Package g3201_3300.s3228_maximum_number_of_operations_to_move_ones_to_the_end

Class Solution

java.lang.Object
g3201_3300.s3228_maximum_number_of_operations_to_move_ones_to_the_end.Solution
public class Solution extends Object
3228 - Maximum Number of Operations to Move Ones to the End.

Medium

You are given a binary string s.

You can perform the following operation on the string any number of times:

  • Choose any index i from the string where i + 1 < s.length such that s[i] == '1' and s[i + 1] == '0'.
  • Move the character s[i] to the right until it reaches the end of the string or another '1'. For example, for s = "010010", if we choose i = 1, the resulting string will be s = “0**001**10”.

Return the maximum number of operations that you can perform.

Example 1:

Input: s = “1001101”

Output: 4

Explanation:

We can perform the following operations:

  • Choose index i = 0. The resulting string is s = “ 001 1101”.
  • Choose index i = 4. The resulting string is s = “0011 01 1”.
  • Choose index i = 3. The resulting string is s = “001**01**11”.
  • Choose index i = 2. The resulting string is s = “00**01**111”.

Example 2:

Input: s = “00111”

Output: 0

Constraints:

  • 1 <= s.length <= 105
  • s[i] is either '0' or '1'.

  • Constructor Details

    • Solution

      public Solution()

  • Method Details

    • maxOperations

      public int maxOperations(String s)