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.

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++

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// C++ implementation to find the
// periodic string with minimum period
#include <bits/stdc++.h>
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;
}

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Python3

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# 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    

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Output:

10101010101010

Time Complexity :O(N)

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Improved By : mohit kumar 29