Java Program to Make All the Array Elements Equal to One by GCD Operations
Last Updated :
28 Jan, 2021
You are given an array A[] of N integers. Your task is to make all the elements of the final array equal to 1. You can perform the below operation any number of times (possibly zero). Choose two indices <i, j>, (0<=i,j<N) and replace Ai and Aj both with the GCD ( Greatest Common Divisor ) of Ai and Aj.
Example :
Input : A[] = {2 , 4 , 6 ,9}
Output: Yes
Explanation: First choose 4 and 9 their GCD will be 1, so now the array is {2, 1, 6, 1}. Now, we can choose 2 and 1, their GCD is 1 so, the array becomes {1,1,6,1}. Finally we can choose any 1 with 6 as their GCD is 1. Thus the final array becomes {1, 1, 1, 1}. So, we can make all the array elements equal to 1.
Input : A[] = {9 , 6 , 15}
Output: No
Naive Approach:
- If we get the GCD of any pair equal to 1, then we can make all array elements 1 by taking that number and 1 one by one.
- So, find any co-prime pair exists or not because the GCD of Co-prime pair is equal to 1.
- If the value of GCD pair is equal to 1, then print “Yes”.
- Else print “No”.
Below is the implementation of the above approach :
Java
// Java program to make all the array elements
// equal to one by GCD operations
import java.io.*;
import java.lang.*;
import java.util.*;
public class Main {
// Function that returns whether it is possible or not
// to make all the array elements equal to one (1)
static boolean possible(int A[])
{
int n = A.length;
// Check all the possible pairs
for (int i = 0; i < n - 1; i++) {
for (int j = i + 1; j < n; j++) {
int gcd = gcd(A[i], A[j]);
// If the gcd is equal to 1 , return true
if (gcd == 1)
return true;
}
}
return false;
}
// Recursive function to return gcd of a and b
static int gcd(int a, int b)
{
if (b == 0)
return a;
return gcd(b, a % b);
}
// Driver Code
public static void main(String[] args)
{
// Given Array
int A[] = { 2, 4, 6, 9 };
boolean answer = possible(A);
System.out.println(answer == true ? "Yes" : "No");
}
}
Time Complexity: O(N^2*log N), where N is the length of an array.
Efficient Approach:
- Calculate the GCD of the whole array.
- If there exists any co-prime pair then their GCD will be 1.
- So after this, any number comes, the GCD will remain 1.
- If at any point in time the GCD becomes 1, then break the loop and print “Yes”.
- If the final value of GCD after the whole iteration is not equal to one, then print “No”.
Below is the implementation of the above approach :
Java
// Java program to make all the array elements
// equal to one by GCD operations
import java.io.*;
import java.lang.*;
import java.util.*;
public class Main {
// Function that returns whether it is possible or not
// to make all the array elements equal to one (1)
static boolean possible(int A[])
{
int n = A.length;
int gcd = 0;
// calculate GCD of the whole array
for (int i = 0; i < n; i++) {
gcd = gcd(gcd, A[i]);
// If the gcd is equal to 1 , return true
if (gcd == 1)
return true;
}
return false;
}
// Recursive function to return gcd of a and b
static int gcd(int a, int b)
{
if (b == 0)
return a;
return gcd(b, a % b);
}
// Driver Code
public static void main(String[] args)
{
// Given Array
int A[] = { 2, 4, 6, 9 };
boolean answer = possible(A);
System.out.println(answer == true ? "Yes" : "No");
}
}
Time Complexity: O(N*logM), where N is the length of an array and M is the smallest element of an array.
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