Hello Everyone,

Given two integer arrays arr1[] and arr2[] sorted in ascending order and an integer k. Find k pairs with smallest sums such that one element of a pair belongs to arr1[] and other element belongs to arr2[]

Examples:

Input : arr1[] = {1, 7, 11}

arr2[] = {2, 4, 6}

k = 3

Output : [1, 2],

[1, 4],

[1, 6]

Explanation: 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]

**Method 1 (Simple)**

- Find all pairs and store their sums. Time complexity of this step is O(n1 * n2) where n1 and n2 are sizes of input arrays.
- Then sort pairs according to sum. Time complexity of this step is O(n1 * n2 * log (n1 * n2))

Overall Time Complexity : O(n1 * n2 * log (n1 * n2))

**Method 2 (Efficient):**

We one by one find k smallest sum pairs, starting from least sum pair. The idea is to keep track of all elements of arr2[] which have been already considered for every element arr1[i1] so that in an iteration we only consider next element. For this purpose, we use an index array index2[] to track the indexes of next elements in the other array. It simply means that which element of second array to be added with the element of first array in each and every iteration. We increment value in index array for the element that forms next minimum value pair.

`// C++ program to prints first k pairs with least sum from two`

`// arrays.`

`#include<bits/stdc++.h>`

`using`

`namespace`

`std;`

`// Function to find k pairs with least sum such`

`// that one elemennt of a pair is from arr1[] and`

`// other element is from arr2[]`

`void`

`kSmallestPair(`

`int`

`arr1[], `

`int`

`n1, `

`int`

`arr2[],`

` `

`int`

`n2, `

`int`

`k)`

`{`

` `

`if`

`(k > n1*n2)`

` `

`{`

` `

`cout << `

`"k pairs don't exist"`

`;`

` `

`return`

`;`

` `

`}`

` `

`// Stores current index in arr2[] for`

` `

`// every element of arr1[]. Initially`

` `

`// all values are considered 0.`

` `

`// Here current index is the index before`

` `

`// which all elements are considered as`

` `

`// part of output.`

` `

`int`

`index2[n1];`

` `

`memset`

`(index2, 0, `

`sizeof`

`(index2));`

` `

`while`

`(k > 0)`

` `

`{`

` `

`// Initialize current pair sum as infinite`

` `

`int`

`min_sum = INT_MAX;`

` `

`int`

`min_index = 0;`

` `

`// To pick next pair, traverse for all elements`

` `

`// of arr1[], for every element, find corresponding`

` `

`// current element in arr2[] and pick minimum of`

` `

`// all formed pairs.`

` `

`for`

`(`

`int`

`i1 = 0; i1 < n1; i1++)`

` `

`{`

` `

`// Check if current element of arr1[] plus`

` `

`// element of array2 to be used gives minimum`

` `

`// sum`

` `

`if`

`(index2[i1] < n2 &&`

` `

`arr1[i1] + arr2[index2[i1]] < min_sum)`

` `

`{`

` `

`// Update index that gives minimum`

` `

`min_index = i1;`

` `

`// update minimum sum`

` `

`min_sum = arr1[i1] + arr2[index2[i1]];`

` `

`}`

` `

`}`

` `

`cout << `

`"("`

`<< arr1[min_index] << `

`", "`

` `

`<< arr2[index2[min_index]] << `

`") "`

`;`

` `

`index2[min_index]++;`

` `

`k--;`

` `

`}`

`}`

`// Driver code`

`int`

`main()`

`{`

` `

`int`

`arr1[] = {1, 3, 11};`

` `

`int`

`n1 = `

`sizeof`

`(arr1) / `

`sizeof`

`(arr1[0]);`

` `

`int`

`arr2[] = {2, 4, 8};`

` `

`int`

`n2 = `

`sizeof`

`(arr2) / `

`sizeof`

`(arr2[0]);`

` `

`int`

`k = 4;`

` `

`kSmallestPair( arr1, n1, arr2, n2, k);`

` `

`return`

`0;`

`}`

**Output:**

(1, 2) (1, 4) (3, 2) (3, 4)

**Time Complexity :** O(k*n1)