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# Python3 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:-

## Python3

 `# Python program to find total ``# count of an element in a range` `# Returns count of element``# in arr[left-1..right-1]``def` `findFrequency(arr, n, left, right, element):` `    ``count ``=` `0``    ``for` `i ``in` `range``(left ``-` `1``, right):``        ``if` `(arr[i] ``=``=` `element):``            ``count ``+``=` `1``    ``return` `count`  `# Driver Code``arr ``=` `[``2``, ``8``, ``6``, ``9``, ``8``, ``6``, ``8``, ``2``, ``11``]``n ``=` `len``(arr)` `# Print frequency of 2 from position 1 to 6``print``(``"Frequency of 2 from 1 to 6 = "``,``        ``findFrequency(arr, n, ``1``, ``6``, ``2``))` `# Print frequency of 8 from position 4 to 9``print``(``"Frequency of 8 from 4 to 9 = "``,``        ``findFrequency(arr, n, ``4``, ``9``, ``8``))``        ` `    ` `# This code is contributed by Anant Agarwal.`

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

• 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 = {1, 8}
map = {2, 5, 7}
map = {3, 6}
ans so on...```
• 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.

• 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’.

• 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 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

## Python3

 `# Python3 program to find total count of an element``from` `collections ``import` `defaultdict as ``dict``from` `bisect ``import` `bisect_left as lower_bound``from` `bisect ``import` `bisect_right as upper_bound` `store ``=` `dict``(``list``)` `# Returns frequency of element``# in arr[left-1..right-1]``def` `findFrequency(arr, n, left, right, element):``    ` `    ``# Find the position of``    ``# first occurrence of element``    ``a ``=` `lower_bound(store[element], left)` `    ``# Find the position of``    ``# last occurrence of element``    ``b ``=` `upper_bound(store[element], right)` `    ``return` `b ``-` `a` `# Driver code``arr ``=` `[``2``, ``8``, ``6``, ``9``, ``8``, ``6``, ``8``, ``2``, ``11``]``n ``=` `len``(arr)` `# Storing the indexes of``# an element in the map``for` `i ``in` `range``(n):``    ``store[arr[i]].append(i ``+` `1``)` `# Print frequency of 2 from position 1 to 6``print``(``"Frequency of 2 from 1 to 6 = "``,``       ``findFrequency(arr, n, ``1``, ``6``, ``2``))` `# Print frequency of 8 from position 4 to 9``print``(``"Frequency of 8 from 4 to 9 = "``,``       ``findFrequency(arr, n, ``4``, ``9``, ``8``))` `# This code is contributed by Mohit Kumar`

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!

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