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Find k pairs with smallest sums in two arrays | Set 2
  • Difficulty Level : Hard
  • Last Updated : 06 Nov, 2020

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: 
 

C++




// C++ program to print the k smallest
// pairs | Set 2
#include <bits/stdc++.h>
using namespace std;
 
typedef struct _pair {
    int first, second;
} _pair;
 
// Function to print the K smallest pairs
void printKPairs(int a1[], int a2[],
                 int size1, int size2, int k)
{
 
    // if k is greater than total pairs
    if (k > (size2 * size1)) {
        cout << "k pairs don't exist\n";
        return;
    }
 
    // _pair _one keeps track of
    // 'first' in a1 and 'second' in a2
    // in _two, _two.first keeps track of
    // element in the a2[] and _two.second in a1[]
    _pair _one, _two;
    _one.first = _one.second = _two.first = _two.second = 0;
 
    int cnt = 0;
 
    // Repeat the above process till
    // all K pairs are printed
    while (cnt < k) {
 
        // when both the pointers are pointing
        // to the same elements (point 3)
        if (_one.first == _two.second
            && _two.first == _one.second) {
            if (a1[_one.first] < a2[_one.second]) {
                cout << "[" << a1[_one.first]
                     << ", " << a2[_one.second] << "] ";
 
                // updates according to step 1
                _one.second = (_one.second + 1) % size2;
                if (_one.second == 0) // see point 2
                    _one.first = (_one.first + 1) % size1;
 
                // updates opposite to step 1
                _two.second = (_two.second + 1) % size2;
                if (_two.second == 0)
                    _two.first = (_two.first + 1) % size2;
            }
            else {
                cout << "[" << a2[_one.second]
                     << ", " << a1[_one.first] << "] ";
 
                // updates according to rule 1
                _one.first = (_one.first + 1) % size1;
                if (_one.first == 0) // see point 2
                    _one.second = (_one.second + 1) % size2;
 
                // updates opposite to rule 1
                _two.first = (_two.first + 1) % size2;
                if (_two.first == 0) // see point 2
                    _two.second = (_two.second + 1) % size1;
            }
        }
        // else update as necessary (point 1)
        else if (a1[_one.first] + a2[_one.second]
                 <= a2[_two.first] + a1[_two.second]) {
            if (a1[_one.first] < a2[_one.second]) {
                cout << "[" << a1[_one.first] << ", "
                     << a2[_one.second] << "] ";
 
                // updating according to rule 1
                _one.second = ((_one.second + 1) % size2);
                if (_one.second == 0) // see point 2
                    _one.first = (_one.first + 1) % size1;
            }
            else {
                cout << "[" << a2[_one.second] << ", "
                     << a1[_one.first] << "] ";
 
                // updating according to rule 1
                _one.first = ((_one.first + 1) % size1);
                if (_one.first == 0) // see point 2
                    _one.second = (_one.second + 1) % size2;
            }
        }
        else if (a1[_one.first] + a2[_one.second]
                 > a2[_two.first] + a1[_two.second]) {
            if (a2[_two.first] < a1[_two.second]) {
                cout << "[" << a2[_two.first] << ", " << a1[_two.second] << "] ";
 
                // updating according to rule 1
                _two.first = ((_two.first + 1) % size2);
                if (_two.first == 0) // see point 2
                    _two.second = (_two.second + 1) % size1;
            }
            else {
                cout << "[" << a1[_two.second]
                     << ", " << a2[_two.first] << "] ";
 
                // updating according to rule 1
                _two.second = ((_two.second + 1) % size1);
                if (_two.second == 0) // see point 2
                    _two.first = (_two.first + 1) % size1;
            }
        }
        cnt++;
    }
}
 
// Driver Code
int main()
{
 
    int a1[] = { 2, 3, 4 };
    int a2[] = { 1, 6, 5, 8 };
    int size1 = sizeof(a1) / sizeof(a1[0]);
    int size2 = sizeof(a2) / sizeof(a2[0]);
    int k = 4;
    printKPairs(a1, a2, size1, size2, k);
    return 0;
}

