Given an array of positive integers, the task is that we find a subset of size greater than one with maximum product.
Input : arr[] = {1, 5, 7, 2, 0};
Output : 5 7
The subset containing 5 and 7 produces maximum
geometric mean
Input : arr[] = { 4, 3 , 5 , 9 , 8 };
Output : 8 9
A Naive Approach is to run two loops and check one by one array elements which give greatest geometric mean (G.M). Time complexity of this solution is O(n*n) and this solution also causes overflow.
An Efficient Solution is based on the fact that the greatest two elements would always produce the greatest mean as the question requires to find a subset of size greater than one.
CPP
// C++ program to find a subset of size 2 or // greater with greatest geometric mean. This // program basically find largest two elements. #include <bits/stdc++.h> using namespace std; void findLargestGM(int arr[], int n) { /* There should be atleast two elements */ if (n < 2) { printf(" Invalid Input "); return; } int first = INT_MIN, second = INT_MIN; for (int i = 0; i < n ; i ++) { /* If current element is smaller than first then update both first and second */ if (arr[i] > first) { second = first; first = arr[i]; } /* If arr[i] is in between first and second then update second */ else if (arr[i] > second) second = arr[i]; } printf("%d %d", second, first); } /* Driver program to test above function */int main() { int arr[] = {12, 13, 17, 10, 34, 1}; int n = sizeof(arr)/sizeof(arr[0]); findLargestGM(arr, n); return 0; } |
Java
// Java program to find a subset of size 2 or // greater with greatest geometric mean. This // program basically find largest two elements. class GFG { static void findLargestGM(int arr[], int n) { // There should be atleast two elements if (n < 2) { System.out.print(" Invalid Input "); } int first = -2147483648, second = -2147483648; for (int i = 0; i < n ; i ++) { /* If current element is smaller than first then update both first and second */ if (arr[i] > first) { second = first; first = arr[i]; } /* If arr[i] is in between first and second then update second */ else if (arr[i] > second) second = arr[i]; } System.out.print(second + " " + first); } // Driver function public static void main(String arg[]) { int arr[] = {12, 13, 17, 10, 34, 1}; int n = arr.length; findLargestGM(arr, n); } } // This code is contributed by Anant Agarwal. |
Python3
# Python3 program to find # a subset of size 2 or # greater with greatest # geometric mean. This # program basically find # largest two elements. import sys def findLargestGM(arr, n): # There should be # atleast two elements if n < 2: print (" Invalid Input ") return first = -sys.maxsize - 1 second = -sys.maxsize - 1 for i in range(0,n): # If current element is # smaller than first # then update both first # and second if arr[i] > first: second = first first = arr[i] # If arr[i] is in between # first and second # then update second elif arr[i] > second: second = arr[i] print ("%d %d"%(second, first)) # Driver program to # test above function arr = [12, 13, 17, 10, 34, 1] n = len(arr) findLargestGM(arr, n) # This code is contributed # by Shreyanshi Arun. |
C#
// C# program to find a subset of size 2 or // greater with greatest geometric mean. This // program basically find largest two elements. using System; class GFG { static void findLargestGM(int []arr, int n) { // There should be atleast two elements if (n < 2) { Console.Write("Invalid Input"); } int first = -2147483648; int second = -2147483648; for (int i = 0; i < n ; i ++) { // If current element is smaller // than first then update both // first and second if (arr[i] > first) { second = first; first = arr[i]; } // If arr[i] is in between first // and second then update second else if (arr[i] > second) second = arr[i]; } Console.Write(second + " " + first); } // Driver code public static void Main() { int []arr = {12, 13, 17, 10, 34, 1}; int n = arr.Length; findLargestGM(arr, n); } } // This code is contributed by Nitin Mittal. |
PHP
<?php // PHP program to find a subset of size 2 or // greater with greatest geometric mean. This // program basically find largest two elements. function findLargestGM($arr, $n) { /* There should be atleast two elements */ if ($n < 2) { echo(" Invalid Input "); return; } $first = PHP_INT_MIN; $second = PHP_INT_MIN; for ($i = 0; $i < $n ; $i++) { /* If current element is smaller than first then update both first and second */ if ($arr[$i] > $first) { $second = $first; $first = $arr[$i]; } /* If arr[i] is in between first and second then update second */ else if ($arr[$i] > $second) $second = $arr[$i]; } echo($second . " " . $first); } /* Driver program to test above function */$arr = array(12, 13, 17, 10, 34, 1); $n = sizeof($arr); findLargestGM($arr, $n); // This code is contributed by Ajit. ?> |
Output:
17 34
Time complexity : O(n)
Space Complexity : O(1)
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