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:- 
 

# 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.

# 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[2] = {1, 8}
  map[8] = {2, 5, 7}
  map[6] = {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[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 

# 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

# 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!
 

kartik

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Article Tags :

• Hash
• Python
• Python Programs
• DSA
• Arrays
• array-range-queries

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