Python – Sort Matrix by K Sized Subarray Maximum Sum

Given Matrix, write a Python program to sort rows by maximum of K sized subarray sum.

Examples:

Input : test_list = [[4, 3, 5, 2, 3], [6, 4, 2, 1, 1], [4, 3, 9, 3, 9], [5, 4, 3, 2, 1]], K = 3 
Output : [[4, 3, 5, 2, 3], [6, 4, 2, 1, 1], [5, 4, 3, 2, 1], [4, 3, 9, 3, 9]] 
Explanation : 12 = 12 = 12 < 21, is order of maximum sum 3 length substring.

Input : test_list = [[4, 3, 5, 2, 3], [4, 3, 9, 3, 9], [5, 4, 3, 2, 1]], K = 3 
Output : [[4, 3, 5, 2, 3], [5, 4, 3, 2, 1], [4, 3, 9, 3, 9]] 
Explanation : 12 = 12 < 21, is order of maximum sum 3 length substring. 

Method #1 : Using max() + sum() + slicing + sort()

In this, maximum of K length subarray is computed using max(), sum() and slicing using external function and inplace sorting is done using sort().

Python3

# Python3 code to demonstrate working of
# Sort Matrix by K Sized Subarray Maximum Sum
# Using max() + sum() + slicing + sort()


def max_ksub(row):

    # getting maximum K length sum
    return max(sum(row[idx: idx + K]) for idx in range(len(row) - K))


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

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

# initializing K
K = 3

# performing inplace sorting
test_list.sort(key=max_ksub)

# printing result
print("The sorted result : " + str(test_list))

Output:

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

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

Method #2 : Using sorted() + lambda + max() + sum() + slicing

In this, we perform task of sorting using sorted() + lambda function which injects comparator logic and avoids calling external function.

Python3

# Python3 code to demonstrate working of
# Sort Matrix by K Sized Subarray Maximum Sum
# Using sorted() + lambda + max() + sum() + slicing

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

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

# initializing K
K = 3

# sorted() performs inplace sort
# lambda function injects comparison logic
res = sorted(test_list, key=lambda row: max(
    sum(row[idx: idx + K]) for idx in range(len(row) - K)))

# printing result
print("The sorted result : " + str(res))

Output:

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

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

Method 3: Using heapq.nlargest() + sum() + enumerate()

Python3

# Python3 code to demonstrate working of
# Sort Matrix by K Sized Subarray Maximum Sum
# Using heapq.nlargest() + sum() + enumerate()

# importing required module
import heapq

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

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

# initializing K
K = 3

# heapq.nlargest() returns the n largest elements from the iterable
# lambda function calculates the sum of K-sized subarrays of a row
# enumerate() adds index to the result of heapq.nlargest()
# sorted() sorts the rows based on their K-sized subarray maximum sum
res = sorted(test_list, key=lambda row: heapq.nlargest(1, ((sum(row[i:i+K]), i) for i in range(len(row) - K + 1)), key=lambda x: x[0])[0][1])

# printing result
print("The sorted result : " + str(res))

Output

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

Time complexity: O(n * k * log(n)), where n is the number of rows and k is the subarray size. 
Auxiliary space: O(n * k).