Added tasks 3477-3480

Author: javadev

README.md

@@ -2088,6 +2088,10 @@
 
 | #    |      Title     | Difficulty  | Tag         | Time, ms | Time, %
 |------|----------------|-------------|-------------|----------|--------
+| 3480 |[Maximize Subarrays After Removing One Conflicting Pair](src/main/kotlin/g3401_3500/s3480_maximize_subarrays_after_removing_one_conflicting_pair)| Hard | Array, Prefix_Sum, Enumeration, Segment_Tree | 48 | 100.00
+| 3479 |[Fruits Into Baskets III](src/main/kotlin/g3401_3500/s3479_fruits_into_baskets_iii)| Medium | Array, Binary_Search, Ordered_Set, Segment_Tree | 53 | 92.86
+| 3478 |[Choose K Elements With Maximum Sum](src/main/kotlin/g3401_3500/s3478_choose_k_elements_with_maximum_sum)| Medium | Array, Sorting, Heap_Priority_Queue | 151 | 100.00
+| 3477 |[Fruits Into Baskets II](src/main/kotlin/g3401_3500/s3477_fruits_into_baskets_ii)| Easy | Array, Binary_Search, Simulation, Segment_Tree | 3 | 100.00
 | 3475 |[DNA Pattern Recognition](src/main/kotlin/g3401_3500/s3475_dna_pattern_recognition)| Medium | Database | 362 | 83.49
 | 3474 |[Lexicographically Smallest Generated String](src/main/kotlin/g3401_3500/s3474_lexicographically_smallest_generated_string)| Hard | String, Greedy, String_Matching | 30 | 100.00
 | 3473 |[Sum of K Subarrays With Length at Least M](src/main/kotlin/g3401_3500/s3473_sum_of_k_subarrays_with_length_at_least_m)| Medium | Array, Dynamic_Programming, Prefix_Sum | 227 | 24.47

src/main/kotlin/g3401_3500/s3477_fruits_into_baskets_ii/readme.md

@@ -0,0 +1,73 @@
+[![](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)
+
+## 3477\. Fruits Into Baskets II
+
+Easy
+
+You are given two arrays of integers, `fruits` and `baskets`, each of length `n`, where `fruits[i]` represents the **quantity** of the <code>i<sup>th</sup></code> type of fruit, and `baskets[j]` represents the **capacity** of the <code>j<sup>th</sup></code> basket.
+
+From left to right, place the fruits according to these rules:
+
+*   Each fruit type must be placed in the **leftmost available basket** with a capacity **greater than or equal** to the quantity of that fruit type.
+*   Each basket can hold **only one** type of fruit.
+*   If a fruit type **cannot be placed** in any basket, it remains **unplaced**.
+
+Return the number of fruit types that remain unplaced after all possible allocations are made.
+
+**Example 1:**
+
+**Input:** fruits = [4,2,5], baskets = [3,5,4]
+
+**Output:** 1
+
+**Explanation:**
+
+*   `fruits[0] = 4` is placed in `baskets[1] = 5`.
+*   `fruits[1] = 2` is placed in `baskets[0] = 3`.
+*   `fruits[2] = 5` cannot be placed in `baskets[2] = 4`.
+
+Since one fruit type remains unplaced, we return 1.
+
+**Example 2:**
+
+**Input:** fruits = [3,6,1], baskets = [6,4,7]
+
+**Output:** 0
+
+**Explanation:**
+
+*   `fruits[0] = 3` is placed in `baskets[0] = 6`.
+*   `fruits[1] = 6` cannot be placed in `baskets[1] = 4` (insufficient capacity) but can be placed in the next available basket, `baskets[2] = 7`.
+*   `fruits[2] = 1` is placed in `baskets[1] = 4`.
+
+Since all fruits are successfully placed, we return 0.
+
+**Constraints:**
+
+*   `n == fruits.length == baskets.length`
+*   `1 <= n <= 100`
+*   `1 <= fruits[i], baskets[i] <= 1000`
+
+## Solution
+
+```kotlin
+class Solution {
+    fun numOfUnplacedFruits(fruits: IntArray, baskets: IntArray): Int {
+        val n = fruits.size
+        var currfruits: Int
+        var count = 0
+        for (i in 0..<n) {
+            currfruits = fruits[i]
+            for (j in 0..<n) {
+                if (baskets[j] >= currfruits) {
+                    count++
+                    baskets[j] = 0
+                    break
+                }
+            }
+        }
+        return n - count
+    }
+}
+```

