Python - Extract list with difference in extreme values greater than K

Given a list of lists. The task is to filter all rows whose difference in min and max values is greater than K.

Examples:

Input : test_list = [[13, 5, 1], [9, 1, 2], [3, 4, 2], [1, 10, 2]], K = 5 
Output : [[9, 1, 2], [1, 10, 2], [13, 5, 1]] 
Explanation : 8, 9, 12 are differences, greater than K.

Input : test_list = [[13, 5, 1], [9, 1, 2], [3, 4, 2], [1, 10, 2]], K = 15 
Output : [] 
Explanation : No list with diff > K. 

Method #1 : Using list comprehension + min() + max()

In this, we perform task of iteration using list comprehension and the task of checking is done using the comparison operator. Values are computed using max() and min().

# Python3 code to demonstrate working of 
# Filter rows with Extreme values greater than K
# Using min() + max() + list comprehension

# initializing list
test_list = [[3, 5, 1], [9, 1, 2], [3, 4, 2], [1, 10, 2]]

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

# initializing K 
K = 5

# max() and min() getting extreme difference
res = [sub for sub in test_list if max(sub) - min(sub) > K]

# printing result 
print("Filtered rows : " + str(res))

Output:

The original list is : [[3, 5, 1], [9, 1, 2], [3, 4, 2], [1, 10, 2]] Filtered rows : [[9, 1, 2], [1, 10, 2]]

Time Complexity: O(n)
Auxiliary Space: O(n)

Method #2 : Using filter() + lambda + min() + max()

In this, we perform task of filtering using filter() and lambda, rest min() and max(), are used to get extreme values difference.

# Python3 code to demonstrate working of 
# Filter rows with Extreme values greater than K
# Using filter() + lambda + min() + max()

# initializing list
test_list = [[3, 5, 1], [9, 1, 2], [3, 4, 2], [1, 10, 2]]

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

# initializing K 
K = 5

# max() and min() getting extreme difference
res = list(filter(lambda sub : max(sub) - min(sub) > K, test_list))

# printing result 
print("Filtered rows : " + str(res))

Output:

The original list is : [[3, 5, 1], [9, 1, 2], [3, 4, 2], [1, 10, 2]] Filtered rows : [[9, 1, 2], [1, 10, 2]]

Time Complexity: O(n) where n is the number of elements in the list “test_list”.  This is because we’re using the built-in filter() + lambda + min() + max() which all has a time complexity of O(n) in the worst case.
Auxiliary Space: O(1), no extra space is required

Method #3: Using a for loop to iterate over the list of lists and append qualifying sublists to a new list

• Initialize the list of lists.
• Initialize the value of K.
• Initialize an empty list to store qualifying sublists.
• Iterate over the list of lists using a for loop.
• Check if the difference between the maximum and minimum values of the current sublist is greater than K.
• If the difference is greater than K, append the current sublist to the result list.
• Print the filtered rows.

# Python3 code to demonstrate working of 
# Filter rows with Extreme values greater than K
# Using for loop

# initializing list
test_list = [[3, 5, 1], [9, 1, 2], [3, 4, 2], [1, 10, 2]]

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

# initializing K
K = 5

# initializing empty list to store qualifying sublists
res = []

# iterating over the list of lists
for sublist in test_list:
    # checking if the difference between the maximum and minimum values of the sublist is greater than K
    if max(sublist) - min(sublist) > K:
        # appending the sublist to the result list
        res.append(sublist)

# printing result 
print("Filtered rows : " + str(res))

The original list is : [[3, 5, 1], [9, 1, 2], [3, 4, 2], [1, 10, 2]]
Filtered rows : [[9, 1, 2], [1, 10, 2]]

Time complexity: O(nm), where n is the number of sublists and m is the length of the longest sublist. The for loop iterates over each sublist once
.Auxiliary space: O(k), where k is the number of qualifying sublists. The result list stores the qualifying sublists.