Given an array arr[] of length N, the task is to find the length of the longest sub-sequence with minimum possible LCM.
Examples:
Input: arr[] = {1, 3, 1}
Output: 2
{1} and {1} are the subsequences
with the minimum possible LCM.
Input: arr[] = {3, 4, 5, 3, 2, 3}
Output: 1
{2} is the required subsequence.
Approach: The minimum possible LCM from the array will be equal to the value of the smallest element in the array. Now, to maximize the length of the resulting subsequence, find the number of elements with a value equal to this smallest 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// minimum possible LCMint maxLen(int* arr, int n){ // Minimum value from the array int min_val = *min_element(arr, arr + n); // To store the frequency of the // minimum element in the array int freq = 0; for (int i = 0; i < n; i++) { // If current element is equal // to the minimum element if (arr[i] == min_val) freq++; } return freq;}// Driver codeint main(){ int arr[] = { 1, 3, 1 }; int n = sizeof(arr) / sizeof(int); cout << maxLen(arr, n); return 0;} |
Java
// Java implementation of the approachimport java.util.Arrays;class GFG {// Function to return the length// of the largest subsequence with// minimum possible LCMstatic int maxLen(int[] arr, int n){ // Minimum value from the array int min_val = Arrays.stream(arr).min().getAsInt(); // To store the frequency of the // minimum element in the array int freq = 0; for (int i = 0; i < n; i++) { // If current element is equal // to the minimum element if (arr[i] == min_val) freq++; } return freq;}// Driver codepublic static void main(String []args){ int arr[] = { 1, 3, 1 }; int n = arr.length; System.out.println(maxLen(arr, n));}}// This code is contributed by PrinciRaj1992 |
Python3
# Python3 implementation of the approach# Function to return the length # of the largest subsequence with # minimum possible LCM def maxLen(arr, n) : # Minimum value from the array min_val = min(arr); # To store the frequency of the # minimum element in the array freq = 0; for i in range(n) : # If current element is equal # to the minimum element if (arr[i] == min_val) : freq += 1; return freq; # Driver code if __name__ == "__main__" : arr = [ 1, 3, 1 ]; n = len(arr); print(maxLen(arr, n)); # This code is contributed by AnkitRai01 |
C#
// C# implementation of the approachusing System;using System.Linq; class GFG {// Function to return the length// of the largest subsequence with// minimum possible LCMstatic int maxLen(int[] arr, int n){ // Minimum value from the array int min_val = arr.Min(); // To store the frequency of the // minimum element in the array int freq = 0; for (int i = 0; i < n; i++) { // If current element is equal // to the minimum element if (arr[i] == min_val) freq++; } return freq;}// Driver codepublic static void Main(String []args){ int []arr = { 1, 3, 1 }; int n = arr.Length; Console.WriteLine(maxLen(arr, n));}}// This code is contributed by 29AjayKumar |
Javascript
<script>// Javascript implementation of the approach// Function to return the length// of the largest subsequence with// minimum possible LCMfunction maxLen(arr, n){ // Minimum value from the array var min_val = arr.reduce((a, b) => Math.min(a,b)) // To store the frequency of the // minimum element in the array var freq = 0; for (var i = 0; i < n; i++) { // If current element is equal // to the minimum element if (arr[i] == min_val) freq++; } return freq;}// Driver codevar arr = [ 1, 3, 1 ];var n = arr.length;document.write( maxLen(arr, n));// This code is contributed by itsok.</script> |
Output:
2
Time Complexity: O(n)
Auxiliary Space: O(1)
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