# Find k ordered pairs in array with minimum difference d

Given an array arr[] and two integers K and D, the task is to find exactly K pairs (arr[i], arr[j]) from the array such that |arr[i] – arr[j]| ≥ D and i != j. If it is impossible to get such pairs then print -1. Note that a single element can only participate in a single pair.

Examples:

Input: arr[] = {4, 6, 10, 23, 14, 7, 2, 20, 9}, K = 4, D = 3
Output:
(2, 10)
(4, 14)
(6, 20)
(7, 23)

Input: arr[] = {2, 10, 4, 6, 12, 5, 7, 3, 1, 9}, K = 5, D = 10
Output : -1

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

Approach: If we had to find only 1 pair then we would have checked only the maximum and minimum element from the array. Similarly, to get K pairs we can compare minimum K elements with the corresponding maximum K elements. Sorting can be used to get the minimum and maximum elements.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to find the required pairs ` `void` `findPairs(``int` `arr[], ``int` `n, ``int` `k, ``int` `d) ` `{ ` ` `  `    ``// There has to be atleast 2*k elements ` `    ``if` `(n < 2 * k) { ` `        ``cout << -1; ` `        ``return``; ` `    ``} ` ` `  `    ``// To store the pairs ` `    ``vector > pairs; ` ` `  `    ``// Sort the given array ` `    ``sort(arr, arr + n); ` ` `  `    ``// For every possible pair ` `    ``for` `(``int` `i = 0; i < k; i++) { ` ` `  `        ``// If the current pair is valid ` `        ``if` `(arr[n - k + i] - arr[i] >= d) { ` ` `  `            ``// Insert it into the pair vector ` `            ``pair<``int``, ``int``> p = make_pair(arr[i], arr[n - k + i]); ` `            ``pairs.push_back(p); ` `        ``} ` `    ``} ` ` `  `    ``// If k pairs are not possible ` `    ``if` `(pairs.size() < k) { ` `        ``cout << -1; ` `        ``return``; ` `    ``} ` ` `  `    ``// Print the pairs ` `    ``for` `(``auto` `v : pairs) { ` `        ``cout << ``"("` `<< v.first << ``", "` `             ``<< v.second << ``")"` `<< endl; ` `    ``} ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 4, 6, 10, 23, 14, 7, 2, 20, 9 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]); ` `    ``int` `k = 4, d = 3; ` ` `  `    ``findPairs(arr, n, k, d); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` `import` `java.util.*; ` ` `  `class` `GFG ` `{ ` `    ``static` `class` `pair  ` `    ``{  ` `        ``int` `first, second;  ` `        ``public` `pair(``int` `first, ``int` `second)  ` `        ``{  ` `            ``this``.first = first;  ` `            ``this``.second = second;  ` `        ``}  ` `    ``} ` ` `  `    ``// Function to find the required pairs ` `    ``static` `void` `findPairs(``int` `arr[], ``int` `n, ` `                          ``int` `k, ``int` `d) ` `    ``{ ` ` `  `        ``// There has to be atleast 2*k elements ` `        ``if` `(n < ``2` `* k) ` `        ``{ ` `            ``System.out.print(-``1``); ` `            ``return``; ` `        ``} ` ` `  `        ``// To store the pairs ` `        ``Vector pairs = ``new` `Vector(); ` ` `  `        ``// Sort the given array ` `        ``Arrays.sort(arr); ` ` `  `        ``// For every possible pair ` `        ``for` `(``int` `i = ``0``; i < k; i++)  ` `        ``{ ` ` `  `            ``// If the current pair is valid ` `            ``if` `(arr[n - k + i] - arr[i] >= d)  ` `            ``{ ` ` `  `                ``// Insert it into the pair vector ` `                ``pair p = ``new` `pair(arr[i],  ` `                                  ``arr[n - k + i]); ` `                ``pairs.add(p); ` `            ``} ` `        ``} ` ` `  `        ``// If k pairs are not possible ` `        ``if` `(pairs.size() < k)  ` `        ``{ ` `            ``System.out.print(-``1``); ` `            ``return``; ` `        ``} ` ` `  `        ``// Print the pairs ` `        ``for` `(pair v : pairs) ` `        ``{ ` `            ``System.out.println(``"("` `+ v.first +  ` `                               ``", "` `+ v.second + ``")"``); ` `        ``} ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `main(String[] args)  ` `    ``{ ` `        ``int` `arr[] = { ``4``, ``6``, ``10``, ``23``, ``14``, ``7``, ``2``, ``20``, ``9` `}; ` `        ``int` `n = arr.length; ` `        ``int` `k = ``4``, d = ``3``; ` `     `  `        ``findPairs(arr, n, k, d); ` `    ``} ` `} ` ` `  `// This code is contributed by 29AjayKumar `

