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> |
(2, 4)
Time Complexity: O(N2)
Auxiliary Space: O(1)
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