Find k pairs with smallest sums in two arrays | Set 2

Given two arrays arr1[] and arr2[] sorted in ascending order and an integer K. The task is to find k pairs with the smallest sums such that one element of a pair belongs to arr1[] and another element belongs to arr2[]. The sizes of arrays may be different. Assume all the elements to be distinct in each array. 
Examples:

Input: a1[] = {1, 7, 11}
       a2[] = {2, 4, 6}
       k = 3
Output: [1, 2], [1, 4], [1, 6]
The first 3 pairs are returned 
from the sequence [1, 2], [1, 4], [1, 6], [7, 2],
[7, 4], [11, 2], [7, 6], [11, 4], [11, 6].
Input: a1[] = { 2, 3, 4 }
       a2[] = { 1, 6, 5, 8 }  
       k = 4
Output: [1, 2] [1, 3] [1, 4] [2, 6] 

An approach with time complexity O(k*n1) has been discussed here. 
Efficient Approach: Since the array is already sorted. The given below algorithm can be followed to solve this problem:  

• The idea is to maintain two pointers, one pointer pointing to one pair in (a1, a2) and the other in (a2, a1). Each time, compare the sums of the elements pointed by the two pairs and print the minimum one. After this, increment the pointer to the element in the printed pair which was larger than the other. This helps to get the next possible k smallest pair.
• Once the pointer has been updated to the element such that it starts pointing to the first element of the array again, update the other pointer to the next value. This update is done cyclically.
• Also, when both the pairs are pointing to the same element, update pointers in both the pairs to avoid extra pair’s printing. Update one pair’s pointer according to rule1 and other’s opposite to rule1. This is done to ensure that all the permutations are considered and no repetitions of pairs are there.

Below is the working of the algorithm for example 1: 
 

a1[] = {1, 7, 11}, a2[] = {2, 4}, k = 3
Let the pairs of pointers be _one, _two
_one.first points to 1, _one.second points to 2 ; 
_two.first points to 2, _two.second points to 1
1st pair: 
Since _one and _two are pointing to same elements, print the pair once and update 
 

• print [1, 2]

then update _one.first to 1, _one.second to 4 (following rule 1) ; 
_two.first points to 2, _two.second points to 7 (opposite to rule 1). 
If rule 1 was followed for both, then both of them would have been pointing to 1 and 4, 
and it is not possible to get all possible permutations.
2nd pair: 
Since a1[_one.first] + a2[_one.second] < a1[_two.second] + a2[_two.first], print them and update 
 

• print [1, 4]

then update _one.first to 1, _one.second to 2 
Since _one.second came to the first element of the array once again, 
therefore _one.first points to 7
Repeat the above process for remaining K pairs

Below is the C++ implementation of the above approach: 

[1, 2] [1, 3] [1, 4] [2, 6]

Time complexity: O(K)

Space Complexity: O(1)

Using Brute Force in Python:

Approach:

One simple approach to solve this problem is to generate all possible pairs of elements from both arrays and compute their sum. Then, we can sort the pairs based on their sum and return the first k pairs. In this program, we define a function k_smallest_pairs which takes in two arrays a1 and a2, and an integer k. We first create a list of all possible pairs by using a nested list comprehension. Then, we sort the pairs based on their sum using a lambda function as the sorting key. Finally, we return the first k pairs from the sorted list using slicing.

We then call the function twice with different inputs to demonstrate its usage and output.

[(1, 2), (1, 4), (1, 6)]
[(2, 1), (3, 1), (4, 1), (2, 5)]

Time complexity: O(mn log mn), where m and n are the lengths of the two arrays.
Space complexity: O(mn), to store all the pairs.