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

• Difficulty Level : Medium
• Last Updated : 30 Mar, 2021

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``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 for minimum distance``      ``if` `(j - i < min_dist)``      ``{` `        ``// Check if one is a``        ``// multiple of other``        ``if` `(a[i] % a[j] == 0 || a[j] % a[i] == 0)``        ``{` `          ``// 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 for minimum distance``                ``if` `(j - i < min_dist) {` `                    ``// Check if one is a``                    ``// multiple of other``                    ``if` `(a[i] % a[j] == ``0``                        ``|| a[j] % a[i] == ``0``) {` `                        ``// 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 for minimum distance``            ``if` `(j ``-` `i < min_dist):` `                ``# Check if one is a``                ``# multiple of other``                ``if` `((a[i] ``%` `a[j] ``=``=` `0``) ``or``                    ``(a[j] ``%` `a[i] ``=``=` `0``)):``        ` `                    ``# 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 for minimum distance``            ``if` `(j - i < min_dist)``            ``{``                ` `                ``// Check if one is a``                ``// multiple of other``                ``if` `(a[i] % a[j] == 0 ||``                    ``a[j] % a[i] == 0)``                ``{``                    ` `                    ``// 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

 ``
Output:
`(2, 4)`

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

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