# Periodic Binary String With Minimum Period and a Given Binary String as Subsequence.

Periodic Binary String : A Binary string is called periodic if it can be written as repetition of a binary string of smaller or same length. For example, 101010 is a periodic binary string with period 10 as we can get the string by repeatedly appending 10 to itself. In general, the string S with period P means, Si is equal to Si + P.

Problem : Given an binary string S, the task is to find the periodic string with minimum possible period with the following additional conditions
1) The given string S should be a subsequence of the result
2) Length of the result should not be more than twice the length of input string.

Examples:

Input:S = “01”
Output:0101
Explanation:The output string has period as 2 and has given string as subsequence.

Input :S = “111”
Output :111
Explanation:The output string has period as 1.

Input:S = “110”
Output: 101010
Explanation:The output string has period as 2 and has given string as subsequence. Please note that 110110 is not answer because the period length is more.

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach:
The main idea depends on two possibility :

1. If the string consists all 1s or all 0s, the answer is the given string S itself having period as 1.
2. If the string consists of dissimilar elements, find the string with period 2 and having length as twice the length of given string S

Below is the implementation of the above approach:

## C++

 `// C++ implementation to find the ` `// periodic string with minimum period ` `#include ` `using` `namespace` `std; ` ` `  `// Function to find the periodic string ` `// with minimum period ` `void` `findPeriodicString(string S) ` `{ ` `    ``int` `l = 2 * S.length(); ` ` `  `    ``int` `count = 0; ` `    ``for` `(``int` `i = 0; i < S.length(); i++) { ` `        ``if` `(S[i] == ``'1'``) ` `            ``count++; ` `    ``} ` ` `  `    ``// Print the string S if it ` `    ``// consists of similar elements ` `    ``if` `(count == S.length() || count == 0) ` `        ``cout << S << ``"\n"``; ` ` `  `    ``// Find the required periodic ` `    ``// string with period 2 ` `    ``else` `{ ` `        ``char` `arr[l]; ` `        ``for` `(``int` `i = 0; i < l; i += 2) { ` `            ``arr[i] = ``'1'``; ` `            ``arr[i + 1] = ``'0'``; ` `        ``} ` ` `  `        ``for` `(``int` `i = 0; i < l; i++) ` `            ``cout << arr[i]; ` `        ``cout << ``"\n"``; ` `    ``} ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``string S = ``"1111001"``; ` `    ``findPeriodicString(S); ` `    ``return` `0; ` `} `

## Python3

 `# Python3 implementation to find the ` `# periodic with minimum period ` ` `  `# Function to find the periodic string ` `# with minimum period ` `def` `findPeriodicString(S): ` `    ``l ``=` `2` `*` `len``(S) ` ` `  `    ``count ``=` `0` `    ``for` `i ``in` `range``(``len``(S)): ` `        ``if` `(S[i] ``=``=` `'1'``): ` `            ``count ``+``=` `1` ` `  `    ``# Print the S if it ` `    ``# consists of similar elements ` `    ``if` `(count ``=``=` `len``(S) ``or` `count ``=``=` `0``): ` `        ``print``(S) ` ` `  `    ``# Find the required periodic ` `    ``# with period 2 ` `    ``else``: ` `        ``arr ``=` `[``'0'``]``*``l ` `        ``for` `i ``in` `range``(``0``, l, ``2``): ` `            ``arr[i] ``=` `'1'` `            ``arr[i ``+` `1``] ``=` `'0'` ` `  `        ``for` `i ``in` `range``(l): ` `            ``print``(arr[i],end``=``"") ` ` `  `# Driver Code ` `if` `__name__ ``=``=` `'__main__'``: ` `    ``S ``=` `"1111001"` `    ``findPeriodicString(S) ` `     `  `# This code is contributed by mohit kumar 29     `

Output:

```10101010101010
```

Time Complexity :O(N) My Personal Notes arrow_drop_up Check out this Author's contributed articles.

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Improved By : mohit kumar 29