Find pair with maximum GCD in an array
We are given an array of positive integers. Find the pair in array with maximum GCD.
Examples:
Input : arr[] : { 1 2 3 4 5 }
Output : 2
Explanation : Pair {2, 4} has GCD 2 which is highest. Other pairs have a GCD of 1.
Input : arr[] : { 2 3 4 8 8 11 12 }
Output : 8
Explanation : Pair {8, 8} has GCD 8 which is highest.
Method 1 (Brute-force): The simplest method to solve this problem is to use two loops to generate all possible pairs of elements of the array and calculate and compare the GCD at the same time. We can use the Extended Euclidean algorithm for efficiently computing GCD of two numbers.
Time Complexity: O(N^2 * log(max(a, b)))
Here, log(max(a, b)) is the time complexity to calculate GCD of a and b.
Method 2 : (Efficient) In this method, we maintain a count array to store the count of divisors of every element. We will traverse the given array and for every element, we will calculate its divisors and increment at the index of count array. The process of computing divisors will take O(sqrt(arr[i])) time, where arr[i] is element in the given array at index i. After the whole traversal, we can simply traverse the count array from last index to index 1. If we found an index with a value greater than 1, then this means that it is a divisor of 2 elements and also the max GCD.
Below is the implementation of above approach :
C++
// C++ Code to find pair with// maximum GCD in an array#include <bits/stdc++.h>using namespace std;// function to find GCD of pair with// max GCD in the arrayint findMaxGCD(int arr[], int n){ // Computing highest element int high = 0; for (int i = 0; i < n; i++) high = max(high, arr[i]); // Array to store the count of divisors // i.e. Potential GCDs int divisors[high + 1] = { 0 }; // Iterating over every element for (int i = 0; i < n; i++) { // Calculating all the divisors for (int j = 1; j <= sqrt(arr[i]); j++) { // Divisor found if (arr[i] % j == 0) { // Incrementing count for divisor divisors[j]++; // Element/divisor is also a divisor // Checking if both divisors are // not same if (j != arr[i] / j) divisors[arr[i] / j]++; } } } // Checking the highest potential GCD for (int i = high; i >= 1; i--) // If this divisor can divide at least 2 // numbers, it is a GCD of at least 1 pair if (divisors[i] > 1) return i; }// Driver codeint main(){ // Array in which pair with max GCD // is to be found int arr[] = { 1, 2, 4, 8, 8, 12 }; // Size of array int n = sizeof(arr) / sizeof(arr[0]); cout << findMaxGCD(arr,n); return 0;} |
Java
// JAVA Code for Find pair with maximum GCD in an arrayclass GFG { // function to find GCD of pair with // max GCD in the array public static int findMaxGCD(int arr[], int n) { // Computing highest element int high = 0; for (int i = 0; i < n; i++) high = Math.max(high, arr[i]); // Array to store the count of divisors // i.e. Potential GCDs int divisors[] =new int[high + 1]; // Iterating over every element for (int i = 0; i < n; i++) { // Calculating all the divisors for (int j = 1; j <= Math.sqrt(arr[i]); j++) { // Divisor found if (arr[i] % j == 0) { // Incrementing count for divisor divisors[j]++; // Element/divisor is also a divisor // Checking if both divisors are // not same if (j != arr[i] / j) divisors[arr[i] / j]++; } } } // Checking the highest potential GCD for (int i = high; i >= 1; i--) // If this divisor can divide at least 2 // numbers, it is a GCD of at least 1 pair if (divisors[i] > 1) return i; return 1; } /* Driver program to test above function */ public static void main(String[] args) { // Array in which pair with max GCD // is to be found int arr[] = { 1, 2, 4, 8, 8, 12 }; // Size of array int n = arr.length; System.out.println(findMaxGCD(arr,n)); } } // This code is contributed by Arnav Kr. Mandal. |
Python
