Given an array arr[] consisting of N integers, the task is to make all array elements equal by selecting any pair of integers from the array and replacing the larger integer from the pair with their absolute difference any number of times. Print the final value of all array elements.
Examples:
Input: arr[] ={2, 3, 4}
Output: 1
Explanation:Â
Step 1: Performing on the pair (2, 3) modifies arr[] = {2, 1, 4}
Step 2: Performing on the pair (2, 4) modifies arr[] = {2, 1, 2}
Step 3: Performing on the pair (2, 1) modifies {1, 1, 2}
Step 4: Performing on the pair (1, 2) modifies arr[] = {1, 1, 1}Input: arr[] = {24, 60}
Output: 12
Approach: From the above problem statement, it can be observed that for any pair (a, b), the absolute difference is subtracted from the maximum element. Then this operation is similar to finding GCD of the pair. Therefore, from this observation, it is clear that all array elements need to be reduced to the GCD of the array. Follow the steps below to solve the problem:
- Initialize a variable gcd as 1.
- Traverse the given array and while traversing update gcd as:
 gcd = gcd(arr[i], gcd), where 0 ? i < N
- After the above step, the value of gcd is the required array element after the given operation is applied to every distinct pair of elements.
Below is the implementation of the above approach:
C++
// C++ Program to implement// the above approach#include <bits/stdc++.h>using namespace std;Â
// Function to return // gcd of a and bint gcd(int a, int b){  // Base Case  if (a == 0)    return b;Â
  // Recursive Call  return gcd(b % a, a);}Â
// Function to find gcd of arrayint findGCD(int arr[], int N){  // Initialise the result  int result = 0;Â
  // Traverse the array arr[]  for (int i = 0; i < N; i++)   {    // Update result as gcd of    // the result and arr[i]    result = gcd(result, arr[i]);Â
    if (result == 1)     {      return 1;    }  }Â
  // Return the resultant GCD    return result;}Â
// Driver Codeint main(){Â Â // Given array arr[]Â Â int arr[] = {2, 3, 4};Â
  int N = sizeof(arr) /           sizeof(arr[0]);Â
  // Function Call  cout << findGCD(arr, N);  return 0;}Â
// This code is contributed by 29AjayKumar |
Java
// Java program for the above approachÂ
public class GCD {Â
    // Function to return gcd of a and b    static int gcd(int a, int b)    {        // Base Case        if (a == 0)            return b;Â
        // Recursive Call        return gcd(b % a, a);    }Â
    // Function to find gcd of array    static int findGCD(int arr[], int N)    {        // Initialise the result        int result = 0;Â
        // Traverse the array arr[]        for (int element : arr) {Â
            // Update result as gcd of            // the result and arr[i]            result = gcd(result, element);Â
            if (result == 1) {                return 1;            }        }Â
        // Return the resultant GCD        return result;    }Â
    // Driver Code    public static void main(String[] args)    {        // Given array arr[]        int arr[] = { 2, 3, 4 };Â
        int N = arr.length;Â
        // Function Call        System.out.println(findGCD(arr, N));    }} |
Python3
# Python3 program for the above approach Â
# Function to return gcd of a and bdef gcd(a, b):         # Base Case    if (a == 0):        return bÂ
    # Recursive call    return gcd(b % a, a)Â
# Function to find gcd of arraydef findGCD(arr, N):         # Initialise the result    result = 0Â
    # Traverse the array arr[]    for element in arr:Â
        # Update result as gcd of        # the result and arr[i]        result = gcd(result, element)Â
        if (result == 1):            return 1Â
    # Return the resultant GCD    return resultÂ
# Driver CodeÂ
# Given array arr[]arr = [ 2, 3, 4 ]Â
N = len(arr)Â
# Function callprint(findGCD(arr, N))Â
# This code is contributed by sanjoy_62 |
C#
// C# program for the above approach using System; Â
class GFG{Â
// Function to return gcd of a and bstatic int gcd(int a, int b){         // Base Case    if (a == 0)        return b;Â
    // Recursive call    return gcd(b % a, a);}Â
// Function to find gcd of arraystatic int findGCD(int[] arr, int N){         // Initialise the result    int result = 0;Â
    // Traverse the array arr[]    foreach(int element in arr)    {Â
        // Update result as gcd of        // the result and arr[i]        result = gcd(result, element);Â
        if (result == 1)        {            return 1;        }    }Â
    // Return the resultant GCD    return result;}Â
// Driver Codepublic static void Main(){Â Â Â Â Â Â Â Â Â // Given array arr[]Â Â Â Â int[] arr = { 2, 3, 4 };Â
    int N = arr.Length;Â
    // Function call    Console.WriteLine(findGCD(arr, N));}}Â
// This code is contributed by sanjoy_62 |
Javascript
<script>Â
// JavaScript program for// the above approachÂ
    // Function to return gcd of a and b    function gcd(a, b)    {        // Base Case        if (a == 0)            return b;          // Recursive Call        return gcd(b % a, a);    }      // Function to find gcd of array    function findGCD(arr, N)    {        // Initialise the result        let result = 0;          // Traverse the array arr[]        for (let element in arr) {              // Update result as gcd of            // the result and arr[i]            result = gcd(result, element);              if (result == 1) {                return 1;            }        }          // Return the resultant GCD        return result;    }Â
// Driver codeÂ
          // Given array arr[]        let arr = [ 2, 3, 4 ];          let N = arr.length;          // Function Call        document.write(findGCD(arr, N));                             </script> |
Output
1
Time Complexity: O(N*logN), where N is the size of the given array.
Auxiliary Space: O(N)
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