Added tasks 3020-3046

Author: javadev

src/main/kotlin/g2801_2900/s2815_max_pair_sum_in_an_array/Solution.kt

@@ -23,7 +23,7 @@ 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/g3001_3100/s3020_find_the_maximum_number_of_elements_in_subset/Solution.kt

@@ -0,0 +1,47 @@
+package g3001_3100.s3020_find_the_maximum_number_of_elements_in_subset
+
+// #Medium #Array #Hash_Table #Enumeration #2024_03_03_Time_626_ms_(82.22%)_Space_57.8_MB_(80.00%)
+
+import kotlin.math.max
+
+class Solution {
+    fun maximumLength(nums: IntArray): Int {
+        return withHashMap(nums)
+    }
+
+    private fun withHashMap(nums: IntArray): Int {
+        val map: MutableMap<Int, Int> = HashMap()
+        for (i in nums) {
+            map[i] = map.getOrDefault(i, 0) + 1
+        }
+        var ans = 0
+        if (map.containsKey(1)) {
+            ans = if (map[1]!! % 2 == 0) {
+                map[1]!! - 1
+            } else {
+                map[1]!!
+            }
+        }
+        for (key in map.keys) {
+            if (key == 1) {
+                continue
+            }
+            val len = findSeries(map, key)
+            ans = max(ans, len)
+        }
+        return ans
+    }
+
+    private fun findSeries(map: Map<Int, Int>, key: Int): Int {
+        val sqr = key * key
+        return if (map.containsKey(sqr)) {
+            if (map[key]!! >= 2) {
+                2 + findSeries(map, sqr)
+            } else {
+                1
+            }
+        } else {
+            1
+        }
+    }
+}

src/main/kotlin/g3001_3100/s3020_find_the_maximum_number_of_elements_in_subset/readme.md

@@ -0,0 +1,32 @@
+3020\. Find the Maximum Number of Elements in Subset
+
+Medium
+
+You are given an array of **positive** integers `nums`.
+
+You need to select a subset of `nums` which satisfies the following condition:
+
+*   You can place the selected elements in a **0-indexed** array such that it follows the pattern: <code>[x, x<sup>2</sup>, x<sup>4</sup>, ..., x<sup>k/2</sup>, x<sup>k</sup>, x<sup>k/2</sup>, ..., x<sup>4</sup>, x<sup>2</sup>, x]</code> (**Note** that `k` can be be any **non-negative** power of `2`). For example, `[2, 4, 16, 4, 2]` and `[3, 9, 3]` follow the pattern while `[2, 4, 8, 4, 2]` does not.
+
+Return _the **maximum** number of elements in a subset that satisfies these conditions._
+
+**Example 1:**
+
+**Input:** nums = [5,4,1,2,2]
+
+**Output:** 3
+
+**Explanation:** We can select the subset {4,2,2}, which can be placed in the array as [2,4,2] which follows the pattern and 2<sup>2</sup> == 4. Hence the answer is 3.
+
+**Example 2:**
+
+**Input:** nums = [1,3,2,4]
+
+**Output:** 1
+
+**Explanation:** We can select the subset {1}, which can be placed in the array as [1] which follows the pattern. Hence the answer is 1. Note that we could have also selected the subsets {2}, {3}, or {4}, there may be multiple subsets which provide the same answer.
+
+**Constraints:**
+
+*   <code>2 <= nums.length <= 10<sup>5</sup></code>
+*   <code>1 <= nums[i] <= 10<sup>9</sup></code>

src/main/kotlin/g3001_3100/s3021_alice_and_bob_playing_flower_game/Solution.kt

@@ -0,0 +1,13 @@
+package g3001_3100.s3021_alice_and_bob_playing_flower_game
+
+// #Medium #Math #2024_03_03_Time_141_ms_(43.24%)_Space_33.2_MB_(86.49%)
+
+class Solution {
+    fun flowerGame(n: Int, m: Int): Long {
+        val nEven = n.toLong() / 2
+        val nOdd = n - nEven
+        val mEven = m.toLong() / 2
+        val mOdd = m - mEven
+        return nEven * mOdd + nOdd * mEven
+    }
+}

