Added tasks 3477-3480

Author: javadev

src/main/kotlin/g3401_3500/s3477_fruits_into_baskets_ii/Solution.kt

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+package g3401_3500.s3477_fruits_into_baskets_ii
+
+// #Easy #2025_03_09_Time_6_ms_(100.00%)_Space_47.78_MB_(100.00%)
+
+class Solution {
+    fun numOfUnplacedFruits(fruits: IntArray, baskets: IntArray): Int {
+        val n = fruits.size
+        var ct = 0
+        val used = BooleanArray(n)
+        for (fruit in fruits) {
+            var flag = true
+            for (i in 0..<n) {
+                if (fruit <= baskets[i] && !used[i]) {
+                    used[i] = true
+                    flag = false
+                    break
+                }
+            }
+            if (flag) {
+                ct++
+            }
+        }
+        return ct
+    }
+}

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

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+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`

src/main/kotlin/g3401_3500/s3478_choose_k_elements_with_maximum_sum/Solution.kt

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+package g3401_3500.s3478_choose_k_elements_with_maximum_sum
+
+// #Medium #2025_03_09_Time_241_ms_(100.00%)_Space_89.04_MB_(100.00%)
+
+import java.util.PriorityQueue
+
+class Solution {
+    fun findMaxSum(nums1: IntArray, nums2: IntArray, k: Int): LongArray {
+        val n = nums1.size
+        val arr = Array<IntArray>(n) { IntArray(3) }
+        for (i in 0..<n) {
+            arr[i][0] = nums1[i]
+            arr[i][1] = nums2[i]
+            arr[i][2] = i
+        }
+        arr.sortWith { a: IntArray, b: IntArray -> a[0] - b[0] }
+        val ans = LongArray(n)
+        val pq = PriorityQueue<Int>()
+        var sum: Long = 0
+        for (i in 0..<n) {
+            if (i > 0 && arr[i - 1][0] == arr[i][0]) {
+                ans[arr[i][2]] = ans[arr[i - 1][2]]
+            } else {
+                ans[arr[i][2]] = sum
+            }
+            pq.add(arr[i][1])
+            sum += arr[i][1].toLong()
+            if (pq.size > k) {
+                sum -= pq.remove()!!.toLong()
+            }
+        }
+        return ans
+    }
+}

src/main/kotlin/g3401_3500/s3478_choose_k_elements_with_maximum_sum/readme.md

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+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`

src/main/kotlin/g3401_3500/s3479_fruits_into_baskets_iii/Solution.kt

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+package g3401_3500.s3479_fruits_into_baskets_iii
+
+// #Medium #2025_03_09_Time_605_ms_(100.00%)_Space_76.36_MB_(100.00%)
+
+import kotlin.math.max
+
+class Solution {
+    fun numOfUnplacedFruits(fruits: IntArray, baskets: IntArray): Int {
+        var unp = 0
+        var max = 0
+        val list: MutableList<Int> = ArrayList<Int>()
+        for (basket in baskets) {
+            list.add(basket)
+            max = max(max, basket)
+        }
+        for (fruit in fruits) {
+            if (fruit > max) {
+                unp++
+                continue
+            }
+            var placed = false
+            for (j in list.indices) {
+                if (list[j] >= fruit) {
+                    list.removeAt(j)
+                    placed = true
+                    break
+                }
+            }
+            if (!placed) {
+                unp++
+            }
+        }
+        return unp
+    }
+}

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

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+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>

src/main/kotlin/g3401_3500/s3480_maximize_subarrays_after_removing_one_conflicting_pair/Solution.kt

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+package g3401_3500.s3480_maximize_subarrays_after_removing_one_conflicting_pair
+
+// #Hard #2025_03_09_Time_768_ms_(100.00%)_Space_170.46_MB_(100.00%)
+
+import java.util.TreeMap
+import kotlin.math.max
+
+class Solution {
+    fun maxSubarrays(n: Int, cp: Array<IntArray>): Long {
+        for (pair in cp) {
+            if (pair[0] > pair[1]) {
+                val temp = pair[0]
+                pair[0] = pair[1]
+                pair[1] = temp
+            }
+        }
+        cp.sortWith { a: IntArray, b: IntArray -> a[0].compareTo(b[0]) }
+        val contributions = LongArray(cp.size)
+        var totalPossible = n.toLong() * (n + 1) / 2
+        val endPointMap = TreeMap<Int, MutableList<Int>>()
+        var currentIndex = cp.size - 1
+        for (start in n downTo 1) {
+            while (currentIndex >= 0 && cp[currentIndex][0] >= start) {
+                val end = cp[currentIndex][1]
+                endPointMap.computeIfAbsent(end) { k: Int -> ArrayList<Int>() }
+                    .add(currentIndex)
+                currentIndex--
+            }
+            if (endPointMap.isEmpty()) {
+                continue
+            }
+            val firstEntry = endPointMap.firstEntry()
+            val smallestEnd: Int = firstEntry.key!!
+            val pairIndex: Int = firstEntry.value!![0]
+            if (firstEntry.value!!.size == 1) {
+                endPointMap.remove(smallestEnd)
+            } else {
+                firstEntry.value!!.removeAt(0)
+            }
+            val covered = n - smallestEnd + 1L
+            totalPossible -= covered
+            if (endPointMap.isEmpty()) {
+                contributions[pairIndex] += covered
+            } else {
+                val nextEnd: Int = endPointMap.firstKey()!!
+                contributions[pairIndex] += (nextEnd - smallestEnd).toLong()
+            }
+            endPointMap.computeIfAbsent(smallestEnd) { k: Int? -> ArrayList<Int>() }
+                .add(pairIndex)
+        }
+        var result = totalPossible
+        for (contribution in contributions) {
+            result = max(result, (totalPossible + contribution))
+        }
+        return result
+    }
+}

