Added tasks 3001-3049

Author: javadev

README.md

@@ -60,7 +60,7 @@ class Solution {
                 continue
             }
             val sum = value.poll() + value.poll()
-            maxSum = max(maxSum.toDouble(), sum.toDouble()).toInt()
+            maxSum = max(maxSum, sum)
         }
         return maxSum
     }

src/main/kotlin/g2801_2900/s2815_max_pair_sum_in_an_array/readme.md

@@ -0,0 +1,83 @@
+[![](https://img.shields.io/github/stars/javadev/LeetCode-in-Kotlin?label=Stars&style=flat-square)](https://github.com/javadev/LeetCode-in-Kotlin)
+[![](https://img.shields.io/github/forks/javadev/LeetCode-in-Kotlin?label=Fork%20me%20on%20GitHub%20&style=flat-square)](https://github.com/javadev/LeetCode-in-Kotlin/fork)
+
+## 3001\. Minimum Moves to Capture The Queen
+
+Medium
+
+There is a **1-indexed** `8 x 8` chessboard containing `3` pieces.
+
+You are given `6` integers `a`, `b`, `c`, `d`, `e`, and `f` where:
+
+*   `(a, b)` denotes the position of the white rook.
+*   `(c, d)` denotes the position of the white bishop.
+*   `(e, f)` denotes the position of the black queen.
+
+Given that you can only move the white pieces, return _the **minimum** number of moves required to capture the black queen_.
+
+**Note** that:
+
+*   Rooks can move any number of squares either vertically or horizontally, but cannot jump over other pieces.
+*   Bishops can move any number of squares diagonally, but cannot jump over other pieces.
+*   A rook or a bishop can capture the queen if it is located in a square that they can move to.
+*   The queen does not move.
+
+**Example 1:**
+
+![](https://assets.leetcode.com/uploads/2023/12/21/ex1.png)
+
+**Input:** a = 1, b = 1, c = 8, d = 8, e = 2, f = 3
+
+**Output:** 2
+
+**Explanation:** We can capture the black queen in two moves by moving the white rook to (1, 3) then to (2, 3).
+
+It is impossible to capture the black queen in less than two moves since it is not being attacked by any of the pieces at the beginning. 
+
+**Example 2:**
+
+![](https://assets.leetcode.com/uploads/2023/12/21/ex2.png)
+
+**Input:** a = 5, b = 3, c = 3, d = 4, e = 5, f = 2
+
+**Output:** 1
+
+**Explanation:** We can capture the black queen in a single move by doing one of the following:
+
+- Move the white rook to (5, 2).
+
+- Move the white bishop to (5, 2). 
+
+**Constraints:**
+
+*   `1 <= a, b, c, d, e, f <= 8`
+*   No two pieces are on the same square.
+
+## Solution
+
+```kotlin
+import kotlin.math.abs
+
+class Solution {
+    fun minMovesToCaptureTheQueen(a: Int, b: Int, c: Int, d: Int, e: Int, f: Int): Int {
+        if (a == e || b == f) {
+            if (a == c && (d > b && d < f || d > f && d < b)) {
+                return 2
+            }
+            if (b == d && (c > a && c < e || c > e && c < a)) {
+                return 2
+            }
+            return 1
+        } else if (abs(c - e) == abs(d - f)) {
+            if (abs(a - c) == abs(b - d) &&
+                abs(e - a) == abs(f - b) &&
+                (a > e && a < c || a > c && a < e)
+            ) {
+                return 2
+            }
+            return 1
+        }
+        return 2
+    }
+}
+```

