Longest sub-sequence with maximum GCD
Last Updated : 23 Dec, 2022
Given an array arr[] of length N, the task is to find the length of the longest sub-sequence with the maximum possible GCD.
Examples:
Input: arr[] = {2, 1, 2}
Output: 2
{2}, {2} and {2, 2} are the subsequences
with the maximum possible GCD.
Input: arr[] = {1, 2, 3}
Output: 1
{3} is the required subsequence.
Approach: The maximum possible GCD from the array will be equal to the value of the largest element in the array. Now, to maximize the length of the resulting subsequence, find the number of elements with a value equal to this largest value in the array, and the count of these elements is the required answer.
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 length // of the largest subsequence with // maximum possible GCD int maxLen(int* arr, int n) { // Maximum value from the array int max_val = *max_element(arr, arr + n); // To store the frequency of the // maximum element in the array int freq = 0; for (int i = 0; i < n; i++) { // If current element is equal // to the maximum element if (arr[i] == max_val) freq++; } return freq; } // Driver code int main() { int arr[] = { 3, 2, 2, 3, 3, 3 }; int n = sizeof(arr) / sizeof(int); cout << maxLen(arr, n); return 0; } // Java implementation of the approach import java.util.Arrays; class GFG { // Function to return the length // of the largest subsequence with // maximum possible GCD static int maxLen(int[] arr, int n) { // Maximum value from the array int max_val = Arrays.stream(arr).max().getAsInt(); // To store the frequency of the // maximum element in the array int freq = 0; for (int i = 0; i < n; i++) { // If current element is equal // to the maximum element if (arr[i] == max_val) freq++; } return freq; } // Driver code public static void main(String []args) { int arr[] = { 3, 2, 2, 3, 3, 3 }; int n = arr.length; System.out.println(maxLen(arr, n)); } } // This code is contributed by Rajput-Ji # Python3 implementation of the approach # Function to return the length # of the largest subsequence with # maximum possible GCD def maxLen(arr, n) : # Maximum value from the array max_val = max(arr); # To store the frequency of the # maximum element in the array freq = 0; for i in range(n) : # If current element is equal # to the maximum element if (arr[i] == max_val) : freq += 1; return freq; # Driver code if __name__ == "__main__" : arr = [ 3, 2, 2, 3, 3, 3 ]; n = len(arr); print(maxLen(arr, n)); # This code is contributed by AnkitRai01
// C# implementation of the approach using System; using System.Linq; class GFG { // Function to return the length // of the largest subsequence with // maximum possible GCD static int maxLen(int[] arr, int n) { // Maximum value from the array int max_val = arr.Max(); // To store the frequency of the // maximum element in the array int freq = 0; for (int i = 0; i < n; i++) { // If current element is equal // to the maximum element if (arr[i] == max_val) freq++; } return freq; } // Driver code public static void Main(String []args) { int []arr = { 3, 2, 2, 3, 3, 3 }; int n = arr.Length; Console.WriteLine(maxLen(arr, n)); } } // This code is contributed by PrinciRaj1992 <script> // Javascript implementation of the approach // Function to return the length // of the largest subsequence with // maximum possible GCD function maxLen(arr, n) { // Maximum value from the array var max_val = arr.reduce((a,b) => Math.max(a,b)); // To store the frequency of the // maximum element in the array var freq = 0; for (var i = 0; i < n; i++) { // If current element is equal // to the maximum element if (arr[i] == max_val) freq++; } return freq; } // Driver code var arr = [3, 2, 2, 3, 3, 3]; var n = arr.length; document.write( maxLen(arr, n)); </script> Time Complexity: O(n), where n is the size of the given array.
Auxiliary Space: O(1), no extra space is required, so it is a constant.
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