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

• Difficulty Level : Expert
• Last Updated : 15 Nov, 2021

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 … DN – 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 ``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

 ``
Output:
`1 3`

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

Auxiliary Space: O(1)

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