Python Program to Sort Matrix Rows by summation of consecutive difference of elements

Given a Matrix, the following article depicts how to sort rows of a matrix on the basis of summation of difference between consecutive elements of a row.

Input : test_list = [[1, 5, 3, 6], [4, 3, 2, 1], [7, 2, 4, 5], [6, 9, 3, 2]], 
Output : [[4, 3, 2, 1], [7, 2, 4, 5], [1, 5, 3, 6], [6, 9, 3, 2]] 
Explanation : 4 < 8 < 9 < 10 is consecutive difference summation.
Input : test_list = [[1, 5, 3, 6], [7, 2, 4, 5], [6, 9, 3, 2]], 
Output : [[7, 2, 4, 5], [1, 5, 3, 6], [6, 9, 3, 2]] 
Explanation : 8 < 9 < 10 is consecutive difference summation. 

Method 1 : Using sort() and abs()

In this, we perform task of in-place sorting using sort() and abs() is used to get the absolute value of the summation of consecutive difference.

Python3

# get abs summation
def diff_sum(row):
    return sum([abs(row[idx + 1] - row[idx]) for idx in range(0, len(row) - 1)])


# initializing list
test_list = [[1, 5, 3, 6], [4, 3, 2, 1], [7, 2, 4, 5], [6, 9, 3, 2]]

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

# performing inplace sort
test_list.sort(key=diff_sum)

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

Output:

The original list is : [[1, 5, 3, 6], [4, 3, 2, 1], [7, 2, 4, 5], [6, 9, 3, 2]]

Sorted Rows : [[4, 3, 2, 1], [7, 2, 4, 5], [1, 5, 3, 6], [6, 9, 3, 2]]

Time Complexity: O(nlogn+mlogm)
Auxiliary Space: O(k)

Method 2 : Using sorted(), lambda, abs() and sum()

In this, the sorting is done using sorted() and lambda function is used to inject conditional statement in sorted(). 

Python3

# initializing list
test_list = [[1, 5, 3, 6], [4, 3, 2, 1], [7, 2, 4, 5], [6, 9, 3, 2]]

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

# performing sort
res = sorted(test_list, key=lambda row: sum(
    [abs(row[idx + 1] - row[idx]) for idx in range(0, len(row) - 1)]))

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

Output:

The original list is : [[1, 5, 3, 6], [4, 3, 2, 1], [7, 2, 4, 5], [6, 9, 3, 2]]

Sorted Rows : [[4, 3, 2, 1], [7, 2, 4, 5], [1, 5, 3, 6], [6, 9, 3, 2]]

Time Complexity: O(nlogn+mlogm)
Auxiliary Space: O(k)