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3228. Maximum Number of Operations to Move Ones to the End

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StringGreedyCounting

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 = "000110".

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 = "0011101".
  • Choose index i = 4. The resulting string is s = "0011011".
  • Choose index i = 3. The resulting string is s = "0010111".
  • Choose index i = 2. The resulting string is s = "0001111".

Example 2:

Input: s = "00111"

Output: 0

 

Constraints:

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

Hints

Hint 1
It is optimal to perform the operation on the lowest index possible each time.
Hint 2
Traverse the string from left to right and perform the operation every time it is possible.
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