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# C++ Program for Range Queries for Frequencies of array elements

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:-

## C++

 `// C++ program to find total count of an element``// in a range``#include``using` `namespace` `std;`` ` `// Returns count of element in arr[left-1..right-1]``int` `findFrequency(``int` `arr[], ``int` `n, ``int` `left,``                         ``int` `right, ``int` `element)``{``    ``int` `count = 0;``    ``for` `(``int` `i=left-1; i<=right; ++i)``        ``if` `(arr[i] == element)``            ``++count;``    ``return` `count;``}`` ` `// Driver Code``int` `main()``{``    ``int` `arr[] = {2, 8, 6, 9, 8, 6, 8, 2, 11};``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);`` ` `    ``// Print frequency of 2 from position 1 to 6``    ``cout << ``"Frequency of 2 from 1 to 6 = "``         ``<< findFrequency(arr, n, 1, 6, 2) << endl;`` ` `    ``// Print frequency of 8 from position 4 to 9``    ``cout << ``"Frequency of 8 from 4 to 9 = "``         ``<< findFrequency(arr, n, 4, 9, 8);`` ` `    ``return` `0;``}`

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

## C++

 `// C++ program to find total count of an element``#include``using` `namespace` `std;`` ` `unordered_map< ``int``, vector<``int``> > store;`` ` `// Returns frequency of element in arr[left-1..right-1]``int` `findFrequency(``int` `arr[], ``int` `n, ``int` `left,``                      ``int` `right, ``int` `element)``{``    ``// Find the position of first occurrence of element``    ``int` `a = lower_bound(store[element].begin(),``                        ``store[element].end(),``                        ``left)``            ``- store[element].begin();`` ` `    ``// Find the position of last occurrence of element``    ``int` `b = upper_bound(store[element].begin(),``                        ``store[element].end(),``                        ``right)``            ``- store[element].begin();`` ` `    ``return` `b-a;``}`` ` `// Driver code``int` `main()``{``    ``int` `arr[] = {2, 8, 6, 9, 8, 6, 8, 2, 11};``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);`` ` `    ``// Storing the indexes of an element in the map``    ``for` `(``int` `i=0; i

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.
Please refer complete article on Range Queries for Frequencies of array elements for more details!