src/main/kotlin/g3001_3100/s3021_alice_and_bob_playing_flower_game/readme.md

@@ -0,0 +1,39 @@
+3021\. Alice and Bob Playing Flower Game
+
+Medium
+
+Alice and Bob are playing a turn-based game on a circular field surrounded by flowers. The circle represents the field, and there are `x` flowers in the clockwise direction between Alice and Bob, and `y` flowers in the anti-clockwise direction between them.
+
+The game proceeds as follows:
+
+1.  Alice takes the first turn.
+2.  In each turn, a player must choose either the clockwise or anti-clockwise direction and pick one flower from that side.
+3.  At the end of the turn, if there are no flowers left at all, the **current** player captures their opponent and wins the game.
+
+Given two integers, `n` and `m`, the task is to compute the number of possible pairs `(x, y)` that satisfy the conditions:
+
+*   Alice must win the game according to the described rules.
+*   The number of flowers `x` in the clockwise direction must be in the range `[1,n]`.
+*   The number of flowers `y` in the anti-clockwise direction must be in the range `[1,m]`.
+
+Return _the number of possible pairs_ `(x, y)` _that satisfy the conditions mentioned in the statement_.
+
+**Example 1:**
+
+**Input:** n = 3, m = 2
+
+**Output:** 3
+
+**Explanation:** The following pairs satisfy conditions described in the statement: (1,2), (3,2), (2,1).
+
+**Example 2:**
+
+**Input:** n = 1, m = 1
+
+**Output:** 0
+
+**Explanation:** No pairs satisfy the conditions described in the statement.
+
+**Constraints:**
+
+*   <code>1 <= n, m <= 10<sup>5</sup></code>

src/main/kotlin/g3001_3100/s3022_minimize_or_of_remaining_elements_using_operations/Solution.kt

@@ -0,0 +1,27 @@
+package g3001_3100.s3022_minimize_or_of_remaining_elements_using_operations
+
+// #Hard #Array #Greedy #Bit_Manipulation #2024_03_03_Time_516_ms_(77.78%)_Space_62.7_MB_(100.00%)
+
+class Solution {
+    fun minOrAfterOperations(nums: IntArray, k: Int): Int {
+        var ans = 0
+        var mask = 0
+        for (j in 30 downTo 0) {
+            mask = mask or (1 shl j)
+            var consecutiveAnd = mask
+            var mergeCount = 0
+            for (i in nums) {
+                consecutiveAnd = consecutiveAnd and i
+                if ((consecutiveAnd or ans) != ans) {
+                    mergeCount++
+                } else {
+                    consecutiveAnd = mask
+                }
+            }
+            if (mergeCount > k) {
+                ans = ans or (1 shl j)
+            }
+        }
+        return ans
+    }
+}

src/main/kotlin/g3001_3100/s3022_minimize_or_of_remaining_elements_using_operations/readme.md

@@ -0,0 +1,53 @@
+3022\. Minimize OR of Remaining Elements Using Operations
+
+Hard
+
+You are given a **0-indexed** integer array `nums` and an integer `k`.
+
+In one operation, you can pick any index `i` of `nums` such that `0 <= i < nums.length - 1` and replace `nums[i]` and `nums[i + 1]` with a single occurrence of `nums[i] & nums[i + 1]`, where `&` represents the bitwise `AND` operator.
+
+Return _the **minimum** possible value of the bitwise_ `OR` _of the remaining elements of_ `nums` _after applying **at most**_ `k` _operations_.
+
+**Example 1:**
+
+**Input:** nums = [3,5,3,2,7], k = 2
+
+**Output:** 3
+
+**Explanation:** Let's do the following operations: 
+1. Replace nums[0] and nums[1] with (nums[0] & nums[1]) so that nums becomes equal to [1,3,2,7]. 
+2. Replace nums[2] and nums[3] with (nums[2] & nums[3]) so that nums becomes equal to [1,3,2]. 
+
+The bitwise-or of the final array is 3. 
+
+It can be shown that 3 is the minimum possible value of the bitwise OR of the remaining elements of nums after applying at most k operations.
+
+**Example 2:**
+
+**Input:** nums = [7,3,15,14,2,8], k = 4
+
+**Output:** 2
+
+**Explanation:** Let's do the following operations: 
+1. Replace nums[0] and nums[1] with (nums[0] & nums[1]) so that nums becomes equal to [3,15,14,2,8]. 
+2. Replace nums[0] and nums[1] with (nums[0] & nums[1]) so that nums becomes equal to [3,14,2,8]. 
+3. Replace nums[0] and nums[1] with (nums[0] & nums[1]) so that nums becomes equal to [2,2,8]. 
+4. Replace nums[1] and nums[2] with (nums[1] & nums[2]) so that nums becomes equal to [2,0]. 
+
+The bitwise-or of the final array is 2. 
+
+It can be shown that 2 is the minimum possible value of the bitwise OR of the remaining elements of nums after applying at most k operations.
+
+**Example 3:**
+
+**Input:** nums = [10,7,10,3,9,14,9,4], k = 1
+
+**Output:** 15
+
+**Explanation:** Without applying any operations, the bitwise-or of nums is 15. It can be shown that 15 is the minimum possible value of the bitwise OR of the remaining elements of nums after applying at most k operations.
+
+**Constraints:**
+
+*   <code>1 <= nums.length <= 10<sup>5</sup></code>
+*   <code>0 <= nums[i] < 2<sup>30</sup></code>
+*   `0 <= k < nums.length`