src/main/kotlin/g3401_3500/s3478_choose_k_elements_with_maximum_sum/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)
+
+## 3478\. Choose K Elements With Maximum Sum
+
+Medium
+
+You are given two integer arrays, `nums1` and `nums2`, both of length `n`, along with a positive integer `k`.
+
+For each index `i` from `0` to `n - 1`, perform the following:
+
+*   Find **all** indices `j` where `nums1[j]` is less than `nums1[i]`.
+*   Choose **at most** `k` values of `nums2[j]` at these indices to **maximize** the total sum.
+
+Return an array `answer` of size `n`, where `answer[i]` represents the result for the corresponding index `i`.
+
+**Example 1:**
+
+**Input:** nums1 = [4,2,1,5,3], nums2 = [10,20,30,40,50], k = 2
+
+**Output:** [80,30,0,80,50]
+
+**Explanation:**
+
+*   For `i = 0`: Select the 2 largest values from `nums2` at indices `[1, 2, 4]` where `nums1[j] < nums1[0]`, resulting in `50 + 30 = 80`.
+*   For `i = 1`: Select the 2 largest values from `nums2` at index `[2]` where `nums1[j] < nums1[1]`, resulting in 30.
+*   For `i = 2`: No indices satisfy `nums1[j] < nums1[2]`, resulting in 0.
+*   For `i = 3`: Select the 2 largest values from `nums2` at indices `[0, 1, 2, 4]` where `nums1[j] < nums1[3]`, resulting in `50 + 30 = 80`.
+*   For `i = 4`: Select the 2 largest values from `nums2` at indices `[1, 2]` where `nums1[j] < nums1[4]`, resulting in `30 + 20 = 50`.
+
+**Example 2:**
+
+**Input:** nums1 = [2,2,2,2], nums2 = [3,1,2,3], k = 1
+
+**Output:** [0,0,0,0]
+
+**Explanation:**
+
+Since all elements in `nums1` are equal, no indices satisfy the condition `nums1[j] < nums1[i]` for any `i`, resulting in 0 for all positions.
+
+**Constraints:**
+
+*   `n == nums1.length == nums2.length`
+*   <code>1 <= n <= 10<sup>5</sup></code>
+*   <code>1 <= nums1[i], nums2[i] <= 10<sup>6</sup></code>
+*   `1 <= k <= n`
+
+## Solution
+
+```kotlin
+import java.util.PriorityQueue
+
+class Solution {
+    fun findMaxSum(nums1: IntArray, nums2: IntArray, k: Int): LongArray {
+        val n = nums1.size
+        val ans = LongArray(n)
+        val ps = Array<Point>(n) { i -> Point(nums1[i], nums2[i], i) }
+        ps.sortWith { p1: Point, p2: Point -> p1.x.compareTo(p2.x) }
+        val pq = PriorityQueue<Int?>()
+        var s: Long = 0
+        var i = 0
+        while (i < n) {
+            var j = i
+            while (j < n && ps[j].x == ps[i].x) {
+                ans[ps[j].i] = s
+                j++
+            }
+            for (p in i..<j) {
+                val cur = ps[p].y
+                if (pq.size < k) {
+                    pq.offer(cur)
+                    s += cur.toLong()
+                } else if (cur > pq.peek()!!) {
+                    s -= pq.poll()!!.toLong()
+                    pq.offer(cur)
+                    s += cur.toLong()
+                }
+            }
+            i = j
+        }
+        return ans
+    }
+
+    private class Point(var x: Int, var y: Int, var i: Int)
+}
+```

