Skip to content
Related Articles

Related Articles

Improve Article

Remove an element to maximize the GCD of the given array

  • Difficulty Level : Hard
  • Last Updated : 07 May, 2021

Given an array arr[] of length N ≥ 2. The task is to remove an element from the given array such that the GCD of the array after removing it is maximized.

Examples: 

Input: arr[] = {12, 15, 18} 
Output:
Remove 12: GCD(15, 18) = 3 
Remove 15: GCD(12, 18) = 6 
Remove 18: GCD(12, 15) = 3

Input: arr[] = {14, 17, 28, 70} 
Output: 14 

Approach:  



  • Idea is to find the GCD value of all the sub-sequences of length (N – 1) and removing the element which is not present in the sub-sequence with that GCD. The maximum GCD found would be the answer.
  • To find the GCD of the sub-sequences optimally, maintain a prefixGCD[] and a suffixGCD[] array using single state dynamic programming.
  • The maximum value of GCD(prefixGCD[i – 1], suffixGCD[i + 1]) is the required answer.

Below is the implementation of the above approach:  

C++




// C++ implementation of the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to return the maximized gcd
// after removing a single element
// from the given array
int MaxGCD(int a[], int n)
{
 
    // Prefix and Suffix arrays
    int Prefix[n + 2];
    int Suffix[n + 2];
 
    // Single state dynamic programming relation
    // for storing gcd of first i elements
    // from the left in Prefix[i]
    Prefix[1] = a[0];
    for (int i = 2; i <= n; i += 1) {
        Prefix[i] = __gcd(Prefix[i - 1], a[i - 1]);
    }
 
    // Initializing Suffix array
    Suffix[n] = a[n - 1];
 
    // Single state dynamic programming relation
    // for storing gcd of all the elements having
    // greater than or equal to i in Suffix[i]
    for (int i = n - 1; i >= 1; i -= 1) {
        Suffix[i] = __gcd(Suffix[i + 1], a[i - 1]);
    }
 
    // If first or last element of
    // the array has to be removed
    int ans = max(Suffix[2], Prefix[n - 1]);
 
    // If any other element is replaced
    for (int i = 2; i < n; i += 1) {
        ans = max(ans, __gcd(Prefix[i - 1], Suffix[i + 1]));
    }
 
    // Return the maximized gcd
    return ans;
}
 
// Driver code
int main()
{
    int a[] = { 14, 17, 28, 70 };
    int n = sizeof(a) / sizeof(a[0]);
 
    cout << MaxGCD(a, n);
 
    return 0;
}

Java




// Java implementation of the above approach
class Test
{
    // Recursive function to return gcd of a and b
    static int gcd(int a, int b)
    {
        if (b == 0)
            return a;
        return gcd(b, a % b);
    }
     
    // Function to return the maximized gcd
    // after removing a single element
    // from the given array
    static int MaxGCD(int a[], int n)
    {
     
        // Prefix and Suffix arrays
        int Prefix[] = new int[n + 2];
        int Suffix[] = new int[n + 2] ;
     
        // Single state dynamic programming relation
        // for storing gcd of first i elements
        // from the left in Prefix[i]
        Prefix[1] = a[0];
        for (int i = 2; i <= n; i += 1)
        {
            Prefix[i] = gcd(Prefix[i - 1], a[i - 1]);
        }
     
        // Initializing Suffix array
        Suffix[n] = a[n - 1];
     
        // Single state dynamic programming relation
        // for storing gcd of all the elements having
        // greater than or equal to i in Suffix[i]
        for (int i = n - 1; i >= 1; i -= 1)
        {
            Suffix[i] = gcd(Suffix[i + 1], a[i - 1]);
        }
     
        // If first or last element of
        // the array has to be removed
        int ans = Math.max(Suffix[2], Prefix[n - 1]);
     
        // If any other element is replaced
        for (int i = 2; i < n; i += 1)
        {
            ans = Math.max(ans, gcd(Prefix[i - 1], Suffix[i + 1]));
        }
     
        // Return the maximized gcd
        return ans;
    }
         
    // Driver code
    public static void main(String[] args)
    {
 
        int a[] = { 14, 17, 28, 70 };
        int n = a.length;
     
        System.out.println(MaxGCD(a, n));
    }
}
 
// This code is contributed by AnkitRai01

Python3




# Python3 implementation of the above approach
import math as mt
 
# Function to return the maximized gcd
# after removing a single element
# from the given array
 
def MaxGCD(a, n):
 
