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 using namespace std; // Function to return the maximized gcd// after removing a single element// from the given arrayint 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 = a;    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, 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 codeint main(){    int a[] = { 14, 17, 28, 70 };    int n = sizeof(a) / sizeof(a);     cout << MaxGCD(a, n);     return 0;}

Java

 // Java implementation of the above approachclass 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 = a;        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, 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 approachimport 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 = a    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, 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 approachusing 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 = a;        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, 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


Output:
14

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

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