src/main/kotlin/g3401_3500/s3479_fruits_into_baskets_iii/readme.md

@@ -0,0 +1,103 @@
+[![](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)
+
+## 3479\. Fruits Into Baskets III
+
+Medium
+
+You are given two arrays of integers, `fruits` and `baskets`, each of length `n`, where `fruits[i]` represents the **quantity** of the <code>i<sup>th</sup></code> type of fruit, and `baskets[j]` represents the **capacity** of the <code>j<sup>th</sup></code> basket.
+
+From left to right, place the fruits according to these rules:
+
+*   Each fruit type must be placed in the **leftmost available basket** with a capacity **greater than or equal** to the quantity of that fruit type.
+*   Each basket can hold **only one** type of fruit.
+*   If a fruit type **cannot be placed** in any basket, it remains **unplaced**.
+
+Return the number of fruit types that remain unplaced after all possible allocations are made.
+
+**Example 1:**
+
+**Input:** fruits = [4,2,5], baskets = [3,5,4]
+
+**Output:** 1
+
+**Explanation:**
+
+*   `fruits[0] = 4` is placed in `baskets[1] = 5`.
+*   `fruits[1] = 2` is placed in `baskets[0] = 3`.
+*   `fruits[2] = 5` cannot be placed in `baskets[2] = 4`.
+
+Since one fruit type remains unplaced, we return 1.
+
+**Example 2:**
+
+**Input:** fruits = [3,6,1], baskets = [6,4,7]
+
+**Output:** 0
+
+**Explanation:**
+
+*   `fruits[0] = 3` is placed in `baskets[0] = 6`.
+*   `fruits[1] = 6` cannot be placed in `baskets[1] = 4` (insufficient capacity) but can be placed in the next available basket, `baskets[2] = 7`.
+*   `fruits[2] = 1` is placed in `baskets[1] = 4`.
+
+Since all fruits are successfully placed, we return 0.
+
+**Constraints:**
+
+*   `n == fruits.length == baskets.length`
+*   <code>1 <= n <= 10<sup>5</sup></code>
+*   <code>1 <= fruits[i], baskets[i] <= 10<sup>9</sup></code>
+
+## Solution
+
+```kotlin
+import kotlin.math.max
+
+class Solution {
+    fun numOfUnplacedFruits(fruits: IntArray, baskets: IntArray): Int {
+        val n = baskets.size
+        var size = 1
+        while (size < n) {
+            size = size shl 1
+        }
+        val seg = IntArray(2 * size)
+        for (i in 0..<n) {
+            seg[size + i] = baskets[i]
+        }
+        for (i in n..<size) {
+            seg[size + i] = 0
+        }
+        for (i in size - 1 downTo 1) {
+            seg[i] = max(seg[i shl 1].toDouble(), seg[i shl 1 or 1].toDouble()).toInt()
+        }
+        var ans = 0
+        for (f in fruits) {
+            if (seg[1] < f) {
+                ans++
+                continue
+            }
+            var idx = 1
+            while (idx < size) {
+                if (seg[idx shl 1] >= f) {
+                    idx = idx shl 1
+                } else {
+                    idx = idx shl 1 or 1
+                }
+            }
+            update(seg, idx - size, 0, size)
+        }
+        return ans
+    }
+
+    private fun update(seg: IntArray, pos: Int, `val`: Int, size: Int) {
+        var i = pos + size
+        seg[i] = `val`
+        i /= 2
+        while (i > 0) {
+            seg[i] = max(seg[i shl 1].toDouble(), seg[i shl 1 or 1].toDouble()).toInt()
+            i /= 2
+        }
+    }
+}
+```