Java




// Java program to print
// the k smallest pairs
// | Set 2
import java.util.*;
class GFG{
 
static class _pair
{
  int first, second;
};
 
// Function to print the K
// smallest pairs
static void printKPairs(int a1[], int a2[],
                        int size1, int size2,
                        int k)
{
  // if k is greater than
  // total pairs
  if (k > (size2 * size1))
  {
    System.out.print("k pairs don't exist\n");
    return;
  }
 
  // _pair _one keeps track of
  // 'first' in a1 and 'second' in a2
  // in _two, _two.first keeps track of
  // element in the a2[] and _two.second
  // in a1[]
  _pair _one = new _pair();
  _pair  _two = new _pair();
  _one.first = _one.second =
  _two.first = _two.second = 0;
 
  int cnt = 0;
 
  // Repeat the above process
  // till all K pairs are printed
  while (cnt < k)
  {
    // when both the pointers are
    // pointing to the same elements
    // (point 3)
    if (_one.first == _two.second &&
        _two.first == _one.second)
    {
      if (a1[_one.first] <
          a2[_one.second])
      {
        System.out.print("[" +  a1[_one.first] +
                         ", " +  a2[_one.second] +
                         "] ");
 
        // updates according to step 1
        _one.second = (_one.second + 1) %
                       size2;
         
        // see point 2
        if (_one.second == 0)
          _one.first = (_one.first + 1) %
                        size1;
 
        // updates opposite to step 1
        _two.second = (_two.second + 1) %
                       size2;
         
        if (_two.second == 0)
          _two.first = (_two.first + 1) %
                        size2;
      }
      else
      {
        System.out.print("[" +  a2[_one.second] +
                         ", " +  a1[_one.first] +
                         "] ");
 
        // updates according to rule 1
        _one.first = (_one.first + 1) %
                      size1;
         
        // see point 2
        if (_one.first == 0)
          _one.second = (_one.second + 1) %
                         size2;
 
        // updates opposite to rule 1
        _two.first = (_two.first + 1) %
                      size2;
         
        // see point 2
        if (_two.first == 0)
           
          _two.second = (_two.second + 1) %
                         size1;
      }
    }
     
    // else update as
    // necessary (point 1)
    else if (a1[_one.first] +
             a2[_one.second] <=
             a2[_two.first] +
             a1[_two.second])
    {
      if (a1[_one.first] <
          a2[_one.second])
      {
        System.out.print("[" +  a1[_one.first] +
                         ", " + a2[_one.second] +
                         "] ");
 
        // updating according to rule 1
        _one.second = ((_one.second + 1) %
                        size2);
         
        // see point 2
        if (_one.second == 0)
          _one.first = (_one.first + 1) %
                        size1;
      }
      else
      {
        System.out.print("[" +  a2[_one.second] +
                         ", " + a1[_one.first] +
                         "] ");
 
        // updating according to rule 1
        _one.first = ((_one.first + 1) %
                       size1);
         
        // see point 2
        if (_one.first == 0)
          _one.second = (_one.second + 1) %
                         size2;
      }
    }
    else if (a1[_one.first] +
             a2[_one.second] >
             a2[_two.first] +
             a1[_two.second])
    {
      if (a2[_two.first] <
          a1[_two.second])
      {
        System.out.print("[" +  a2[_two.first] +
                         ", " +  a1[_two.second] +
                         "] ");
 
        // updating according to rule 1
        _two.first = ((_two.first + 1) %
                       size2);
         
        // see point 2
        if (_two.first == 0)
          _two.second = (_two.second + 1) %
                         size1;
      }
      else {
        System.out.print("[" +  a1[_two.second] +
                         ", " +  a2[_two.first] +
                         "] ");
 
        // updating according to rule 1
        _two.second = ((_two.second + 1) %
                        size1);
         
        // see point 2
        if (_two.second == 0)
          _two.first = (_two.first + 1) %
                        size1;
      }
    }
    cnt++;
  }
}
 
// Driver Code
public static void main(String[] args)
{
  int a1[] = {2, 3, 4};
  int a2[] = {1, 6, 5, 8};
  int size1 = a1.length;
  int size2 = a2.length;
  int k = 4;
  printKPairs(a1, a2,
              size1, size2, k);
}
}
 
