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
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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 <bits/stdc++.h>using namespace std;// Function to find the required pairsvoid 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 codeint 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 approachimport 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 pairsdef 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 codearr = [4, 6, 10, 23, 14, 7, 2, 20, 9]n = len(arr)k = 4d = 3findPairs(arr, n, k, d)# This code is contributed by mohit kumar 29 |
C#
// C# implementation of the approachusing 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 pairsfunction 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 leto 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 codelet 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> |
(2, 10) (4, 14) (6, 20) (7, 23)
