# Find the Good String

## Problem Statement

Given an Integer N and and String S of length 4N containing all 0's.

The task is to fill the String S with N number of 1's in such a way

• For any two pair of 1's  inserted in the string,the GCD  of the  position of  that two 1's inserted ≠ 1.
• For any two position  where 1's will be inserted ,they should not be divisible by each other.

Print any Binary String which satisfies above 2 conditions

Examples:

Input:N=3,

S=000000000000

Output: 000101000100

Explanation:In the above string we have to insert N number of 1's which in this case is 3 1's are inserted in the position(1 based indexing) 4,6,10. For any two pair of (4,6,10),any two number's GCD ≠ 1 and no two number between (4,6,10) divides each other. So this is an Valid output.

Input:N=2,S=00000000

Output: 00010100

Approach:The question can be solved using observation that the pattern 4n,4n-2,4n-4.... upto N terms will always satisfy the Above two conditions.

## Solution

Below is the Java implementation of the above approach:

Code

import java.io.*;

class Solution {
public static void findString(char arr[], int n)
{
int strLen = 4 * n;
// filling N 1's in string in pattern
// 4N,4N-2,4N-4....upto N terms
for (int i = 1; i <= n; i++) {
arr[strLen - 1] = '1';
strLen -= 2;
}
// printing the string
System.out.println(arr);
}
// Driver Code
public static void main(String[] args)
{
int n = 3;
char arr[] = new char[4 * n];
Arrays.fill(arr, '0');
// function call
findString(arr, n);
}
}

Output:

000000010101

Time Complexity: O(n)

Auxiliary Space: O(1)

Feel free to comment for any improvement and suggestion in comments section

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