// This code is contributed by gauravrajput1

Python3




# Python3 program to print the k smallest
# pairs | Set 2
 
# Function to print the K smallest pairs
def printKPairs(a1, a,size1, size2, k):
 
    # if k is greater than total pairs
    if (k > (size2 * size1)):
        print("k pairs don't exist\n")
        return
 
    # _pair _one keeps track of
    # 'first' in a1 and 'second' in a2
    # in _two, _two[0] keeps track of
    # element in the a2and _two[1] in a1[]
    _one, _two = [0, 0], [0, 0]
 
    cnt = 0
 
    # Repeat the above process till
    # all K pairs are printed
    while (cnt < k):
 
        # when both the pointers are pointing
        # to the same elements (po3)
        if (_one[0] == _two[1]
            and _two[0] == _one[1]):
            if (a1[_one[0]] < a2[_one[1]]):
                print("[", a1[_one[0]], ", ",
                        a2[_one[1]],"] ", end=" ")
 
                # updates according to step 1
                _one[1] = (_one[1] + 1) % size2
                if (_one[1] == 0): #see po2
                    _one[0] = (_one[0] + 1) % size1
 
                # updates opposite to step 1
                _two[1] = (_two[1] + 1) % size2
                if (_two[1] == 0):
                    _two[0] = (_two[0] + 1) % size2
 
            else:
                print("[",a2[_one[1]]
                    ,", ",a1[_one[0]],"] ",end=" ")
 
                # updates according to rule 1
                _one[0] = (_one[0] + 1) % size1
                if (_one[0] == 0): #see po2
                    _one[1] = (_one[1] + 1) % size2
 
                # updates opposite to rule 1
                _two[0] = (_two[0] + 1) % size2
                if (_two[0] == 0): #see po2
                    _two[1] = (_two[1] + 1) % size1
 
        # else update as necessary (po1)
        elif (a1[_one[0]] + a2[_one[1]]
                <= a2[_two[0]] + a1[_two[1]]):
            if (a1[_one[0]] < a2[_one[1]]):
                print("[",a1[_one[0]],", ",
                    a2[_one[1]],"] ",end=" ")
 
                # updating according to rule 1
                _one[1] = ((_one[1] + 1) % size2)
                if (_one[1] == 0): # see po2
                    _one[0] = (_one[0] + 1) % size1
            else:
                print("[",a2[_one[1]],", ",
                    a1[_one[0]],"] ", end=" ")
 
                # updating according to rule 1
                _one[0] = ((_one[0] + 1) % size1)
                if (_one[0] == 0): # see po2
                    _one[1] = (_one[1] + 1) % size2
 
        elif (a1[_one[0]] + a2[_one[1]]
                > a2[_two[0]] + a1[_two[1]]):
            if (a2[_two[0]] < a1[_two[1]]):
                print("[",a2[_two[0]],", ",a1[_two[1]],"] ",end=" ")
 
                # updating according to rule 1
                _two[0] = ((_two[0] + 1) % size2)
                if (_two[0] == 0): #see po2
                    _two[1] = (_two[1] + 1) % size1
 
            else:
                print("[",a1[_two[1]]
                    ,", ",a2[_two[0]],"] ",end=" ")
 
                # updating according to rule 1
                _two[1] = ((_two[1] + 1) % size1)
                if (_two[1] == 0): #see po2
                    _two[0] = (_two[0] + 1) % size1
 
        cnt += 1
 
# Driver Code
if __name__ == '__main__':
 
    a1= [2, 3, 4]
    a2= [1, 6, 5, 8]
    size1 = len(a1)
    size2 = len(a2)
    k = 4
    printKPairs(a1, a2, size1, size2, k)
 
# This code is contributed by mohit kumar 29

C#




// C# program to print
// the k smallest pairs
// | Set 2
using System;
class GFG{
 
public class _pair
{
  public int first,
             second;
};
 
