Range Frequency Queries

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

思路：reverted index，先收集每个元素出现的index，每个元素肯定是个递增的序列，那么可以用binarysearch或者用treemap，logn去搜，[1，4，7，9，10，24] 里面全部是index，然后用left去搜第一个大于left的index，在新array里面占第i，用right去搜最后一个小于right的index，在新array里面占第j，然后个数就是 j - i + 1; 这题要明确，treemap里面存的是什么；

class RangeFreqQuery {
                                //原来index， 在新list里面占的位置 index
    private HashMap<Integer, TreeMap<Integer, Integer>> hashmap;
    public RangeFreqQuery(int[] arr) {
        this.hashmap = new HashMap<>();
        for(int i = 0; i < arr.length; i++) {
            if(!hashmap.containsKey(arr[i])) {
                hashmap.put(arr[i], new TreeMap<Integer, Integer>());
            }
            hashmap.get(arr[i]).put(i, hashmap.get(arr[i]).size());
        }
    }
    
    public int query(int left, int right, int value) {
        if(!hashmap.containsKey(value)) {
            return 0;
        }
        TreeMap<Integer, Integer> nums = hashmap.get(value);
        Integer a = nums.ceilingKey(left);
        Integer b = nums.floorKey(right);
        if(a == null || b == null) {
            return 0;
        }
        return nums.get(b) - nums.get(a) + 1;
    }
}

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

标签：RangeFreqQuery,arr,right,int,value,Range,Frequency,Queries,left
来源： https://blog.csdn.net/u013325815/article/details/121469611