Given an array of n non-negative integers. The task is to find frequency of a particular element in the arbitrary range of array[]. The range is given as positions (not 0 based indexes) in array. There can be multiple queries of given type. 
Examples: 
 

Input  : arr[] = {2, 8, 6, 9, 8, 6, 8, 2, 11};
         left = 2, right = 8, element = 8
         left = 2, right = 5, element = 6      
Output : 3
         1
The element 8 appears 3 times in arr[left-1..right-1]
The element 6 appears 1 time in arr[left-1..right-1]

Naive approach: is to traverse from left to right and update count variable whenever we find the element. 
Below is the code of Naive approach:- 
 

Output: 

 Frequency of 2 from 1 to 6 = 1
 Frequency of 8 from 4 to 9 = 2

Time complexity of this approach is O(right – left + 1) or O(n) 
Auxiliary space: O(1)
An Efficient approach is to use hashing. In C++, we can use unordered_map
 

1. At first, we will store the position in map[] of every distinct element as a vector like that 
    

  int arr[] = {2, 8, 6, 9, 8, 6, 8, 2, 11};
  map[2] = {1, 8}
  map[8] = {2, 5, 7}
  map[6] = {3, 6} 
  ans so on...

1.  
2. As we can see that elements in map[] are already in sorted order (Because we inserted elements from left to right), the answer boils down to find the total count in that hash map[] using binary search like method. 
    
3. In C++ we can use lower_bound which will returns an iterator pointing to the first element in the range [first, last] which has a value not less than ‘left’. and upper_bound returns an iterator pointing to the first element in the range [first,last) which has a value greater than ‘right’. 
    
4. After that we just need to subtract the upper_bound() and lower_bound() result to get the final answer. For example, suppose if we want to find the total count of 8 in the range from [1 to 6], then the map[8] of lower_bound() function will return the result 0 (pointing to 2) and upper_bound() will return 2 (pointing to 7), so we need to subtract the both the result like 2 – 0 = 2 . 
    

Below is the code of above approach 
 

Output: 
 

Frequency of 2 from 1 to 6 = 1
Frequency of 8 from 4 to 9 = 2

This approach will be beneficial if we have a large number of queries of an arbitrary range asking the total frequency of particular element.
Time complexity: O(log N) for single query.
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