Check if it is possible to rearrange a binary string with alternate 0s and 1s

Given a binary string of length at-least two. We need to check if it is possible to rearrange a binary string such that there alternate 0’s and 1’s. If possible then output is YES, otherwise output is NO.

Examples:

Input : 1011
Output : NO
We can’t rearrange the string such that it has alternate 0s and 1s.

Input : 1100
Output : YES
There are exactly two ways to rearrange the string, and they are 0101 or 1010 .



We can place all 0’s in even position and all 1’s in odd position or we can place all 0’s in odd position and all 1’s in even position. If the length of the string is even then to satisfy the given condition the count of 1’s and 0’s must be equal. If the length of the string is odd then to satisfy the given condition the absolute difference of count
of 1’s and 0’s must be one.

Below is the implementation of above approach:

C++

// CPP program to check if we can rearrange a
// string such that it has alternate 0s and 1s.
#include <bits/stdc++.h>
using namespace std;
  
// function to check the binary string
bool is_possible(string s)
{
    // length of string
    int l = s.length();
  
    int one = 0, zero = 0;
  
    for (int i = 0; i < l; i++) {
  
        // count zero's
        if (s[i] == '0')
            zero++;
  
        // count one's
        else
            one++;
    }
  
    // if length is even
    if (l % 2 == 0) 
        return (one == zero);
  
    // if length is odd
    else 
        return (abs(one - zero) == 1);
}
  
// Driver code
int main()
{
    string s = "100110";
    if (is_possible(s))
      cout << "Yes";
    else
      cout << "No";
    return 0;
}

Python 3

# Python3 program to check if 
# we can rearrange a
# string such that it has alternate 
# 0s and 1s.
  
# function to check the binary string
def is_possible(s):
  
    # length of string
    l = len(s)
  
    one = 0
    zero = 0
  
    for i in range(0,l) :
  
        # count zero's
        if (s[i] == '0'):
            zero += 1
  
        # count one's
        else:
            one += 1
  
    # if length is even
    if (l % 2 == 0) :
        return (one == zero)
  
    # if length is odd
    else:
        return (abs(one - zero) == 1)
  
# Driver code
if __name__ == "__main__":
    s = "100110"
    if (is_possible(s)):
        print("Yes")
    else:
        print("No")
  
# This code is contributed by
# ChitraNayal

Output:

Yes

Time Complexity: O(l), where l is length of the binary string.



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Improved By : ChitraNayal



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