src/main/kotlin/g3401_3500/s3480_maximize_subarrays_after_removing_one_conflicting_pair/readme.md

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+3480\. Maximize Subarrays After Removing One Conflicting Pair
+
+Hard
+
+You are given an integer `n` which represents an array `nums` containing the numbers from 1 to `n` in order. Additionally, you are given a 2D array `conflictingPairs`, where `conflictingPairs[i] = [a, b]` indicates that `a` and `b` form a conflicting pair.
+
+Remove **exactly** one element from `conflictingPairs`. Afterward, count the number of non-empty subarrays of `nums` which do not contain both `a` and `b` for any remaining conflicting pair `[a, b]`.
+
+Return the **maximum** number of subarrays possible after removing **exactly** one conflicting pair.
+
+**Example 1:**
+
+**Input:** n = 4, conflictingPairs = [[2,3],[1,4]]
+
+**Output:** 9
+
+**Explanation:**
+
+*   Remove `[2, 3]` from `conflictingPairs`. Now, `conflictingPairs = [[1, 4]]`.
+*   There are 9 subarrays in `nums` where `[1, 4]` do not appear together. They are `[1]`, `[2]`, `[3]`, `[4]`, `[1, 2]`, `[2, 3]`, `[3, 4]`, `[1, 2, 3]` and `[2, 3, 4]`.
+*   The maximum number of subarrays we can achieve after removing one element from `conflictingPairs` is 9.
+
+**Example 2:**
+
+**Input:** n = 5, conflictingPairs = [[1,2],[2,5],[3,5]]
+
+**Output:** 12
+
+**Explanation:**
+
+*   Remove `[1, 2]` from `conflictingPairs`. Now, `conflictingPairs = [[2, 5], [3, 5]]`.
+*   There are 12 subarrays in `nums` where `[2, 5]` and `[3, 5]` do not appear together.
+*   The maximum number of subarrays we can achieve after removing one element from `conflictingPairs` is 12.
+
+**Constraints:**
+
+*   <code>2 <= n <= 10<sup>5</sup></code>
+*   `1 <= conflictingPairs.length <= 2 * n`
+*   `conflictingPairs[i].length == 2`
+*   `1 <= conflictingPairs[i][j] <= n`
+*   `conflictingPairs[i][0] != conflictingPairs[i][1]`

src/test/kotlin/g3401_3500/s3477_fruits_into_baskets_ii/SolutionTest.kt

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+package g3401_3500.s3477_fruits_into_baskets_ii
+
+import org.hamcrest.CoreMatchers.equalTo
+import org.hamcrest.MatcherAssert.assertThat
+import org.junit.jupiter.api.Test
+
+internal class SolutionTest {
+    @Test
+    fun numOfUnplacedFruits() {
+        assertThat<Int>(
+            Solution().numOfUnplacedFruits(intArrayOf(4, 2, 5), intArrayOf(3, 5, 4)),
+            equalTo<Int>(1),
+        )
+    }
+
+    @Test
+    fun numOfUnplacedFruits2() {
+        assertThat<Int>(
+            Solution().numOfUnplacedFruits(intArrayOf(3, 6, 1), intArrayOf(6, 4, 7)),
+            equalTo<Int>(0),
+        )
+    }
+}

src/test/kotlin/g3401_3500/s3478_choose_k_elements_with_maximum_sum/SolutionTest.kt

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+package g3401_3500.s3478_choose_k_elements_with_maximum_sum
+
+import org.hamcrest.CoreMatchers.equalTo
+import org.hamcrest.MatcherAssert.assertThat
+import org.junit.jupiter.api.Test
+
+internal class SolutionTest {
+    @Test
+    fun findMaxSum() {
+        assertThat<LongArray>(
+            Solution()
+                .findMaxSum(intArrayOf(4, 2, 1, 5, 3), intArrayOf(10, 20, 30, 40, 50), 2),
+            equalTo<LongArray>(longArrayOf(80L, 30L, 0L, 80L, 50L)),
+        )
+    }
+
+    @Test
+    fun findMaxSum2() {
+        assertThat<LongArray>(
+            Solution().findMaxSum(intArrayOf(2, 2, 2, 2), intArrayOf(3, 1, 2, 3), 1),
+            equalTo<LongArray>(longArrayOf(0L, 0L, 0L, 0L)),
+        )
+    }
+}