Formatted question description: https://leetcode.ca/all/2080.html

# 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

Companies:
Quora

Related Topics:
Array, Binary Search

Initialization: For each unique value in A, store its corresponding indices in a map m.

Query:

• If value doesn’t exist in the original array, return 0.
• Otherwise, let v = m[value] – the array of all the indices of value in the original array
• Let j be the first index that v[j] > right.
• Let i be the first index that v[i] >= left.
• The answer is j - i.
• // OJ: https://leetcode.com/problems/range-frequency-queries/
// Time:
//      RangeFreqQuery: O(N)
//      query: O(log(N))
// Space: O(N)
class RangeFreqQuery {
unordered_map<int, vector<int>> m; // map from a value to all its indices
public:
RangeFreqQuery(vector<int>& A) {
for (int i = 0; i < A.size(); ++i) m[A[i]].push_back(i);
}
int query(int left, int right, int value) {
if (m.count(value) == 0) return 0; // value doesn't exist in the original array
auto &v = m[value]; // v is the array of all the indices of value in the original array
int j = upper_bound(begin(v), end(v), right) - begin(v); // Find the first index j that v[j] > right.
int i = lower_bound(begin(v), end(v), left) - begin(v); // Find the first index i that v[i] >= left.
return j - i; // The answer is j - i
}
};

• // Todo

• # 2080. Range Frequency Queries
# https://leetcode.com/problems/range-frequency-queries/

class RangeFreqQuery:
​
def __init__(self, arr: List[int]):
self.mp = collections.defaultdict(list)
for i, x in enumerate(arr):
self.mp[x].append(i)
​
def query(self, left: int, right: int, value: int) -> int:
a = bisect.bisect_left(self.mp[value], left)
b = bisect.bisect_right(self.mp[value], right)

return b - a

​
​
# Your RangeFreqQuery object will be instantiated and called as such:
# obj = RangeFreqQuery(arr)
# param_1 = obj.query(left,right,value)



## Discuss

https://leetcode.com/problems/range-frequency-queries/discuss/1589028/C%2B%2B-Straightforward-Binary-Search