 
    # Prefix and Suffix arrays
    Prefix=[0 for i in range(n + 2)]
    Suffix=[0 for i in range(n + 2)]
 
    # Single state dynamic programming relation
    # for storing gcd of first i elements
    # from the left in Prefix[i]
    Prefix[1] = a[0]
    for i in range(2,n+1):
        Prefix[i] = mt.gcd(Prefix[i - 1], a[i - 1])
 
    # Initializing Suffix array
    Suffix[n] = a[n - 1]
 
    # Single state dynamic programming relation
    # for storing gcd of all the elements having
    # greater than or equal to i in Suffix[i]
    for i in range(n-1,0,-1):
        Suffix[i] =mt.gcd(Suffix[i + 1], a[i - 1])
 
    # If first or last element of
    # the array has to be removed
    ans = max(Suffix[2], Prefix[n - 1])
 
    # If any other element is replaced
    for i in range(2,n):
        ans = max(ans, mt.gcd(Prefix[i - 1], Suffix[i + 1]))
 
    # Return the maximized gcd
    return ans
 
# Driver code
 
a=[14, 17, 28, 70]
n = len(a)
 
print(MaxGCD(a, n))
 
# This code is contributed by mohit kumar 29

C#




// C# implementation of the above approach
using System;
 
class GFG
{
     
    // Recursive function to return gcd of a and b
    static int gcd(int a, int b)
    {
        if (b == 0)
            return a;
        return gcd(b, a % b);
    }
     
    // Function to return the maximized gcd
    // after removing a single element
    // from the given array
    static int MaxGCD(int []a, int n)
    {
     
        // Prefix and Suffix arrays
        int []Prefix = new int[n + 2];
        int []Suffix = new int[n + 2] ;
     
        // Single state dynamic programming relation
        // for storing gcd of first i elements
        // from the left in Prefix[i]
        Prefix[1] = a[0];
        for (int i = 2; i <= n; i += 1)
        {
            Prefix[i] = gcd(Prefix[i - 1], a[i - 1]);
        }
     
        // Initializing Suffix array
        Suffix[n] = a[n - 1];
     
        // Single state dynamic programming relation
        // for storing gcd of all the elements having
        // greater than or equal to i in Suffix[i]
        for (int i = n - 1; i >= 1; i -= 1)
        {
            Suffix[i] = gcd(Suffix[i + 1], a[i - 1]);
        }
     
        // If first or last element of
        // the array has to be removed
        int ans = Math.Max(Suffix[2], Prefix[n - 1]);
     
        // If any other element is replaced
        for (int i = 2; i < n; i += 1)
        {
            ans = Math.Max(ans, gcd(Prefix[i - 1], Suffix[i + 1]));
        }
     
        // Return the maximized gcd
        return ans;
    }
         
    // Driver code
    static public void Main ()
    {
         
        int []a = { 14, 17, 28, 70 };
        int n = a.Length;
     
        Console.Write(MaxGCD(a, n));
    }
}
 
// This code is contributed by ajit.

Javascript




<script>
 
// Javascript implementation of the above approach
 
// Recursive function to return gcd of a and b 
function gcd(a, b)
{
    if (b == 0)
        return a;
         
    return gcd(b, a % b);
}
     
// Function to return the maximized gcd
// after removing a single element
// from the given array
function MaxGCD(a, n)
{
     
    // Prefix and Suffix arrays
    let Prefix = new Array(n + 2);
    let Suffix = new Array(n + 2);
 
    // Single state dynamic programming relation
    // for storing gcd of first i elements
    // from the left in Prefix[i]
    Prefix[1] = a[0];
    for(let i = 2; i <= n; i += 1)
    {
        Prefix[i] = gcd(Prefix[i - 1], a[i - 1]);
    }
 
    // Initializing Suffix array
    Suffix[n] = a[n - 1];
 
    // Single state dynamic programming relation
    // for storing gcd of all the elements having
    // greater than or equal to i in Suffix[i]
    for(let i = n - 1; i >= 1; i -= 1)
    {
        Suffix[i] = gcd(Suffix[i + 1], a[i - 1]);
    }
 
    // If first or last element of
    // the array has to be removed
    let ans = Math.max(Suffix[2], Prefix[n - 1]);
 
    // If any other element is replaced
    for(let i = 2; i < n; i += 1)
    {
        ans = Math.max(ans, gcd(Prefix[i - 1],
                                Suffix[i + 1]));
    }
     
    // Return the maximized gcd
    return ans;
}
 
// Driver code
let a = [ 14, 17, 28, 70 ];
let n = a.length;
 
document.write(MaxGCD(a, n));
 
// This code is contributed by rishavmahato348
 
</script>
Output: 
14

 

Time Complexity: O(N * log(M)) where M is the maximum element from the array.
 

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.




My Personal Notes arrow_drop_up
Recommended Articles
Page :