LeetCode in Kotlin

1851. Minimum Interval to Include Each Query

Hard

You are given a 2D integer array intervals, where intervals[i] = [lefti, righti] describes the ith interval starting at lefti and ending at righti (inclusive). The size of an interval is defined as the number of integers it contains, or more formally righti - lefti + 1.

You are also given an integer array queries. The answer to the jth query is the size of the smallest interval i such that lefti <= queries[j] <= righti. If no such interval exists, the answer is -1.

Return an array containing the answers to the queries.

Example 1:

Input: intervals = [[1,4],[2,4],[3,6],[4,4]], queries = [2,3,4,5]

Output: [3,3,1,4]

Explanation: The queries are processed as follows:

• Query = 2: The interval [2,4] is the smallest interval containing 2. The answer is 4 - 2 + 1 = 3.

• Query = 3: The interval [2,4] is the smallest interval containing 3. The answer is 4 - 2 + 1 = 3.

• Query = 4: The interval [4,4] is the smallest interval containing 4. The answer is 4 - 4 + 1 = 1.

• Query = 5: The interval [3,6] is the smallest interval containing 5. The answer is 6 - 3 + 1 = 4.

Example 2:

Input: intervals = [[2,3],[2,5],[1,8],[20,25]], queries = [2,19,5,22]

Output: [2,-1,4,6]

Explanation: The queries are processed as follows:

• Query = 2: The interval [2,3] is the smallest interval containing 2. The answer is 3 - 2 + 1 = 2.

• Query = 19: None of the intervals contain 19. The answer is -1.

• Query = 5: The interval [2,5] is the smallest interval containing 5. The answer is 5 - 2 + 1 = 4.

• Query = 22: The interval [20,25] is the smallest interval containing 22. The answer is 25 - 20 + 1 = 6.

Constraints:

• 1 <= intervals.length <= 105
• 1 <= queries.length <= 105
• intervals[i].length == 2
• 1 <= lefti <= righti <= 107
• 1 <= queries[j] <= 107

Solution

import java.util.PriorityQueue

class Solution {
    fun minInterval(intervals: Array<IntArray>, queries: IntArray): IntArray {
        val numQuery = queries.size
        val queriesWithIndex = Array(numQuery) { IntArray(2) }
        for (i in 0 until numQuery) {
            queriesWithIndex[i] = intArrayOf(queries[i], i)
        }
        intervals.sortWith { a: IntArray, b: IntArray -> a[0].compareTo(b[0]) }
        queriesWithIndex.sortWith { a: IntArray, b: IntArray -> a[0].compareTo(b[0]) }
        val minHeap = PriorityQueue({ a: IntArray, b: IntArray -> (a[1] - a[0]).compareTo(b[1] - b[0]) })
        val result = IntArray(numQuery)
        var j = 0
        for (i in queries.indices) {
            val queryVal = queriesWithIndex[i][0]
            val queryIndex = queriesWithIndex[i][1]
            while (j < intervals.size && intervals[j][0] <= queryVal) {
                minHeap.add(intervals[j])
                j++
            }
            while (minHeap.isNotEmpty() && minHeap.peek()[1] < queryVal) {
                minHeap.remove()
            }
            result[queryIndex] = if (minHeap.isEmpty()) -1 else minHeap.peek()[1] - minHeap.peek()[0] + 1
        }
        return result
    }
}

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