// Function to print the K
// smallest pairs
static void printKPairs(int []a1, int []a2,
                        int size1, int size2,
                        int k)
{
  // if k is greater than
  // total pairs
  if (k > (size2 * size1))
  {
    Console.Write("k pairs don't exist\n");
    return;
  }
 
  // _pair _one keeps track of
  // 'first' in a1 and 'second' in a2
  // in _two, _two.first keeps track of
  // element in the a2[] and _two.second
  // in a1[]
  _pair _one = new _pair();
  _pair  _two = new _pair();
  _one.first = _one.second =
  _two.first = _two.second = 0;
 
  int cnt = 0;
 
  // Repeat the above process
  // till all K pairs are printed
  while (cnt < k)
  {
    // when both the pointers are
    // pointing to the same elements
    // (point 3)
    if (_one.first == _two.second &&
        _two.first == _one.second)
    {
      if (a1[_one.first] <
          a2[_one.second])
      {
        Console.Write("[" +  a1[_one.first] +
                      ", " +  a2[_one.second] +
                      "] ");
 
        // updates according to step 1
        _one.second = (_one.second + 1) %
                        size2;
 
        // see point 2
        if (_one.second == 0)
          _one.first = (_one.first + 1) %
                         size1;
 
        // updates opposite to step 1
        _two.second = (_two.second + 1) %
                        size2;
 
        if (_two.second == 0)
          _two.first = (_two.first + 1) %
                         size2;
      }
      else
      {
        Console.Write("[" + a2[_one.second] +
                      ", " + a1[_one.first] +
                      "] ");
 
        // updates according to rule 1
        _one.first = (_one.first + 1) %
                       size1;
 
        // see point 2
        if (_one.first == 0)
          _one.second = (_one.second + 1) %
                          size2;
 
        // updates opposite to rule 1
        _two.first = (_two.first + 1) %
                       size2;
 
        // see point 2
        if (_two.first == 0)
          _two.second = (_two.second + 1) %
                          size1;
      }
    }
 
    // else update as
    // necessary (point 1)
    else if (a1[_one.first] +
             a2[_one.second] <=
             a2[_two.first] +
             a1[_two.second])
    {
      if (a1[_one.first] <
          a2[_one.second])
      {
        Console.Write("[" + a1[_one.first] +
                      ", " + a2[_one.second] +
                      "] ");
 
        // updating according to rule 1
        _one.second = ((_one.second + 1) %
                         size2);
         
        // see point 2
        if (_one.second == 0)
          _one.first = (_one.first + 1) %
                         size1;
      }
      else
      {
        Console.Write("[" + a2[_one.second] +
                      ", " + a1[_one.first] +
                      "] ");
 
        // updating according to rule 1
        _one.first = ((_one.first + 1) %
                        size1);
 
        // see point 2
        if (_one.first == 0)
          _one.second = (_one.second + 1) %
                          size2;
      }
    }
    else if (a1[_one.first] +
             a2[_one.second] >
             a2[_two.first] +
             a1[_two.second])
    {
      if (a2[_two.first] <
          a1[_two.second])
      {
        Console.Write("[" + a2[_two.first] +
                      ", " + a1[_two.second] +
                      "] ");
 
        // updating according to rule 1
        _two.first = ((_two.first + 1) %
                        size2);
 
        // see point 2
        if (_two.first == 0)
          _two.second = (_two.second + 1) %
                          size1;
      }
      else {
        Console.Write("[" + a1[_two.second] +
                      ", " + a2[_two.first] +
                      "] ");
 
        // updating according to rule 1
        _two.second = ((_two.second + 1) %
                         size1);
 
        // see point 2
        if (_two.second == 0)
          _two.first = (_two.first + 1) %
                         size1;
      }
    }
    cnt++;
  }
}
 
// Driver Code
public static void Main(String[] args)
{
  int []a1 = {2, 3, 4};
  int []a2 = {1, 6, 5, 8};
  int size1 = a1.Length;
  int size2 = a2.Length;
  int k = 4;
  printKPairs(a1, a2,
              size1,
              size2, k);
}
}
 
// This code is contributed by gauravrajput1
Output: 
[1, 2] [1, 3] [1, 4] [2, 6]







 

Time complexity: O(K)

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