Given an array of positive distinct integers arr[], the task is to find the final number obtained by performing the following operation on the elements of the array:
Operation: Take two unequal numbers and replace the larger number with their difference until all numbers become equal.
Examples:
Input: arr[] = {5, 2, 3}
Output: 1
5 – 3 = 2, arr[] = {2, 2, 3}
3 – 2 = 1, arr[] = {2, 2, 1}
2 – 1 = 1, arr[] = {2, 1, 1}
2 – 1 = 1, arr[] = {1, 1, 1}
Input: arr[] = {3, 9, 6, 36}
Output: 3
Naive approach: Since final answer will always be distinct, one can just sort the array and replace the largest term with the difference of the two largest elements and repeat the process until all the numbers become equal.
C++
#include <algorithm> #include <iostream> int finalNumber( int arr[], int n) { std::sort(arr, arr + n); // Sort the array in ascending order while (arr[n - 1] != arr[0]) { // Keep looping until all // elements are equal int diff = arr[n - 1] - arr[0]; arr[n - 1] = diff; std::sort(arr, arr + n); // Sort the array again after // changing the last element } return arr[0]; } int main() { int arr[] = { 3, 9, 6, 36 }; int n = sizeof (arr) / sizeof (arr[0]); // Get the size of the array std::cout << finalNumber(arr, n) << std::endl; return 0; } |
Java
import java.util.Arrays; public class Main { public static int finalNumber( int [] arr) { int n = arr.length; Arrays.sort(arr); while (arr[n - 1 ] != arr[ 0 ]) { int diff = arr[n - 1 ] - arr[ 0 ]; arr[n - 1 ] = diff; Arrays.sort(arr); } return arr[ 0 ]; } public static void main(String[] args) { int [] arr2 = { 3 , 9 , 6 , 36 }; System.out.println(finalNumber(arr2)); } } |
Python3
def final_number(arr): n = len (arr) # Get the length of the input array arr.sort() # Sort the array using sort function while arr[n - 1 ] ! = arr[ 0 ]: # While the largest and smallest elements in the array are not equal diff = arr[n - 1 ] - arr[ 0 ] # Calculate the difference between the largest and smallest elements arr[n - 1 ] = diff # Replace the largest element with the difference arr.sort() # Sort the array again return arr[ 0 ] # Return the smallest element in the array if __name__ = = '__main__' : arr = [ 3 , 9 , 6 , 36 ] # Define the input array print (final_number(arr)) # Call the final_number function and print the result |
Javascript
function finalNumber(arr) { const n = arr.length; // Get the length of the input array arr.sort((a, b) => a - b); // Sort the array using sort function while (arr[n - 1] !== arr[0]) { // While the largest and smallest elements in the array are not equal const diff = arr[n - 1] - arr[0]; // Calculate the difference between the largest and smallest elements arr[n - 1] = diff; // Replace the largest element with the difference arr.sort((a, b) => a - b); // Sort the array again } return arr[0]; // Return the smallest element in the array } const arr = [3, 9, 6, 36]; // Define the input array console.log(finalNumber(arr)); // Call the finalNumber function and print the result |
C#
// C# implementation of the approach using System; using System.Linq; class Program { static int FinalNumber( int [] arr, int n) { Array.Sort(arr); // Sort the array in ascending order while (arr[n - 1] != arr[0]) // Keep looping until all elements are equal { int diff = arr[n - 1] - arr[0]; arr[n - 1] = diff; Array.Sort(arr); // Sort the array again after changing the last element } return arr[0]; } static void Main( string [] args) { int [] arr = { 3, 9, 6, 36 }; int n = arr.Length; Console.WriteLine(FinalNumber(arr, n)); } } // Contributed by adityasharmadev01 |
Output :
3
Efficient approach: From Euclidean’s algorithm, it is known that gcd(a, b) = gcd(a – b, b). This can be extended to gcd(A1, A2, A3, …, An) = gcd(A1 – A2, A2, A3, …, An).
Also, let’s say that after applying the given operation, the final number obtained be K. Hence, from the extended algorithm, it can be said that gcd(A1, A2, A3, …, An) = gcd(K, K, …, n times). Since gcd(K, K, …, n times) = K, the solution of the given problem can be found
by finding the gcd of all the elements of the array.
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 final number // obtained after performing the // given operation int finalNum( int arr[], int n) { // Find the gcd of the array elements int result = 0; for ( int i = 0; i < n; i++) { result = __gcd(result, arr[i]); } return result; } // Driver code int main() { int arr[] = { 3, 9, 6, 36 }; int n = sizeof (arr) / sizeof (arr[0]); cout << finalNum(arr, n); return 0; } |
Java
// Java implementation of the approach import java.io.*; class GFG { // Function to return the final number // obtained after performing the // given operation static int finalNum( int arr[], int n) { // Find the gcd of the array elements int result = 0 ; for ( int i = 0 ; i < n; i++) { result = __gcd(result, arr[i]); } return result; } static int __gcd( int a, int b) { return b == 0 ? a:__gcd(b, a % b); } // Driver code public static void main(String[] args) { int arr[] = { 3 , 9 , 6 , 36 }; int n = arr.length; System.out.print(finalNum(arr, n)); } } // This code is contributed by 29AjayKumar |
Python3
# Python3 implementation of the approach from math import gcd as __gcd # Function to return the final number # obtained after performing the # given operation def finalNum(arr, n): # Find the gcd of the array elements result = arr[ 0 ] for i in arr: result = __gcd(result, i) return result # Driver code arr = [ 3 , 9 , 6 , 36 ] n = len (arr) print (finalNum(arr, n)) # This code is contributed by Mohit Kumar |
C#
// C# implementation of the approach using System; class GFG { // Function to return the readonly number // obtained after performing the // given operation static int finalNum( int []arr, int n) { // Find the gcd of the array elements int result = 0; for ( int i = 0; i < n; i++) { result = __gcd(result, arr[i]); } return result; } static int __gcd( int a, int b) { return b == 0 ? a : __gcd(b, a % b); } // Driver code public static void Main(String[] args) { int []arr = { 3, 9, 6, 36 }; int n = arr.Length; Console.Write(finalNum(arr, n)); } } // This code is contributed by 29AjayKumar |
Javascript
<script> // javascript implementation of the approach // Function to return the final number // obtained after performing the // given operation function finalNum(arr , n) { // Find the gcd of the array elements var result = 0; for (i = 0; i < n; i++) { result = __gcd(result, arr[i]); } return result; } function __gcd(a , b) { return b == 0 ? a : __gcd(b, a % b); } // Driver code var arr = [ 3, 9, 6, 36 ]; var n = arr.length; document.write(finalNum(arr, n)); // This code contributed by aashish1995 </script> |
3
Time Complexity: O(N*logN), as we are using a loop to traverse N times so it will cost us O(N) time and _gcd function will take logN time.
Auxiliary Space: O(1), as we are not using any extra space.
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