src/main/kotlin/g3001_3100/s3001_minimum_moves_to_capture_the_queen/readme.md

@@ -0,0 +1,86 @@
+[![](https://img.shields.io/github/stars/javadev/LeetCode-in-Kotlin?label=Stars&style=flat-square)](https://github.com/javadev/LeetCode-in-Kotlin)
+[![](https://img.shields.io/github/forks/javadev/LeetCode-in-Kotlin?label=Fork%20me%20on%20GitHub%20&style=flat-square)](https://github.com/javadev/LeetCode-in-Kotlin/fork)
+
+## 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>
+
+## Solution
+
+```kotlin
+import kotlin.math.min
+
+class Solution {
+    fun maximumSetSize(nums1: IntArray, nums2: IntArray): Int {
+        val uniq1 = HashSet<Int>()
+        val uniq2 = HashSet<Int>()
+        for (i in nums1.indices) {
+            uniq1.add(nums1[i])
+            uniq2.add(nums2[i])
+        }
+        var common = 0
+        if (uniq1.size <= uniq2.size) {
+            for (u in uniq1) {
+                if (uniq2.contains(u)) {
+                    common++
+                }
+            }
+        } else {
+            for (u in uniq2) {
+                if (uniq1.contains(u)) {
+                    common++
+                }
+            }
+        }
+        val half = nums1.size / 2
+        val from1 = min(uniq1.size - common, half)
+        val from2 = min(uniq2.size - common, half)
+        val takeFromCommon1 = half - from1
+        val takeFromCommon2 = half - from2
+        return from1 + from2 + min(takeFromCommon1 + takeFromCommon2, common)
+    }
+}
+```

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

@@ -0,0 +1,161 @@
+[![](https://img.shields.io/github/stars/javadev/LeetCode-in-Kotlin?label=Stars&style=flat-square)](https://github.com/javadev/LeetCode-in-Kotlin)
+[![](https://img.shields.io/github/forks/javadev/LeetCode-in-Kotlin?label=Fork%20me%20on%20GitHub%20&style=flat-square)](https://github.com/javadev/LeetCode-in-Kotlin/fork)
+
+## 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`
+
+## Solution
+
+```kotlin
+import kotlin.math.max
+
+class Solution {
+    fun maxPartitionsAfterOperations(s: String, k: Int): Int {
+        if (k == ALPHABET_SIZE) {
+            return 1
+        }
+        val n = s.length
+        val ansr = IntArray(n)
+        val usedr = IntArray(n)
+        var used = 0
+        var cntUsed = 0
+        var ans = 1
+        for (i in n - 1 downTo 0) {
+            val ch = s[i].code - 'a'.code
+            if ((used and (1 shl ch)) == 0) {
+                if (cntUsed == k) {
+                    cntUsed = 0
+                    used = 0
+                    ans++
+                }
+                used = used or (1 shl ch)
+                cntUsed++
+            }
+            ansr[i] = ans
+            usedr[i] = used
+        }
+        var ansl = 0
+        ans = ansr[0]
+        var l = 0
+        while (l < n) {
+            used = 0
+            cntUsed = 0
+            var usedBeforeLast = 0
+            var usedTwiceBeforeLast = 0
+            var last = -1
+            var r = l
+            while (r < n) {
+                val ch = s[r].code - 'a'.code
+                if ((used and (1 shl ch)) == 0) {
+                    if (cntUsed == k) {
+                        break
+                    }
+                    usedBeforeLast = used
+                    last = r
+                    used = used or (1 shl ch)
+                    cntUsed++
+                } else if (cntUsed < k) {
+                    usedTwiceBeforeLast = usedTwiceBeforeLast or (1 shl ch)
+                }
+                r++
+            }
+            if (cntUsed == k) {
+                if (last - l > Integer.bitCount(usedBeforeLast)) {
+                    ans = max(ans, (ansl + 1 + ansr[last]))
+                }
+                if (last + 1 < r) {
+                    if (last + 2 >= n) {
+                        ans = max(ans, (ansl + 1 + 1))
+                    } else {
+                        if (Integer.bitCount(usedr[last + 2]) == k) {
+                            val canUse = ((1 shl ALPHABET_SIZE) - 1) and used.inv() and usedr[last + 2].inv()
+                            ans = if (canUse > 0) {
+                                max(ans, (ansl + 1 + 1 + ansr[last + 2]))
+                            } else {
+                                max(ans, (ansl + 1 + ansr[last + 2]))
+                            }
+                            val l1 = s[last + 1].code - 'a'.code
+                            if ((usedTwiceBeforeLast and (1 shl l1)) == 0) {
+                                ans = max(ans, (ansl + 1 + ansr[last + 1]))
+                            }
+                        } else {
+                            ans = max(ans, (ansl + 1 + ansr[last + 2]))
+                        }
+                    }
+                }
+            }
+            l = r
+            ansl++
+        }
+        return ans
+    }
+
+    companion object {
+        private const val ALPHABET_SIZE = 'z'.code - 'a'.code + 1
+    }
+}
+```