Python – Count % K elements

Checking a number/element by a condition is a common problem one faces and is done in almost every program. Sometimes we also require to get the totals that match the particular condition to have a distinguish which to not match for further utilization. Lets discuss certain ways in which the task of checking count of numbers divisible to K can be achieved. 

Method #1 : Using sum() + generator expression This method uses the trick of adding 1 to the sum whenever the generator expression returns true. By the time list gets exhausted, summation of count of numbers matching % K is returned. 

Python3

# Python 3 code to demonstrate
# Count % K elements
# using sum() + generator expression

# initializing list
test_list = [3, 5, 1, 6, 7, 9]

# printing original list
print ("The original list is : " + str(test_list))

# initializing K
K = 3

# using sum() + generator expression
# to get count of elements matching condition
# Count % K elements
res = sum(1 for i in test_list if i % K != 0)

# printing result
print ("The number of % K elements: " + str(res))

Output : 

The original list is : [3, 5, 1, 6, 7, 9]
The number of % K elements: 3

Time Complexity: O(n), where n is the length of the list test_list 
Auxiliary Space: O(1) additional space is not needed

Method #2 : Using sum() + map() map() does the task almost similar to the generator expression, difference is just the internal data structure employed by it is different hence more efficient. 

Python3

# Python 3 code to demonstrate
# Count % K elements
# using sum()+ map()

# initializing list
test_list = [3, 5, 1, 6, 7, 9]

# printing original list
print ("The original list is : " + str(test_list))

# initializing K
K = 3

# using sum()+ map()
# to get count of elements matching condition
# Count % K elements
res = sum(map(lambda i: i % K != 0, test_list))

# printing result
print ("The number of % K elements: " + str(res))

Output : 

The original list is : [3, 5, 1, 6, 7, 9]
The number of % K elements: 3

Time Complexity: O(n*n), where n is the number of elements in the list “test_list”.
Auxiliary Space: O(1), constant extra space is required

Method #3 : Using filter()
The filter() function returns the elements of an iterable that match the condition specified.

Python3

# Python 3 code to demonstrate
# Count % K elements
# using filter()

# initializing list
test_list = [3, 5, 1, 6, 7, 9]

# printing original list
print ("The original list is : " + str(test_list))

# initializing K
K = 3

# using filter()
# to get count of elements matching condition
# Count % K elements
res = len(list(filter(lambda i: i % K != 0, test_list)))

# printing result
print ("The number of % K elements: " + str(res))
#This code is contributed by Edula Vinay Kumar Reddy

Output

The original list is : [3, 5, 1, 6, 7, 9]
The number of % K elements: 3

Time Complexity: O(n) where n is the number of elements in the string list. The filter() is used to perform the task and it takes O(n) time.
Auxiliary Space: O(n) additional space of size n is created where n is the number of elements in the test list.