## Python3

 `# Python3 implementation of the approach ` ` `  `# Function to find the required pairs ` `def` `findPairs(arr, n, k, d): ` ` `  `    ``# There has to be atleast 2*k elements ` `    ``if` `(n < ``2` `*` `k): ` `        ``print``(``"-1"``) ` `        ``return` ` `  `    ``# To store the pairs ` `    ``pairs``=``[] ` ` `  `    ``# Sort the given array ` `    ``arr``=``sorted``(arr) ` ` `  `    ``# For every possible pair ` `    ``for` `i ``in` `range``(k): ` ` `  `        ``# If the current pair is valid ` `        ``if` `(arr[n ``-` `k ``+` `i] ``-` `arr[i] >``=` `d): ` ` `  `            ``# Insert it into the pair vector ` `            ``pairs.append([arr[i], arr[n ``-` `k ``+` `i]]) ` ` `  ` `  `    ``# If k pairs are not possible ` `    ``if` `(``len``(pairs) < k): ` `        ``print``(``"-1"``) ` `        ``return` ` `  `    ``# Print the pairs ` `    ``for` `v ``in` `pairs: ` `        ``print``(``"("``,v[``0``],``", "``,v[``1``],``")"``) ` ` `  `# Driver code ` ` `  `arr ``=` `[``4``, ``6``, ``10``, ``23``, ``14``, ``7``, ``2``, ``20``, ``9``] ` `n ``=` `len``(arr) ` `k ``=` `4` `d ``=` `3` ` `  `findPairs(arr, n, k, d) ` ` `  `# This code is contributed by mohit kumar 29 `

## C#

 `// C# implementation of the approach  ` `using` `System; ` `using` `System.Collections.Generic; ` ` `  `class` `GFG ` `{ ` `    ``public` `class` `pair  ` `    ``{  ` `        ``public` `int` `first, second;  ` `        ``public` `pair(``int` `first, ``int` `second)  ` `        ``{  ` `            ``this``.first = first;  ` `            ``this``.second = second;  ` `        ``}  ` `    ``} ` ` `  `    ``// Function to find the required pairs ` `    ``static` `void` `findPairs(``int` `[]arr, ``int` `n, ` `                          ``int` `k, ``int` `d) ` `    ``{ ` ` `  `        ``// There has to be atleast 2*k elements ` `        ``if` `(n < 2 * k) ` `        ``{ ` `            ``Console.Write(-1); ` `            ``return``; ` `        ``} ` ` `  `        ``// To store the pairs ` `        ``List pairs = ``new` `List(); ` ` `  `        ``// Sort the given array ` `        ``Array.Sort(arr); ` ` `  `        ``// For every possible pair ` `        ``for` `(``int` `i = 0; i < k; i++)  ` `        ``{ ` ` `  `            ``// If the current pair is valid ` `            ``if` `(arr[n - k + i] - arr[i] >= d)  ` `            ``{ ` ` `  `                ``// Insert it into the pair vector ` `                ``pair p = ``new` `pair(arr[i],  ` `                                  ``arr[n - k + i]); ` `                ``pairs.Add(p); ` `            ``} ` `        ``} ` ` `  `        ``// If k pairs are not possible ` `        ``if` `(pairs.Count < k)  ` `        ``{ ` `            ``Console.Write(-1); ` `            ``return``; ` `        ``} ` ` `  `        ``// Print the pairs ` `        ``foreach` `(pair v ``in` `pairs) ` `        ``{ ` `            ``Console.WriteLine (``"("` `+ v.first +  ` `                               ``", "` `+ v.second + ``")"``); ` `        ``} ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `Main(String[] args)  ` `    ``{ ` `        ``int` `[]arr = { 4, 6, 10, 23,  ` `                      ``14, 7, 2, 20, 9 }; ` `        ``int` `n = arr.Length; ` `        ``int` `k = 4, d = 3; ` `     `  `        ``findPairs(arr, n, k, d); ` `    ``} ` `} ` ` `  `// This code is contributed by PrinciRaj1992  `

Output:

```(2, 10)
(4, 14)
(6, 20)
(7, 23)
```

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