# 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}

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 `

Output:

```1 3
```

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

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