3228. Maximum Number of Operations to Move Ones to the End
Description
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 wherei + 1 < s.length
such 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 bes = "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 iss = "0011101"
. - Choose index
i = 4
. The resulting string iss = "0011011"
. - Choose index
i = 3
. The resulting string iss = "0010111"
. - Choose index
i = 2
. The resulting string iss = "0001111"
.
Example 2:
Input: s = "00111"
Output: 0
Constraints:
1 <= s.length <= 105
s[i]
is either'0'
or'1'
.
Solutions
Solution 1: Greedy
We use a variable $\textit{ans}$ to record the answer and another variable $\textit{cnt}$ to count the current number of $1$s.
Then, we iterate through the string $s$. If the current character is $1$, then we increment $\textit{cnt}$. Otherwise, if there is a previous character and the previous character is $1$, then the previous $\textit{cnt}$ number of $1$s can be moved backward, and we add $\textit{cnt}$ to the answer.
Finally, we return the answer.
The time complexity is $O(n)$, where $n$ is the length of the string $s$. The space complexity is $O(1)$.
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