Closest pair in an Array such that one number is multiple of the other

Given an array arr[] of integers of size N, the task is to find the closest pair in the given array such that one element is the multiple of the other. If no such pair exists then print -1.

Note: Closest pair means the difference between the index of any two elements must be minimum.

Examples:

Input: arr[] = {2, 3, 4, 5, 6}
Output: 2 4
Explanation:
The only possible pairs are (2, 4), (2, 6), (3, 6) out of which the pair which have minimum distance between them is (2, 4). 

Input: arr[] = { 2, 3, 6, 4, 5 }
Output: 3 6
Explanation:
The only possible pairs are (2, 4), (2, 6), (3, 6) out of which the pair which have minimum distance between them is (3, 6).

Approach: The idea is to generate all possible pairs of the given array and check if there exists any pair of elements in the array if one element is the multiple of other and update the required minimum distance with the current pair. After the above operation print, the pair have the minimum distance among them and one is a multiple of the other. If no such pair exists then print -1.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include<bits/stdc++.h>
using namespace std;

// Function to find the minimum
// distance pair where one is the
// multiple of the other
void findPair(int a[], int n)
{

  // Initialize the variables
  int min_dist = INT_MAX;
  int index_a = -1, index_b = -1;

  // Iterate for all the elements
  for (int i = 0; i < n; i++)
  {

    // Loop to make pairs
    for (int j = i + 1; j < n; j++)
    {
        // Check if one is a
        // multiple of other
          // and have minimum distance
        if ((a[i] % a[j] == 0 || a[j] % a[i] == 0) and j-i<min_dist)
        {

          // Update the distance
          min_dist = j - i;

          // Store indexes
          index_a = i;
          index_b = j;
        }
    }
  }

  // If no such pair exists
  if (index_a == -1)
  {
    cout << ("-1");
  }

  // Print the answer
  else
  {
    cout << "(" <<  a[index_a]
       <<  ", " <<  a[index_b]  << ")";
  }
}

// Driver Code
int main()
{
  // Given array arr[]
  int a[] = { 2, 3, 4, 5, 6 };
  int n = sizeof(a)/sizeof(int);

  // Function Call
  findPair(a, n);
}

// This code is contributed by rock_cool

Java

// Java program for the above approach
import java.util.*;
class GFG {

    // Function to find the minimum
    // distance pair where one is the
    // multiple of the other
    public static void
    findPair(int a[], int n)
    {

        // Initialize the variables
        int min_dist = Integer.MAX_VALUE;
        int index_a = -1, index_b = -1;

        // Iterate for all the elements
        for (int i = 0; i < n; i++) {

            // Loop to make pairs
            for (int j = i + 1; j < n; j++) {

                    // Check if one is a
                    // multiple of other
                    if ((a[i] % a[j] == 0
                        || a[j] % a[i] == 0) && j-i<min_dist) {

                        // Update the distance
                        min_dist = j - i;

                        // Store indexes
                        index_a = i;
                        index_b = j;
                    }
            }
        }

        // If no such pair exists
        if (index_a == -1) {
            System.out.println("-1");
        }

        // Print the answer
        else {
            System.out.print(
                "(" + a[index_a]
                + ", "
                + a[index_b] + ")");
        }
    }

    // Driver Code
    public static void
        main(String[] args)
    {
        // Given array arr[]
        int a[] = { 2, 3, 4, 5, 6 };
        int n = a.length;

        // Function Call
        findPair(a, n);
    }
}

Python3

# Python3 program for the above approach
import sys

# Function to find the minimum
# distance pair where one is the
# multiple of the other
def findPair(a, n):

    # Initialize the variables
    min_dist = sys.maxsize
    index_a = -1
    index_b = -1

    # Iterate for all the elements
    for i in range(n):

        # Loop to make pairs
        for j in range(i + 1, n):

          # Check if one is a
          # multiple of other
          if (((a[i] % a[j] == 0) or
              (a[j] % a[i] == 0)) and j-i<min_dist):

            # Update the distance
            min_dist = j - i

            # Store indexes
            index_a = i
            index_b = j

    # If no such pair exists
    if (index_a == -1):
        print("-1")

    # Print the answer
    else:
        print("(", a[index_a],
             ", ", a[index_b], ")")

# Driver Code

# Given array arr[]
a = [ 2, 3, 4, 5, 6 ]

n = len(a)

# Function call
findPair(a, n)

# This code is contributed by sanjoy_62

C#

// C# program for the above approach
using System;

class GFG{

// Function to find the minimum
// distance pair where one is the
// multiple of the other
public static void findPair(int []a, int n)
{

    // Initialize the variables
    int min_dist = int.MaxValue;
    int index_a = -1, index_b = -1;

    // Iterate for all the elements
    for(int i = 0; i < n; i++)
    {

        // Loop to make pairs
        for(int j = i + 1; j < n; j++)
        {

            // Check if one is a
            // multiple of other
            if ((a[i] % a[j] == 0 ||
                a[j] % a[i] == 0) && j-i<min_dist)
            {

              // Update the distance
              min_dist = j - i;

              // Store indexes
              index_a = i;
              index_b = j;
            }
        }
    }

    // If no such pair exists
    if (index_a == -1)
    {
        Console.WriteLine("-1");
    }

    // Print the answer
    else
    {
        Console.Write("(" + a[index_a] +
                     ", " + a[index_b] + ")");
    }
}

// Driver Code
public static void Main(String[] args)
{

    // Given array []arr
    int []a = { 2, 3, 4, 5, 6 };

    int n = a.Length;

    // Function Call
    findPair(a, n);
}
}

// This code is contributed by 29AjayKumar

Javascript

<script>

// Javascript program for the above approach

// Function to find the minimum
// distance pair where one is the
// multiple of the other
function findPair(a, n)
{

    // Initialize the variables
    let min_dist = Number.MAX_VALUE;
    let index_a = -1, index_b = -1;

    // Iterate for all the elements
    for(let i = 0; i < n; i++)
    {

        // Loop to make pairs
        for(let j = i + 1; j < n; j++)
        {


                // Check if one is a
                // multiple of other
                if ((a[i] % a[j] == 0 ||
                    a[j] % a[i] == 0) && j-i<min_dist)
                {

                // Update the distance
                min_dist = j - i;

                // Store indexes
                index_a = i;
                index_b = j;
                }
        }
    }

    // If no such pair exists
    if (index_a == -1)
    {
        document.write("-1");
    }

    // Print the answer
    else
    {
        document.write("(" + a[index_a] +
                      ", " + a[index_b] + ")");
    }
}

// Driver code

// Given array arr[]
let a = [ 2, 3, 4, 5, 6 ];
let n = a.length;

// Function Call
findPair(a, n);

// This code is contributed by divyesh072019

</script>

Output:

(2, 4)

Time Complexity: O(N²)
Auxiliary Space: O(1)