LeetCode in Kotlin

2080. Range Frequency Queries

Medium

Design a data structure to find the frequency of a given value in a given subarray.

The frequency of a value in a subarray is the number of occurrences of that value in the subarray.

Implement the RangeFreqQuery class:

• RangeFreqQuery(int[] arr) Constructs an instance of the class with the given 0-indexed integer array arr.
• int query(int left, int right, int value) Returns the frequency of value in the subarray arr[left...right].

A subarray is a contiguous sequence of elements within an array. arr[left...right] denotes the subarray that contains the elements of nums between indices left and right (inclusive).

Example 1:

Input

["RangeFreqQuery", "query", "query"]
[[[12, 33, 4, 56, 22, 2, 34, 33, 22, 12, 34, 56]], [1, 2, 4], [0, 11, 33]]

Output: [null, 1, 2]

Explanation:

RangeFreqQuery rangeFreqQuery = new RangeFreqQuery([12, 33, 4, 56, 22, 2, 34, 33, 22, 12, 34, 56]);
rangeFreqQuery.query(1, 2, 4); // return 1. The value 4 occurs 1 time in the subarray [33, 4]
rangeFreqQuery.query(0, 11, 33); // return 2. The value 33 occurs 2 times in the whole array. 

Constraints:

• 1 <= arr.length <= 105
• 1 <= arr[i], value <= 104
• 0 <= left <= right < arr.length
• At most 105 calls will be made to query

Solution

import java.util.Collections

class RangeFreqQuery(arr: IntArray) {
    private val map: MutableMap<Int, MutableList<Int>>

    init {
        map = HashMap()
        for (i in arr.indices) {
            if (!map.containsKey(arr[i])) {
                map[arr[i]] = ArrayList()
            }
            map[arr[i]]!!.add(i)
        }
    }

    fun query(left: Int, right: Int, value: Int): Int {
        if (!map.containsKey(value)) {
            return 0
        }
        val list: List<Int> = map[value]!!
        var s = Collections.binarySearch(list, left)
        var e = Collections.binarySearch(list, right)
        if (s < 0) {
            s = (s + 1) * -1
        }
        if (e < 0) {
            e = (e + 2) * -1
        }
        return e - s + 1
    }
}

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