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

Last Updated : 23 Jul, 2025

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# 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)

Analysis of Algorithms

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