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

Last Updated : 02 Aug, 2022

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:

Idea is to find the GCD value of all the sub-sequences of length (N – 1) and removing the element which has to be replaced in the sub-sequence as it can be replaced with any other element already in the subsequence. 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. Also note that suffixGCD[1] and prefixGCD[N - 2] also need to be compared in case the first or the last element has to be replaced.

Below is the implementation of the above approach:

C++

// 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]);
    printf("%d", MaxGCD(a, n));
    return 0;
}

// C implementation of the approach
#include <stdio.h>

// Find maximum between two numbers.
int max(int num1, int num2)
{
    return (num1 > num2) ? num1 : num2;
}

// Find minimum between two numbers.
int min(int num1, int num2)
{
    return (num1 > num2) ? num2 : num1;
}

// Function to return gcd of a and b
int gcd(int a, int b)
{
    int result = min(a, b); // Finding minimum of a nd b
    while (result > 0) {
        if (a % result == 0 && b % result == 0) {
            break;
        }
        result--;
    }
    return result; // return gcd of a nd b
}

// 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]);
    printf("%d", MaxGCD(a, n));
    return 0;
}

// This code is contributed by Sania Kumari Gupta

Java

// 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

Python3

# 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

// 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

JavaScript

<script>

// Javascript implementation of the approach

function gcd(a, b) {
  if (b==0) {
    return a;
  }

  return gcd(b, a % b);
}
// Function to return the maximum
// possible gcd after replacing
// a single element
function MaxGCD(a, n)
{

    // Prefix and Suffix arrays
    var Prefix = Array(n + 2).fill(0);
    var Suffix = Array(n + 2).fill(0);

    var i;
    // Single state dynamic programming relation
    // for storing gcd of first i elements
    // from the left in Prefix[i]
    Prefix[1] = a[0];
    for (i = 2; i <= n; i++) {
        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 = n - 1; i >= 1; i--) {
        Suffix[i] = gcd(Suffix[i + 1], a[i - 1]);
    }

    // If first or last element of
    // the array has to be replaced
    var ans = Math.max(Suffix[2], Prefix[n - 1]);

    // If any other element is replaced
    for (i = 2; i < n; i++) {
        ans = Math.max(ans, gcd(Prefix[i - 1], 
        Suffix[i + 1]));
    }

    // Return the maximized gcd
    return ans;
}

// Driver code
    var a = [6, 7, 8];
    n = a.length;

    document.write(MaxGCD(a, n));

</script>

Output:

Time Complexity: O(n * max(a, b))

Auxiliary Space: O(n)

Maximize count of distinct elements possible in an Array from the given operation

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

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