Find k ordered pairs in array with minimum difference d

Last Updated : 23 Jun, 2022

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

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 <bits/stdc++.h>
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<pair<int, int> > 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<pair> pairs = new Vector<pair>();
 
        // 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<pair> pairs = new List<pair>();
 
        // 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

   <script>
 
// JavaScript implementation of the approach
 
// Function to find the required pairs
function findPairs(arr, n, k, d) {
 
    // There has to be atleast 2*k elements
    if (n < 2 * k) {
        document.write(-1);
        return;
    }
 
    // To store the pairs
    let pairs = [];
 
    // Sort the given array
    arr.sort((a, b) => a - b);
 
    // For every possible pair
    for (let 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
            let p = [arr[i], arr[n - k + i]];
            pairs.push(p);
        }
    }
 
    // If k pairs are not possible
    if (pairs.length < k) {
        document.write(-1);
        return;
    }
 
    // Print the pairs
    for (let v of pairs) {
        document.write("(" + v[0] + ", " + v[1] + ")" + "<br>");
    }
}
 
// Driver code
 
let arr = [4, 6, 10, 23, 14, 7, 2, 20, 9];
let n = arr.length;
let k = 4, d = 3;
 
findPairs(arr, n, k, d);
 
 
// This code is contributed by gfgking
 
</script>
 
  

Output:

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

Time Complexity: O(n*log(n))
Auxiliary Space: O(n)

C++ Program to Find a pair with the given difference

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Practice Tags :

Find k ordered pairs in array with minimum difference d

C++

Java

Python3

C#

Javascript

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