Remove an element to maximize the GCD of the given array

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: 6
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:

CPP

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// 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;
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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.

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Output:

14

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



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