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

26 May, 2021
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 = 3Let 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 11st 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 7Repeat the above process for remaining K pairs
Below is the C++ implementation of the above approach:  
C++
Java
Python3
C#
Javascript
// 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 pairsvoid 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 Codeint 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 program to print// the k smallest pairs// | Set 2import java.util.*;class GFG{ static class _pair{  int first, second;}; // Function to print the K// smallest pairsstatic 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 Codepublic 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 program to print the k smallest# pairs | Set 2 # Function to print the K smallest pairsdef 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 Codeif __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# program to print// the k smallest pairs// | Set 2using System;class GFG{ public class _pair{  public int first,             second;}; // Function to print the K// smallest pairsstatic 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 Codepublic 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
<script>// Javascript program to print// the k smallest pairs// | Set 2 // Function to print the K// smallest pairsfunction printKPairs(a1,a2,size1,size2,k){    // if k is greater than  // total pairs  if (k > (size2 * size1))  {    document.write("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 a2[] and _two[1]  // in a1[]  let _one = [0,0];  let  _two = [0,0];    let 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[0] == _two[1] &&        _two[0] == _one[1])    {      if (a1[_one[0]] <          a2[_one[1]])      {        document.write("[" +  a1[_one[0]] +                         ", " +  a2[_one[1]] +                         "] ");          // updates according to step 1        _one[1] = (_one[1] + 1) %                       size2;                  // see point 2        if (_one[1] == 0)          _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      {        document.write("[" +  a2[_one[1]] +                         ", " +  a1[_one[0]] +                         "] ");          // updates according to rule 1        _one[0] = (_one[0] + 1) %                      size1;                  // see point 2        if (_one[0] == 0)          _one[1] = (_one[1] + 1) %                         size2;          // updates opposite to rule 1        _two[0] = (_two[0] + 1) %                      size2;                  // see point 2        if (_two[0] == 0)                      _two[1] = (_two[1] + 1) %                         size1;      }    }          // else update as    // necessary (point 1)    else if (a1[_one[0]] +             a2[_one[1]] <=             a2[_two[0]] +             a1[_two[1]])    {      if (a1[_one[0]] <          a2[_one[1]])      {        document.write("[" +  a1[_one[0]] +                         ", " + a2[_one[1]] +                         "] ");          // updating according to rule 1        _one[1] = ((_one[1] + 1) %                        size2);                  // see point 2        if (_one[1] == 0)          _one[0] = (_one[0] + 1) %                        size1;      }      else      {        document.write("[" +  a2[_one[1]] +                         ", " + a1[_one[0]] +                         "] ");          // updating according to rule 1        _one[0] = ((_one[0] + 1) %                       size1);                  // see point 2        if (_one[0] == 0)          _one[1] = (_one[1] + 1) %                         size2;      }    }    else if (a1[_one[0]] +             a2[_one[1]] >             a2[_two[0]] +             a1[_two[1]])    {      if (a2[_two[0]] <          a1[_two[1]])      {        document.write("[" +  a2[_two[0]] +                         ", " +  a1[_two[1]] +                         "] ");          // updating according to rule 1        _two[0] = ((_two[0] + 1) %                       size2);                  // see point 2        if (_two[0] == 0)          _two[1] = (_two[1] + 1) %                         size1;      }      else {        document.write("[" +  a1[_two[1]] +                         ", " +  a2[_two[0]] +                         "] ");          // updating according to rule 1        _two[1] = ((_two[1] + 1) %                        size1);                  // see point 2        if (_two[1] == 0)          _two[0] = (_two[0] + 1) %                        size1;      }    }    cnt++;  }} // Driver Codelet a1=[2, 3, 4];let a2=[1, 6, 5, 8];let size1 = a1.length;let size2 = a2.length;let k = 4;printKPairs(a1, a2,              size1, size2, k);     // This code is contributed by unknown2108</script>
[1, 2] [1, 3] [1, 4] [2, 6]
 
Time complexity: O(K)
mohit kumar 29
Code_Mech
GauravRajput1
unknown2108
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