Find k ordered pairs in array with minimum difference d

• Last Updated : 08 Jun, 2021

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]| â‰¥ 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

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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`

Javascript

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

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