# Python program to Find pair with# maximum GCD in an arrayimport math# function to find GCD of pair with# max GCD in the arraydef findMaxGCD(arr, n) : # Computing highest element high = 0 i = 0 while i < n : high = max(high, arr[i]) i = i + 1 # Array to store the count of divisors # i.e. Potential GCDs divisors = [0] * (high + 1) # Iterating over every element i = 0 while i < n : # Calculating all the divisors j = 1 while j <= math.sqrt(arr[i]) : # Divisor found if (arr[i] % j == 0) : # Incrementing count for divisor divisors[j]= divisors[j]+1 # Element/divisor is also a divisor # Checking if both divisors are # not same if (j != arr[i] / j) : divisors[arr[i] / j] = divisors[arr[i] / j] + 1 j = j + 1 i = i + 1 # Checking the highest potential GCD i = high while i >= 1 : # If this divisor can divide at least 2 # numbers, it is a GCD of at least 1 pair if (divisors[i] > 1) : return i i = i - 1 return 1# Driver code# Array in which pair with max GCD# is to be foundarr = [ 1, 2, 4, 8, 8, 12 ] # Size of arrayn = len(arr)print findMaxGCD(arr,n)# This code is contributed by Nikita Tiwari. |
C#
// C# Code for Find pair with// maximum GCD in an arrayusing System;class GFG { // Function to find GCD of pair // with max GCD in the array public static int findMaxGCD(int []arr, int n) { // Computing highest element int high = 0; for (int i = 0; i < n; i++) high = Math.Max(high, arr[i]); // Array to store the count of // divisors i.e. Potential GCDs int []divisors =new int[high + 1]; // Iterating over every element for (int i = 0; i < n; i++) { // Calculating all the divisors for (int j = 1; j <= Math.Sqrt(arr[i]); j++) { // Divisor found if (arr[i] % j == 0) { // Incrementing count // for divisor divisors[j]++; // Element / divisor is also // a divisor Checking if both // divisors are not same if (j != arr[i] / j) divisors[arr[i] / j]++; } } } // Checking the highest potential GCD for (int i = high; i >= 1; i--) // If this divisor can divide at // least 2 numbers, it is a // GCD of at least 1 pair if (divisors[i] > 1) return i; return 1; } // Driver Code public static void Main(String []args) { // Array in which pair with // max GCD is to be found int []arr = {1, 2, 4, 8, 8, 12}; // Size of array int n = arr.Length; Console.WriteLine(findMaxGCD(arr,n)); }}// This code is contributed by vt_m. |
PHP
<?php// PHP Code for Find pair with// maximum GCD in an array// Function to find GCD// of pair with max GCD// in the arrayfunction findMaxGCD($arr, $n){ // Computing highest element $high = 0; for ($i = 0; $i < $n; $i++) $high = max($high, $arr[$i]); // Array to store the // count of divisors // i.e. Potential GCDs $divisors = array_fill(0, $high + 1, 0); // Iterating over every element for ($i = 0; $i < $n; $i++) { // Calculating all // the divisors for ($j = 1; $j <= (int)(sqrt($arr[$i])); $j++) { // Divisor found if ($arr[$i] % $j == 0) { // Incrementing count // for divisor $divisors[$j]++; // Element/divisor is also // a divisor Checking if // both divisors are not same if ($j != (int)($arr[$i] / $j)) $divisors[(int)($arr[$i] / $j)]++; } } } // Checking the highest // potential GCD for ($i = $high; $i >= 1; $i--) // If this divisor can divide // at least 2 numbers, it is // a GCD of at least 1 pair if ($divisors[$i] > 1) return $i;}// Driver code// Array in which pair// with max GCD is to// be found$arr = array( 1, 2, 4, 8, 8, 12 );// Size of array$n = sizeof($arr);echo findMaxGCD($arr,$n);// This code is contributed by mits?> |
Javascript
<script>// JavaScript Code for Find pair with// maximum GCD in an array // function to find GCD of pair with// max GCD in the arrayfunction findMaxGCD(arr , n){ // Computing highest element var high = 0; for (var i = 0; i < n; i++) high = Math.max(high, arr[i]); // Array to store the count of divisors // i.e. Potential GCDs var divisors = Array.from({length: high + 1}, (_, i) => 0); // Iterating over every element for (var i = 0; i < n; i++) { // Calculating all the divisors for (var j = 1; j <= Math.sqrt(arr[i]); j++) { // Divisor found if (arr[i] % j == 0) { // Incrementing count for divisor divisors[j]++; // Element/divisor is also a divisor // Checking if both divisors are // not same if (j != arr[i] / j) divisors[arr[i] / j]++; } } } // Checking the highest potential GCD for (var i = high; i >= 1; i--) // If this divisor can divide at least 2 // numbers, it is a GCD of at least 1 pair if (divisors[i] > 1) return i; return 1;}/* Driver program to test above function */ // Array in which pair with max GCD// is to be found var arr = [ 1, 2, 4, 8, 8, 12 ]; // Size of array var n = arr.length; document.write(findMaxGCD(arr,n));// This code contributed by shikhasingrajput</script> |
Output:
8
Time Complexity: O(N * sqrt(arr[i]) + H) , where arr[i] denotes the element of the array and H denotes the largest number of the array.
