Find k pairs with smallest sums in two arrays

3.6

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)

  1. 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.
  2. 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++

// 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;
}

Java

// Java code to print first k pairs with least
// sum from two arrays.
import java.io.*;
 
class KSmallestPair
{
    // Function to find k pairs with least sum such
    // that one elemennt of a pair is from arr1[] and
    // other element is from arr2[]
    static void kSmallestPair(int arr1[], int n1, int arr2[],
                                            int n2, int k)
    {
        if (k > n1*n2)
        {
            System.out.print("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[] = new int[n1];
     
        while (k > 0)
        {
            // Initialize current pair sum as infinite
            int min_sum = Integer.MAX_VALUE;
            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]];
                }
            }
     
            System.out.print("(" + arr1[min_index] + ", " +
                            arr2[index2[min_index]]+ ") ");
     
            index2[min_index]++;
            k--;
        }
    }

    // Driver code
    public static void main (String[] args)
    {
        int arr1[] = {1, 3, 11};
        int n1 = arr1.length;
     
        int arr2[] = {2, 4, 8};
        int n2 = arr2.length;
     
        int k = 4;
        kSmallestPair( arr1, n1, arr2, n2, k);
    }
}
/*This code is contributed by Prakriti Gupta*/


Output:

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

Time Complexity : O(k*n1)

Reference :
http://sudhansu-codezone.blogspot.in/2012/02/triplets-in-array-with-sum-0.html

This article is contributed by Sahil Chhabra . If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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