Maximum possible GCD after replacing at most one element in the given array

Given an array arr[] of size N > 1. The task is to find the maximum possible GCD of the array by replacing at most one element.

Examples:

Input: arr[] = {6, 7, 8}
Output: 2
Replace 7 with 2 and gcd(6, 2, 8) = 2
which is maximum possible.

Input: arr[] = {12, 18, 30}
Output: 6

Approach:



Below is the implementation of the above approach:

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to return the maximum
// possible gcd after replacing
// a single element
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
    // index 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 replaced
    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[] = { 6, 7, 8 };
    int n = sizeof(a) / sizeof(a[0]);
  
    cout << MaxGCD(a, n);
  
    return 0;
}
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// Java implementation of the approach
class GFG 
{
  
// Function to return the maximum
// possible gcd after replacing
// a single element
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
    // index 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 replaced
    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;
}
  
static int __gcd(int a, int b) 
    return b == 0 ? a : __gcd(b, a % b);     
}
  
// Driver code
public static void main(String[] args) 
{
    int a[] = { 6, 7, 8 };
    int n = a.length;
  
    System.out.println(MaxGCD(a, n));
}
}
  
// This code is contributed by PrinciRaj1992
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# Python3 implementation of the approach 
from math import gcd as __gcd
  
# Function to return the maximum 
# possible gcd after replacing 
# a single element 
def MaxGCD(a, n) :
  
    # Prefix and Suffix arrays 
    Prefix = [0] * (n + 2); 
    Suffix = [0] * (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] = __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 
    # index greater than or equal to i in Suffix[i] 
    for i in range(n - 1, 0, -1) :
        Suffix[i] = __gcd(Suffix[i + 1], a[i - 1]); 
  
    # If first or last element of 
    # the array has to be replaced 
    ans = max(Suffix[2], Prefix[n - 1]); 
  
    # If any other element is replaced 
    for i in range(2, n) :
        ans = max(ans, __gcd(Prefix[i - 1], 
                             Suffix[i + 1])); 
  
    # Return the maximized gcd 
    return ans; 
  
# Driver code 
if __name__ == "__main__" :
  
    a = [ 6, 7, 8 ]; 
    n = len(a); 
  
    print(MaxGCD(a, n)); 
  
# This code is contributed by AnkitRai01
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// C# implementation of the approach
using System;
      
class GFG 
{
  
// Function to return the maximum
// possible gcd after replacing
// a single element
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
    // index 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 replaced
    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;
}
  
static int __gcd(int a, int b) 
    return b == 0 ? a : __gcd(b, a % b);     
}
  
// Driver code
public static void Main(String[] args) 
{
    int []a = { 6, 7, 8 };
    int n = a.Length;
  
    Console.WriteLine(MaxGCD(a, n));
}
}
  
// This code is contributed by 29AjayKumar
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Output:
2

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