src/main/kotlin/g3001_3100/s3024_type_of_triangle/Solution.kt

@@ -0,0 +1,18 @@
+package g3001_3100.s3024_type_of_triangle
+
+// #Easy #Array #Math #Sorting #2024_03_03_Time_163_ms_(81.03%)_Space_34.7_MB_(93.10%)
+
+class Solution {
+    fun triangleType(nums: IntArray): String {
+        if (nums[0] + nums[1] > nums[2] && nums[1] + nums[2] > nums[0] && nums[2] + nums[0] > nums[1]) {
+            return if (nums[0] == nums[1] && nums[1] == nums[2]) {
+                "equilateral"
+            } else if (nums[0] == nums[1] || nums[0] == nums[2] || nums[2] == nums[1]) {
+                "isosceles"
+            } else {
+                "scalene"
+            }
+        }
+        return "none"
+    }
+}

src/main/kotlin/g3001_3100/s3024_type_of_triangle/readme.md

@@ -0,0 +1,42 @@
+3024\. Type of Triangle
+
+Easy
+
+You are given a **0-indexed** integer array `nums` of size `3` which can form the sides of a triangle.
+
+*   A triangle is called **equilateral** if it has all sides of equal length.
+*   A triangle is called **isosceles** if it has exactly two sides of equal length.
+*   A triangle is called **scalene** if all its sides are of different lengths.
+
+Return _a string representing_ _the type of triangle that can be formed_ _or_ `"none"` _if it **cannot** form a triangle._
+
+**Example 1:**
+
+**Input:** nums = [3,3,3]
+
+**Output:** "equilateral"
+
+**Explanation:** Since all the sides are of equal length, therefore, it will form an equilateral triangle.
+
+**Example 2:**
+
+**Input:** nums = [3,4,5]
+
+**Output:** "scalene"
+
+**Explanation:** 
+
+nums[0] + nums[1] = 3 + 4 = 7, which is greater than nums[2] = 5. 
+
+nums[0] + nums[2] = 3 + 5 = 8, which is greater than nums[1] = 4. 
+
+nums[1] + nums[2] = 4 + 5 = 9, which is greater than nums[0] = 3. 
+
+Since the sum of the two sides is greater than the third side for all three cases, therefore, it can form a triangle. 
+
+As all the sides are of different lengths, it will form a scalene triangle.
+
+**Constraints:**
+
+*   `nums.length == 3`
+*   `1 <= nums[i] <= 100`

src/main/kotlin/g3001_3100/s3025_find_the_number_of_ways_to_place_people_i/Solution.kt

@@ -0,0 +1,26 @@
+package g3001_3100.s3025_find_the_number_of_ways_to_place_people_i
+
+// #Medium #Array #Math #Sorting #Enumeration #Geometry
+// #2024_03_03_Time_252_ms_(44.12%)_Space_44.1_MB_(73.53%)
+
+class Solution {
+    fun numberOfPairs(points: Array<IntArray>): Int {
+        points.sortWith { a: IntArray, b: IntArray -> if (a[0] == b[0]) b[1] - a[1] else a[0] - b[0] }
+        var cnt = 0
+        for (i in points.indices) {
+            var bot = Int.MIN_VALUE
+            var top = points[i][1]
+            for (j in i + 1 until points.size) {
+                val y1 = points[j][1]
+                if (y1 <= top && y1 > bot) {
+                    cnt++
+                    bot = y1
+                    if (y1 == top) {
+                        top--
+                    }
+                }
+            }
+        }
+        return cnt
+    }
+}