Python3 Program for Range Queries for Frequencies of array elements
Last Updated :
29 Jan, 2022
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
def findFrequency(arr, n, left, right, element):
count = 0
for i in range (left - 1 , right):
if (arr[i] = = element):
count + = 1
return count
arr = [ 2 , 8 , 6 , 9 , 8 , 6 , 8 , 2 , 11 ]
n = len (arr)
print ( "Frequency of 2 from 1 to 6 = " ,
findFrequency(arr, n, 1 , 6 , 2 ))
print ( "Frequency of 8 from 4 to 9 = " ,
findFrequency(arr, n, 4 , 9 , 8 ))
|
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
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 )
def findFrequency(arr, n, left, right, element):
a = lower_bound(store[element], left)
b = upper_bound(store[element], right)
return b - a
arr = [ 2 , 8 , 6 , 9 , 8 , 6 , 8 , 2 , 11 ]
n = len (arr)
for i in range (n):
store[arr[i]].append(i + 1 )
print ( "Frequency of 2 from 1 to 6 = " ,
findFrequency(arr, n, 1 , 6 , 2 ))
print ( "Frequency of 8 from 4 to 9 = " ,
findFrequency(arr, n, 4 , 9 , 8 ))
|
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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