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3493. Maximum Number Of Operations To Move Ones To The End

String Greedy Counting

Problem - 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 = "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'.

Solutions

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class Solution:
    def maxOperations(self, s: str) -> int:
        result = count = 0
        for i, char in enumerate(s):
            if char == "1":
                count += 1
            elif i and s[i - 1] == "1":
                result += count
        return result

Submission Stats:

  • Runtime: 62 ms (58.08%)
  • Memory: 18 MB (81.44%)