array #Hash_Table #Counting #2024_02_26_Time_1_ms_(99.76%)_Space_41.6_MB_(98.97%)
+
+import java.util.ArrayList;
+import java.util.List;
+
+public class Solution {
+    public int maxFrequencyElements(int[] nums) {
+        if (nums.length == 1) {
+            return 1;
+        }
+        List<Integer> list = new ArrayList<>();
+        int co = 0;
+        int prev = 0;
+        for (int num : nums) {
+            if (list.contains(num)) {
+                continue;
+            }
+            list.add(num);
+            if (list.size() == nums.length) {
+                break;
+            }
+            int c = 0;
+            for (int i : nums) {
+                if (num == i) {
+                    c++;
+                }
+            }
+            if (c > 1) {
+                if (c > prev) {
+                    co = c;
+                    prev = c;
+                } else if (c == prev) {
+                    co = c + co;
+                }
+            }
+        }
+        if (co == 0) {
+            return nums.length;
+        }
+        return co;
+    }
+}

src/main/java/g3001_3100/s3005_count_elements_with_maximum_frequency/readme.md

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+3005\. Count Elements With Maximum Frequency
+
+Easy
+
+You are given an array `nums` consisting of **positive** integers.
+
+Return _the **total frequencies** of elements in_ `nums` _such that those elements all have the **maximum** frequency_.
+
+The **frequency** of an element is the number of occurrences of that element in the array.
+
+**Example 1:**
+
+**Input:** nums = [1,2,2,3,1,4]
+
+**Output:** 4
+
+**Explanation:** The elements 1 and 2 have a frequency of 2 which is the maximum frequency in the array. So the number of elements in the array with maximum frequency is 4.
+
+**Example 2:**
+
+**Input:** nums = [1,2,3,4,5]
+
+**Output:** 5
+
+**Explanation:** All elements of the array have a frequency of 1 which is the maximum. So the number of elements in the array with maximum frequency is 5.
+
+**Constraints:**
+
+*   `1 <= nums.length <= 100`
+*   `1 <= nums[i] <= 100`