Auxiliary Space: O(high), high is the maximum element in the array
Method 3 (Most Efficient): This approach is based on the idea of Sieve Of Eratosthenes.
First let’s solve a simpler problem, given a value X we have to tell whether a pair has a GCD equal to X. This can be done by checking that how many elements in the array are multiples of X. If the number of such multiples is greater than 1, then X will be a GCD of some pair.
Now for pair with maximum GCD, we maintain a count array of the original array. Our method is based on the above problem with Sieve-like approach for loop. Below is the step by step algorithm of this approach:
- Iterate ‘i’ from MAX (maximum array element) to 1.
- Iterate ‘j’ from ‘i’ to MAX. We will check if the count array is 1 at index ‘j’.
- Increment the index ‘j’ everytime with ‘i’. This way, we can check for
i, 2i, 3i, and so on. - If we get 1 two times at count array that means 2 multiples of i exists. This makes it the highest GCD.
Below is the implementation of above approach :
C++
// C++ Code to// Find pair with// maximum GCD in// an array#include <bits/stdc++.h>using namespace std;// function to find// GCD of pair with// max GCD in the// arrayint findMaxGCD(int arr[], int n){ // Calculating MAX in array int high = 0; for (int i = 0; i < n; i++) high = max(high, arr[i]); // Maintaining count array int count[high + 1] = {0}; for (int i = 0; i < n; i++) count[arr[i]]++; // Variable to store the // multiples of a number int counter = 0; // Iterating from MAX to 1 // GCD is always between // MAX and 1. The first // GCD found will be the // highest as we are // decrementing the potential // GCD for (int i = high; i >= 1; i--) { int j = i; counter = 0; // Iterating from current // potential GCD // till it is less than // MAX while (j <= high) { // A multiple found if(count[j] >=2) return j; else if (count[j] == 1) counter++; // Incrementing potential // GCD by itself // To check i, 2i, 3i.... j += i; // 2 multiples found, // max GCD found if (counter == 2) return i; } }}// Driver codeint main(){ // Array in which pair // with max GCD is to // be found int arr[] = { 1, 2, 4, 8, 8, 12 }; // Size of array int n = sizeof(arr) / sizeof(arr[0]); cout << findMaxGCD(arr, n); return 0;} |
Java
// Java Code to// Find pair with// maximum GCD in// an arrayclass GFG { // function to find // GCD of pair with // max GCD in the // array public static int findMaxGCD(int arr[], int n) { // Calculating MAX in // array int high = 0; for (int i = 0; i < n; i++) high = Math.max(high, arr[i]); // Maintaining count array int count[]=new int[high + 1]; for (int i = 0; i < n; i++) count[arr[i]]++; // Variable to store // the multiples of // a number int counter = 0; // Iterating from MAX // to 1 GCD is always // between MAX and 1 // The first GCD found // will be the highest // as we are decrementing // the potential GCD for (int i = high; i >= 1; i--) { int j = i; // Iterating from current // potential GCD till it // is less than MAX while (j <= high) { // A multiple found if (count[j]>0) counter+=count[j]; // Incrementing potential // GCD by itself // To check i, 2i, 3i.... j += i; // 2 multiples found, // max GCD found if (counter == 2) return i; } counter=0; } return 1; } /* Driver program to test above function */ public static void main(String[] args) { // Array in which pair // with max GCD is to // be found int arr[] = {1, 2, 4, 8, 8, 12}; // Size of array int n = arr.length; System.out.println(findMaxGCD(arr,n)); }}// This code is contributed by Arnav Kr. Mandal. |
Python3
# Python3 Code to# Find pair with# maximum GCD in# an array# function to find# GCD of pair with# max GCD in the# arraydef findMaxGCD(arr, n) : # Calculating MAX in # array high = 0 for i in range(0, n) : high = max(high, arr[i]) # Maintaining count array count = [0] * (high + 1) for i in range(0, n) : count[arr[i]]+=1 # Variable to store the # multiples of a number counter = 0 # Iterating from MAX # to 1 GCD is always # between MAX and 1 # The first GCD found # will be the highest # as we are decrementing # the potential GCD for i in range(high, 0, -1) : j = i # Iterating from current # potential GCD till it # is less than MAX while (j <= high) : # A multiple found if (count[j] >0) : counter+=count[j] # Incrementing potential # GCD by itself # To check i, 2i, 3i.... j += i # 2 multiples found, # max GCD found if (counter == 2) : return i counter=0 # Driver code# Array in which pair# with max GCD is to# be foundarr = [1, 2, 4, 8, 8, 12]# Size of arrayn = len(arr)print(findMaxGCD(arr, n))#This code is contributed by Nikita Tiwari. |
C#
