Skip to content
Related Articles

Related Articles

Find N – 1 pairs from given array such that GCD of all pair-sums is greater than 1

View Discussion
Improve Article
Save Article
Like Article
  • Last Updated : 29 Mar, 2022

Given an array arr[] of 2 * N integers, the task is to find a set of N – 1 pairs such that the GCD of all pair sums is greater than 1.

Examples:

Input: arr[] = {1, 2, 3, 4, 5, 6}
Output:
1 3
2 4
Explanation: The given array has 3 * 2 elements. The pair (1, 3) and (2, 4) has sum of elements as 4 and 6 respectively. Also, GCD(4, 6) > 1. Hence, (1, 3) and (2, 4) are the required pairs. Other possible pairs are (1, 5) and (2, 6), (3, 5) and (2, 5), etc..

Input: arr[] = {1, 1, 1, 2, 3, 4}
Output:
1 3
2 4

 

Approach: The given problem is an observation-based problem. The idea is the create N – 1 pairs such that their sum is even. Also, for a pair to have an even sum, either both the integers should be even or both should be odd. Below are the steps to follow:

  • Initialize two vectors Odd and Even to store odd and even integers respectively.
  • Traverse the given array arr[] and insert odd integers in Odd vector and even integers in Even vector.
  • Pair the consecutive elements of both the vectors Odd and Even and print the N – 1 first pair of integers.

Below is the implementation of the above approach:

C++




// C++ Program of the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find N - 1 pairs from
// given array such that GCD of all
// pair-sums is greater than 1
void findPairs(int arr[], int N)
{
    // Stores even and odd
    // integers respectively
    vector<int> even, odd;
 
    // Loop to iterate arr[]
    for (int i = 0; i < 2 * N; i++) {
        if (arr[i] & 1)
            odd.push_back(arr[i]);
        else
            even.push_back(arr[i]);
    }
 
    // Stores the required pairs
    vector<pair<int, int> > ans;
 
    // Insert all possible pairs
    // from odd elements
    for (int i = 0; i + 1 < odd.size(); i += 2)
        ans.push_back({ odd[i], odd[i + 1] });
 
    // Insert all possible pairs
    // from even elements
    for (int i = 0; i + 1 < even.size(); i += 2)
        ans.push_back({ even[i], even[i + 1] });
 
    // Print Answer
    for (int i = 0; i < N - 1; i++) {
        cout << ans[i].first << " " << ans[i].second
             << endl;
    }
}
 
// Driver Code
int main()
{
    int arr[] = { 1, 2, 3, 4, 5, 6 };
    int N = 3;
    findPairs(arr, N);
 
    return 0;
}

Java




// JAVA Program of the above approach
import java.util.*;
class Pair {
  int x;
  int y;
 
  // Constructor
  public Pair(int x, int y)
  {
    this.x = x;
    this.y = y;
  }
}
class GFG
{
 
  // Function to find N - 1 pairs from
  // given array such that GCD of all
  // pair-sums is greater than 1
  public static void findPairs(int arr[], int N)
  {
 
    // Stores even and odd
    // integers respectively
    ArrayList<Integer> even = new ArrayList<Integer>();
    ArrayList<Integer> odd = new ArrayList<Integer>();
 
    // Loop to iterate arr[]
    for (int i = 0; i < 2 * N; i++) {
      if (arr[i] % 2 == 1)
        odd.add(arr[i]);
      else
        even.add(arr[i]);
    }
 
    // Stores the required pairs
    Pair[] ans1 = new Pair[N];
    Pair[] ans2 = new Pair[N];
 
    // Insert all possible pairs
    // from odd elements
    for (int i = 0; i + 1 < odd.size(); i += 2) {
      ans1[i] = new Pair(odd.get(i), odd.get(i + 1));
    }
 
    // Insert all possible pairs
    // from even elements
    for (int i = 0; i + 1 < even.size(); i += 2) {
      ans2[i]
        = new Pair(even.get(i), even.get(i + 1));
    }
 
    // Print Answer
    for (int i = 0; i < N - 1; i++) {
      System.out.println(ans1[i].x + " " + ans1[i].y);
      System.out.println(ans2[i].x + " " + ans2[i].y);
      break;
    }
  }
 
  // Driver Code
  public static void main(String[] args)
  {
    int[] arr = new int[] { 1, 2, 3, 4, 5, 6 };
    int N = 3;
    findPairs(arr, N);
  }
}
 
