Python – Sort row by K multiples

Given a Matrix, perform row sorting by number of multiple of K present in row.

Input : test_list = [[3, 4, 8, 1], [12, 32, 4, 16], [1, 2, 3, 4], [9, 7, 5]], K = 4 
Output : [[9, 7, 5], [1, 2, 3, 4], [3, 4, 8, 1], [12, 32, 4, 16]] 
Explanation : 0 < 1 < 2 < 4, multiple of 4 occurrence order.

Input : test_list = [[3, 4, 8, 1], [12, 32, 4, 16], [1, 2, 3, 4], [9, 7, 5]], K = 2 
Output : [[9, 7, 5], [1, 2, 3, 4], [3, 4, 8, 1], [12, 32, 4, 16]] 
Explanation : 0 < 2 = 2 < 4, multiple of 2 occurrence order. 

Method #1 : Using sort() + % operator + len()

In this, we test for multiple using % operator and then compute count by getting length of filtered elements, provided to key to sort() which performs inplace sorting of rows.

Python3

# Python3 code to demonstrate working of
# Sort row by K multiples
# Using sort() + % operator + len()

# checking for multiples count


def k_mul(row):
    return len([ele for ele in row if ele % K == 0])


# initializing list
test_list = [[3, 4, 8, 1], [12, 32, 4, 16], [1, 2, 3, 4], [9, 7, 5]]

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

# initializing K
K = 4

# performing sort
test_list.sort(key=k_mul)

# printing result
print("Sorted result : " + str(test_list))

Output:

The original list is : [[3, 4, 8, 1], [12, 32, 4, 16], [1, 2, 3, 4], [9, 7, 5]] 
Sorted result : [[9, 7, 5], [1, 2, 3, 4], [3, 4, 8, 1], [12, 32, 4, 16]]

Time Complexity: O(nlogn*mlogm)
Auxiliary Space: O(1)

Method #2 : Using sorted() + lambda + len()

In this, sorting is done using sorted(), len() is used to get length of all the multiples of K as in above method. The lambda function provides single statement alternative to perform logical injection.

Python3

# Python3 code to demonstrate working of
# Sort row by K multiples
# Using sorted() + lambda + len()

# initializing list
test_list = [[3, 4, 8, 1], [12, 32, 4, 16], [1, 2, 3, 4], [9, 7, 5]]

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

# initializing K
K = 4

# performing sort using sorted()
# lambda avoiding external function call
res = sorted(test_list, key=lambda row: len(
    [ele for ele in row if ele % K == 0]))

# printing result
print("Sorted result : " + str(res))

Output:

The original list is : [[3, 4, 8, 1], [12, 32, 4, 16], [1, 2, 3, 4], [9, 7, 5]] 
Sorted result : [[9, 7, 5], [1, 2, 3, 4], [3, 4, 8, 1], [12, 32, 4, 16]]

Time Complexity: O(n*logn), where n is the length of the input list. This is because we’re using the built-in sorted() function which has a time complexity of O(nlogn) in the worst case.
Auxiliary Space: O(n), as we’re using additional space other than the input list itself.