// C# Code to find pair with// maximum GCD in an arrayusing System;class GFG { // function to find GCD // of pair with max // max GCD in the array public static int findMaxGCD(int []arr, int n) { // Calculating Max // in array int high = 0; for (int i = 0; i < n; i++) high = Math.Max(high, arr[i]); // Maintaining count array int []count=new int[high + 1]; for (int i = 0; i < n; i++) count[arr[i]]++; // Variable to store // the multiples of // a number int counter = 0; // Iterating from MAX // to 1 GCD is always // between MAX and 1 // The first GCD found // will be the highest // as we are decrementing // the potential GCD for (int i = high; i >= 1; i--) { int j = i; // Iterating from current // potential GCD till it // is less than MAX while (j <= high) { // A multiple found if (count[j]>0) counter+=count[j]; // Incrementing potential // GCD by itself // To check i, 2i, 3i.... j += i; // 2 multiples found, // max GCD found if (counter == 2) return i; } counter=0; } return 1; } // Driver Code public static void Main(String []args) { // Array in which pair // with max GCD is to // be found int []arr = {1, 2, 4, 8, 8, 12}; // Size of array int n = arr.Length; Console.WriteLine(findMaxGCD(arr,n)); }}// This code is contributed by vt_m. |
PHP
<?php// PHP Code to Find pair with maximum// GCD in an array// function to find GCD of pair with// max GCD in the arrayfunction findMaxGCD($arr, $n){ // Calculating MAX in array $high = 0; for ($i = 0; $i < $n; $i++) $high = max($high, $arr[$i]); // Maintaining count array $count = array_fill(0, $high + 1, 0); for ($i = 0; $i < $n; $i++) $count[$arr[$i]]++; // Variable to store the multiples // of a number $counter = 0; // Iterating from MAX to 1 GCD is always // between MAX and 1. The first GCD found // will be the highest as we are decrementing // the potential GCD for ($i = $high; $i >= 1; $i--) { $j = $i; $counter = 0; // Iterating from current potential GCD // till it is less than MAX while ($j <= $high) { // A multiple found if($count[$j] >= 2) return $j; else if ($count[$j] == 1) $counter++; // Incrementing potential GCD by itself // To check i, 2i, 3i.... $j += $i; // 2 multiples found, max GCD found if ($counter == 2) return $i; } }}// Driver code// Array in which pair with max GCD// is to be found$arr = array( 1, 2, 4, 8, 8, 12 );// Size of array$n = count($arr);print(findMaxGCD($arr, $n));// This code is contributed by mits?> |
Javascript
<script>// javascript Code to// Find pair with// maximum GCD in// an array // function to find // GCD of pair with // max GCD in the // array function findMaxGCD(arr , n) { // Calculating MAX in // array var high = 0; for (let i = 0; i < n; i++) high = Math.max(high, arr[i]); // Maintaining count array var count = Array(high + 1).fill(0); for (let i = 0; i < n; i++) count[arr[i]]++; // Variable to store // the multiples of // a number var counter = 0; // Iterating from MAX // to 1 GCD is always // between MAX and 1 // The first GCD found // will be the highest // as we are decrementing // the potential GCD for (let i = high; i >= 1; i--) { var j = i; // Iterating from current // potential GCD till it // is less than MAX while (j <= high) { // A multiple found if (count[j] > 0) counter += count[j]; // Incrementing potential // GCD by itself // To check i, 2i, 3i.... j += i; // 2 multiples found, // max GCD found if (counter == 2) return i; } counter = 0; } return 1; } /* Driver program to test above function */ // Array in which pair // with max GCD is to // be found var arr = [ 1, 2, 4, 8, 8, 12 ]; // Size of array var n = arr.length; document.write(findMaxGCD(arr, n));// This code is contributed by aashish1995</script> |
Output:
8
Time Complexity: The time complexity of this approach is till an open problem known as the Dirichlet divisor problem.
Time Complexity: O(high2) , high is the maximum element in the array
Auxiliary Space: O(high), high is the maximum element in the array
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