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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:
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:

In this approach, we iterate through each element of the array and remove it to get the remaining array. We then calculate the GCD of the remaining array and keep track of the maximum GCD found so far. Finally, we return the maximum GCD.

• Define a function gcd to calculate the greatest common divisor of two integers using the Euclidean algorithm.
• Define a function MaxGCD which takes an array a and its length n as input and returns the maximum possible GCD after removing one element from the array.
• Initialize a variable ans to 0.
• Iterate over the array and remove each element in turn. To do this, create a temporary array temp of size n-1, and copy all elements from a except the current element to temp.
• Calculate the GCD of the elements in temp using the gcd function. Set this value to a variable res.
• Update the value of ans to be the maximum of its current value and res.
• Return the value of ans as the final answer.

C++

 `#include ``using` `namespace` `std;` `int` `gcd(``int` `a, ``int` `b) {``    ``if``(b == 0) ``return` `a;``    ``return` `gcd(b, a%b);``}` `int` `MaxGCD(``int` `a[], ``int` `n) {``    ``int` `ans = 0;` `    ``// Try removing each element and calculate GCD of the remaining array``    ``for``(``int` `i = 0; i < n; i++) {``        ``int` `temp[n-1];``        ``int` `k = 0;``        ``for``(``int` `j = 0; j < n; j++) {``            ``if``(j != i) {``                ``temp[k++] = a[j];``            ``}``        ``}``        ``int` `res = temp[0];``        ``for``(``int` `j = 1; j < n-1; j++) {``            ``res = gcd(res, temp[j]);``        ``}``        ``ans = max(ans, res);``    ``}` `    ``return` `ans;``}` `int` `main() {``    ``int` `a[] = {14, 17, 28, 70};``    ``int` `n = ``sizeof``(a)/``sizeof``(a[0]);``    ``cout << MaxGCD(a, n) << endl;``    ``return` `0;``}`

Java

 `import` `java.util.*;` `public` `class` `Main {``    ``public` `static` `void` `main(String[] args) {``        ``int``[] a = {``14``, ``17``, ``28``, ``70``};``        ``int` `n = a.length;` `        ``// Call the MaxGCD function and print the result``        ``System.out.println(MaxGCD(a, n));``    ``}` `    ``// Function to calculate the greatest common divisor (GCD) of two numbers using Euclidean algorithm``    ``public` `static` `int` `gcd(``int` `a, ``int` `b) {``        ``if` `(b == ``0``)``            ``return` `a;``        ``return` `gcd(b, a % b);``    ``}` `    ``// Function to find the maximum GCD of all possible subarrays by removing one element at a time``    ``public` `static` `int` `MaxGCD(``int``[] a, ``int` `n) {``        ``int` `ans = ``0``;` `        ``// Try removing each element and calculate GCD of the remaining array``        ``for` `(``int` `i = ``0``; i < n; i++) {``            ``// Create a temporary array to hold the elements of the original array except for the one at index i``            ``int``[] temp = ``new` `int``[n - ``1``];``            ``int` `k = ``0``;``            ``for` `(``int` `j = ``0``; j < n; j++) {``                ``// Skip the element at index i while copying elements to the temporary array``                ``if` `(j != i) {``                    ``temp[k++] = a[j];``                ``}``            ``}` `            ``// Calculate the GCD of the elements in the temporary array``            ``int` `res = temp[``0``];``            ``for` `(``int` `j = ``1``; j < n - ``1``; j++) {``                ``res = gcd(res, temp[j]);``            ``}` `            ``// Update the maximum GCD found so far``            ``ans = Math.max(ans, res);``        ``}` `        ``// Return the maximum GCD of all subarrays obtained by removing one element at a time``        ``return` `ans;``    ``}``}`

Python3

 `def` `gcd(a, b):``    ``# Euclidean algorithm to calculate GCD``    ``if` `b ``=``=` `0``:``        ``return` `a``    ``return` `gcd(b, a ``%` `b)` `def` `max_gcd(arr):``    ``n ``=` `len``(arr)``    ``ans ``=` `0` `    ``# Try removing each element and calculate GCD of the remaining array``    ``for` `i ``in` `range``(n):``        ``temp ``=` `arr[:i] ``+` `arr[i``+``1``:]  ``# Create a modified array by removing element at index i``        ``res ``=` `temp[``0``]``        ``for` `j ``in` `range``(``1``, n ``-` `1``):``            ``res ``=` `gcd(res, temp[j])``        ``ans ``=` `max``(ans, res)` `    ``return` `ans` `if` `__name__ ``=``=` `"__main__"``:``    ``a ``=` `[``14``, ``17``, ``28``, ``70``]``    ``print``(max_gcd(a))`

C#

 `using` `System;` `namespace` `MaxGCD``{``    ``class` `Program``    ``{``        ``// Function to calculate the greatest common divisor (GCD)``        ``static` `int` `Gcd(``int` `a, ``int` `b)``        ``{``            ``if` `(b == 0)``                ``return` `a;``            ``return` `Gcd(b, a % b);``        ``}` `        ``// Function to calculate the maximum GCD after removing each element``        ``static` `int` `MaxGCD(``int``[] a, ``int` `n)``        ``{``            ``int` `ans = 0;` `            ``// Try removing each element and calculate GCD of the remaining array``            ``for` `(``int` `i = 0; i < n; i++)``            ``{``                ``int``[] temp = ``new` `int``[n - 1];``                ``int` `k = 0;``                ``for` `(``int` `j = 0; j < n; j++)``                ``{``                    ``if` `(j != i)``                    ``{``                        ``temp[k++] = a[j];``                    ``}``                ``}``                ``int` `res = temp[0];``                ``for` `(``int` `j = 1; j < n - 1; j++)``                ``{``                    ``res = Gcd(res, temp[j]);``                ``}``                ``ans = Math.Max(ans, res);``            ``}` `            ``return` `ans;``        ``}` `        ``// Main function``        ``static` `void` `Main(``string``[] args)``        ``{``            ``int``[] a = { 14, 17, 28, 70 };``            ``int` `n = a.Length;` `            ``// Function call``            ``Console.WriteLine(MaxGCD(a, n));``        ``}``    ``}``}`

Output

```14

```

Time Complexity: O(n^2 * log(max(a))), where n is the size of the array and max(a) is the maximum value in the array. This is because the function gcd has a time complexity of O(log(max(a))) and the function MaxGCD has two nested loops, each of which iterates n times.

Auxiliary Space: O(n^2) because the function MaxGCD creates a new array of size n-1 for each element of the input array, resulting in n^2 total elements in all created arrays.

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 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

 ``

Output

```14

```

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

Auxiliary Space: O(N)