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"` `);` ` ` `}` `}` |

**Output**

Yes

**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"` `);` ` ` `}` `}` |

**Output**

Yes

**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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