Python Program to Filter Rows with a specific Pair Sum

Given Matrix, the following program shows how to extract all rows which have a pair such that their sum is equal to a specific number, here denoted as K.

Input : test_list = [[1, 5, 3, 6], [4, 3, 2, 1], [7, 2, 4, 5], [6, 9, 3, 2]], k = 8 
Output : [[1, 5, 3, 6], [6, 9, 3, 2]] 
Explanation : 5 + 3 = 8 and 6 + 2 = 8. 

Input : test_list = [[1, 5, 4, 6], [4, 3, 2, 1], [7, 2, 4, 5], [6, 9, 3, 7]], k = 8 
Output : [] 
Explanation : No list with 8 as pair summation. 

Method 1: Using loop and list comprehension

In this we perform the task of getting all the pair whose sum is equal to a given value using external function and filtering of lists is done using list comprehension.

Python3

# get pair sum
def pair_sum(x, k):

    # checking pair sum
    for idx in range(len(x)):
        for ix in range(idx + 1, len(x)):
            if x[idx] + x[ix] == k:
                return True
    return False


# 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))

# initializing K
k = 8

# checking for pair sum
res = [ele for ele in test_list if pair_sum(ele, k)]

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

Output

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

Time Complexity: O(n*m)
Auxiliary Space: O(k)

Method 2 : Using filter() and lambda

In this, we perform the task of filtering using filter() and lambda function. 

Python3

# get pair sum
def pair_sum(x, k):

    # checking pair sum
    for idx in range(len(x)):
        for ix in range(idx + 1, len(x)):
            if x[idx] + x[ix] == k:
                return True
    return False


# 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))

# initializing K
k = 8

# checking for pair sum
# filtering using filter() and lambda
res = list(filter(lambda ele: pair_sum(ele, k), test_list))

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

Output:

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

Filtered Rows : [[1, 5, 3, 6], [6, 9, 3, 2]]

Time Complexity: O(m*n) where m and n is the number of rows and columns in the list 
Auxiliary Space: O(k) additional space of size k is created 

Method #3:Using itertools.filterfalse() method

Python3

import itertools
# get pair sum
def pair_sum(x, k):

    # checking pair sum
    for idx in range(len(x)):
        for ix in range(idx + 1, len(x)):
            if x[idx] + x[ix] == k:
                return True
    return False


# 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))

# initializing K
k = 8

# checking for pair sum
# filtering using itertools.filterfalse() and lambda
res = list(itertools.filterfalse(lambda ele: not pair_sum(ele, k), test_list))

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

Output

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

Time Complexity: O(N*N)
Auxiliary Space: O(N*N)

Method #4: Using sets

Use sets to solve this problem. We can create an empty set called seen and iterate through the list of numbers. For each number num, we check if k – num is in seen. If it is, then we have found a pair that adds up to k. If it is not, then we add num to seen. If we iterate through the entire list and do not find a pair, then we return False.

Approach:

1. Define the function pair_sum(x, k) that takes in a list x and an integer k.
2. Create an empty set called seen.
3. Iterate through the list of numbers x.
4. For each number num, check if k – num is in seen. If it is, then return True as we have found a pair that adds up to k.
5. If k – num is not in seen, then add num to seen.
6. If we iterate through the entire list and do not find a pair, then return False.
7. Call the pair_sum function with the test_list and k variables.
8. Print the result.

Python3

# get pair sum
def pair_sum(x, k):
    seen = set()
    for num in x:
        if k - num in seen:
            return True
        seen.add(num)
    return False

# 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))

# initializing K
k = 8

# checking for pair sum
res = [ele for ele in test_list if pair_sum(ele, k)]

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

Output

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

Time complexity: O(n^2) where n is the length of the longest list in test_list. 
Auxiliary space: O(n) scenario where n is the length of the longest list in test_list, as we are storing each element in the set seen.

Method 5: Using the two-pointer technique.

Steps for this approach:

• Initialize two pointers left and right at the beginning and end of the list respectively.
• While left is less than right, do the following:
• Compute the sum of x[left] and x[right].
• If the sum is equal to k, return True.
• If the sum is less than k, increment left.
• If the sum is greater than k, decrement right.
• If the loop completes without finding a pair that sums to k, return False.

Python3

def pair_sum_using_two_pointer(x, k):
    x_sorted = sorted(x)
    left, right = 0, len(x_sorted) - 1
    while left < right:
        curr_sum = x_sorted[left] + x_sorted[right]
        if curr_sum == k:
            return True
        elif curr_sum < k:
            left += 1
        else:
            right -= 1
    return False

# 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))

# initializing K
k = 8

# checking for pair sum using two-pointer approach
res = [ele for ele in test_list if pair_sum_using_two_pointer(ele, k)]

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

Output

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

Time complexity: O(n*m(log m)) where n is the number of rows and m is the maximum number of columns in the rows (due to the sorting step in the two-pointer approach)
Auxiliary Space: O(n) (to store the filtered rows)