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) = 3Input: 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 <bits/stdc++.h>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 ansif __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)); } }} |
Javascript
function gcd( a, b) { if(b == 0) return a; return gcd(b, a%b);}function MaxGCD(a, n) { let ans = 0; // Try removing each element and calculate GCD of the remaining array for(let i = 0; i < n; i++) { let temp=new Array(n-1); let k = 0; for(let j = 0; j < n; j++) { if(j != i) { temp[k++] = a[j]; } } let res = temp[0]; for(let j = 1; j < n-1; j++) { res = gcd(res, temp[j]); } ans = Math.max(ans, res); } return ans;}let a = [14, 17, 28, 70];let n = a.length;document.write(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 <bits/stdc++.h>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[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 codeint 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 approachimport math as mt# Function to return the maximized gcd# after removing a single element# from the given arraydef 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 codea=[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[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
<script>// Javascript implementation of the above approach// Recursive function to return gcd of a and b function gcd(a, 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 arrayfunction MaxGCD(a, n){ // Prefix and Suffix arrays let Prefix = new Array(n + 2); let Suffix = new Array(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(let 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(let 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 let ans = Math.max(Suffix[2], Prefix[n - 1]); // If any other element is replaced for(let i = 2; i < n; i += 1) { ans = Math.max(ans, gcd(Prefix[i - 1], Suffix[i + 1])); } // Return the maximized gcd return ans;}// Driver codelet a = [ 14, 17, 28, 70 ];let n = a.length;document.write(MaxGCD(a, n));// This code is contributed by rishavmahato348</script> |
Output
14
Time Complexity: O(N * log(M)) where M is the maximum element from the array.
Auxiliary Space: O(N)
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