// This code is contributed by Taranpreet

Python3




# Python code for the above approach
 
# Function to find N - 1 pairs from
# given array such that GCD of all
# pair-sums is greater than 1
def findPairs(arr, N):
 
    # Stores even and odd
    # integers respectively
    even = []
    odd = [];
 
    # Loop to iterate arr[]
    for i in range(2 * N):
        if (arr[i] & 1):
            odd.append(arr[i]);
        else:
            even.append(arr[i]);
     
 
    # Stores the required pairs
    ans = [];
 
    # Insert all possible pairs
    # from odd elements
    i = 0;
    while(i + 1 < len(odd)):
        ans.append({ "first": odd[i], "second": odd[i + 1] });
        i += 2
 
    # Insert all possible pairs
    # from even elements
    i = 0
    while(i + 1 < len(odd)):
        ans.append({ "first": even[i], "second": even[i + 1] });
        i += 2
 
    # Print Answer
    for i in range(N - 1):
        for y in ans[i].values():
            print(y, end=" ")
        print("")
 
# Driver Code
arr = [1, 2, 3, 4, 5, 6];
N = 3;
findPairs(arr, N);
 
# This code is contributed by Saurabh Jaiswal

C#




// C# Program of the above approach
using System;
using System.Collections.Generic;
 
public class Pair {
  public int x;
  public int y;
 
  // Constructor
  public Pair(int x, int y) {
    this.x = x;
    this.y = y;
  }
}
 
public class GFG
{
 
  // Function to find N - 1 pairs from
  // given array such that GCD of all
  // pair-sums is greater than 1
  public static void findPairs(int []arr, int N) {
 
    // Stores even and odd
    // integers respectively
    List<int> even = new List<int>();
    List<int> odd = new List<int>();
 
    // Loop to iterate []arr
    for (int i = 0; i < 2 * N; i++) {
      if (arr[i] % 2 == 1)
        odd.Add(arr[i]);
      else
        even.Add(arr[i]);
    }
 
    // Stores the required pairs
    Pair[] ans1 = new Pair[N];
    Pair[] ans2 = new Pair[N];
 
    // Insert all possible pairs
    // from odd elements
    for (int i = 0; i + 1 < odd.Count; i += 2) {
      ans1[i] = new Pair(odd[i], odd[i + 1]);
    }
 
    // Insert all possible pairs
    // from even elements
    for (int i = 0; i + 1 < even.Count; i += 2) {
      ans2[i] = new Pair(even[i], even[i + 1]);
    }
 
    // Print Answer
    for (int i = 0; i < N - 1; i++) {
      Console.WriteLine(ans1[i].x + " " + ans1[i].y);
      Console.WriteLine(ans2[i].x + " " + ans2[i].y);
      break;
    }
  }
 
  // Driver Code
  public static void Main(String[] args) {
    int[] arr = new int[] { 1, 2, 3, 4, 5, 6 };
    int N = 3;
    findPairs(arr, N);
  }
}
 
// This code is contributed by Rajput-Ji

Javascript




<script>
       // JavaScript code for the above approach
 
       // Function to find N - 1 pairs from
       // given array such that GCD of all
       // pair-sums is greater than 1
       function findPairs(arr, N)
       {
        
           // Stores even and odd
           // integers respectively
           let even = [], odd = [];
 
           // Loop to iterate arr[]
           for (let i = 0; i < 2 * N; i++) {
               if (arr[i] & 1)
                   odd.push(arr[i]);
               else
                   even.push(arr[i]);
           }
 
           // Stores the required pairs
           let ans = [];
 
           // Insert all possible pairs
           // from odd elements
           for (let i = 0; i + 1 < odd.length; i += 2)
               ans.push({ first: odd[i], second: odd[i + 1] });
 
           // Insert all possible pairs
           // from even elements
           for (let i = 0; i + 1 < even.length; i += 2)
               ans.push({ first: even[i], second: even[i + 1] });
 
           // Print Answer
           for (let i = 0; i < N - 1; i++) {
               document.write(ans[i].first + " " + ans[i].second
                   + "<br>");
           }
       }
 
       // Driver Code
       let arr = [1, 2, 3, 4, 5, 6];
       let N = 3;
       findPairs(arr, N);
 
      // This code is contributed by Potta Lokesh
   </script>

 
 

Output

1 3
2 4

 

Time Complexity: O(N)
Auxiliary Space: O(N)

 


My Personal Notes arrow_drop_up
Recommended Articles
Page :

Start Your Coding Journey Now!