Class Solution
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public final class Solution3228 - 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
ifrom the string wherei + 1 < s.lengthsuch thats[i] == '1'ands[i + 1] == '0'.Move the character
s[i]to the right until it reaches the end of the string or another'1'. For example, fors = "010010", if we choosei = 1, the resulting string will be <code>s = "0**<ins>001</ins>**10"</code>.
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 <code>s = "<ins> 001 </ins>1101"</code>.Choose index
i = 4. The resulting string is <code>s = "0011<ins> 01 </ins>1"</code>.Choose index
i = 3. The resulting string is <code>s = "001**<ins>01</ins>**11"</code>.Choose index
i = 2. The resulting string is <code>s = "00**<ins>01</ins>**111"</code>.
Example 2:
Input: s = "00111"
Output: 0
Constraints:
<code>1 <= s.length <= 10<sup>5</sup></code>
s[i]is either'0'or'1'.
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Constructor Summary
Constructors Constructor Description Solution()
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Method Summary
Modifier and Type Method Description final IntegermaxOperations(String s)-
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Method Detail
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maxOperations
final Integer maxOperations(String s)
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