2080. Range Frequency Queries

Problem

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 <= 10^5

• 1 <= arr[i], value <= 10^4

• 0 <= left <= right < arr.length

• At most 10^5 calls will be made to query

Solution (Java)

class RangeFreqQuery {

    private Map<Integer, List<Integer>> map;

    public RangeFreqQuery(int[] arr) {
        map = new HashMap<>();
        for (int i = 0; i < arr.length; i++) {
            if (!map.containsKey(arr[i])) {
                map.put(arr[i], new ArrayList<>());
            }
            map.get(arr[i]).add(i);
        }
    }

    public int query(int left, int right, int value) {
        if (!map.containsKey(value)) {
            return 0;
        }
        List<Integer> list = map.get(value);
        int s = Collections.binarySearch(list, left);
        int e = Collections.binarySearch(list, right);
        if (s < 0) {
            s = (s + 1) * -1;
        }
        if (e < 0) {
            e = (e + 2) * -1;
        }
        return e - s + 1;
    }
}

/**
 * Your RangeFreqQuery object will be instantiated and called as such:
 * RangeFreqQuery obj = new RangeFreqQuery(arr);
 * int param_1 = obj.query(left,right,value);
 */

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).