Find the Kth pair in ordered list of all possible sorted pairs of the Array

Given an array arr[] containing N integers and a number K, the task is to find the K-th pair in the ordered list of all possible N² sorted pairs of the array arr[].  

A pair (p1, q1) is lexicographically smaller than the pair (p2, q2) only if p1 ? p2 and q1 < q2.

Examples:

Input: arr[] = {2, 1}, K = 4 
Output: {2, 2} 
Explanation:
The sorted sequence for the given array is {1, 1}, {1, 2}, {2, 1}, {2, 2}. So the 4th pair is {2, 2}.
Input: arr[] = {3, 1, 5}, K = 2 
Output: {1, 3}  

Approach: Naturally, K-th sorted pair from all possible set of pairs will be {arr[K/N], arr[K%N]}. But, this method works only if all the elements in the array are unique. Therefore, the following steps are followed to make the array behave like a unique array:  

• Let the array arr[] be {X, X, X, … D₁, D₂, D₃ … D_{N – T}}.
• Here, let’s assume the number of repeating elements in the array to be T and the element which is being repeated be X. So, the number of distinct elements in the array is (N – T).
• Now, from the first N * T pairs out of N² pairs of elements, the first T² elements will always be {X, X}.
• The next T elements will be {X, D₂} and the next T elements will be {X, D₂} and so on.
• So, if we need to find the K-th element, subtract N * T from K and skip the first T same elements.
• Repeat the above process until K becomes less than N * T.
• At this step, the first element in the pair would be the counter variable ‘i’. The second element would be the remaining K-th element from the remaining elements which is K / T. So, the required answer is {arr[i], arr[K/T]}.

Below is the implementation of the above approach: 

C++

// C++ program to find the K-th pair
// in a lexicographically sorted array

#include <bits/stdc++.h>
using namespace std;

// Function to find the k-th pair
void kthpair(int n, int k, int arr[])
{
    int i, t;

    // Sorting the array
    sort(arr, arr + n);

    --k;

    // Iterating through the array
    for (i = 0; i < n; i += t) {

        // Finding the number of same elements
        for (t = 1; arr[i] == arr[i + t]; ++t)
            ;

        // Checking if N*T is less than the
        // remaining K. If it is, then arr[i]
        // is the first element in the required
        // pair
        if (t * n > k)
            break;

        k = k - t * n;
    }

    // Printing the K-th pair
    cout << arr[i] << ' ' << arr[k / t];
}

// Driver code
int main()
{

    int n = 3, k = 2;
    int arr[n] = { 3, 1, 5 };
    kthpair(n, k, arr);
}

Java

// Java program to find the K-th pair
// in a lexicographically sorted array
import java.util.*;
class GFG{

// Function to find the k-th pair
static void kthpair(int n, int k,
                    int arr[])
{
    int i, t = 0;

    // Sorting the array
    Arrays.sort(arr);

    --k;

    // Iterating through the array
    for (i = 0; i < n; i += t)
    {

        // Finding the number of same elements
        for (t = 1; arr[i] == arr[i + t]; ++t)
            ;

        // Checking if N*T is less than the
        // remaining K. If it is, then arr[i]
        // is the first element in the required
        // pair
        if (t * n > k)
            break;

        k = k - t * n;
    }

    // Printing the K-th pair
    System.out.print(arr[i] + " " +
                     arr[k / t]);
}

// Driver code
public static void main(String[] args)
{
    int n = 3, k = 2;
    int arr[] = { 3, 1, 5 };
    kthpair(n, k, arr);
}
}

// This code is contributed by 29AjayKumar

Python3

# Python3 program to find the K-th pair
# in a lexicographically sorted array

# Function to find the k-th pair
def kthpair(n, k, arr):

    # Sorting the array
    arr.sort()
    k -= 1

    # Iterating through the array
    i = 0
    while (i < n):

        # Finding the number of same elements
        t = 1
        while (arr[i] == arr[i + t]):
            t += 1

        # Checking if N*T is less than the
        # remaining K. If it is, then arr[i]
        # is the first element in the required
        # pair
        if (t * n > k):
            break
        k = k - t * n

        i += t

    # Printing the K-th pair
    print(arr[i], " ", arr[k // t])

# Driver code
if __name__ == "__main__":

    n, k = 3, 2
    arr = [ 3, 1, 5 ]

    kthpair(n, k, arr)

# This code is contributed by chitranayal

C#

// C# program to find the K-th pair
// in a lexicographically sorted array
using System;

class GFG{

// Function to find the k-th pair
static void kthpair(int n, int k,
                    int[] arr)
{
    int i, t = 0;

    // Sorting the array
    Array.Sort(arr);

    --k;

    // Iterating through the array
    for(i = 0; i < n; i += t)
    {

       // Finding the number of same elements
       for(t = 1; arr[i] == arr[i + t]; ++t);

          // Checking if N*T is less than the
          // remaining K. If it is, then arr[i]
          // is the first element in the required
          // pair
          if (t * n > k)
              break;
          k = k - t * n;
    }

    // Printing the K-th pair
    Console.Write(arr[i] + " " + arr[k / t]);
}

// Driver code
static public void Main ()
{
    int n = 3, k = 2;
    int[] arr = { 3, 1, 5 };

    kthpair(n, k, arr);
}
}

// This code is contributed by ShubhamCoder

Javascript

<script>

// Java program to find the K-th pair
// in a lexicographically sorted array


// Function to find the k-th pair
function kthpair(n,k,arr)
{
    let i, t = 0;

    // Sorting the array
    arr.sort();

    --k;

    // Iterating through the array
    for (i = 0; i < n; i += t)
    {

        // Finding the number of same elements
        for (t = 1; arr[i] == arr[i + t]; ++t)
            ;

        // Checking if N*T is less than the
        // remaining K. If it is, then arr[i]
        // is the first element in the required
        // pair
        if (t * n > k)
            break;

        k = k - t * n;
    }

    // Printing the K-th pair
    document.write(arr[i] + " " +
                     arr[k / t]);
}

// Driver code

    let n = 3, k = 2;
   let arr =[ 3, 1, 5 ];
    kthpair(n, k, arr);



//contributed by 171fa07058
</script>

Output:

1 3

Time Complexity: O(N * log(N)), where N is the size of the array.

Auxiliary Space: O(1)