You are given a binary string s.
You can perform the following operation on the string any number of times:
i from the string where i + 1 < s.length such that s[i] == '1' and s[i + 1] == '0'.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:
i = 0. The resulting string is s = "0011101".i = 4. The resulting string is s = "0011011".i = 3. The resulting string is s = "0010111".i = 2. The resulting string is s = "0001111".Example 2:
Input: s = "00111"
Output: 0
Constraints:
1 <= s.length <= 105s[i] is either '0' or '1'.Problem Overview: Given a binary string s, determine the maximum number of operations required to move all 1s to the end. An operation effectively moves a 1 to the right across a 0. The goal is to count how many such moves can occur before all 1s are positioned after every 0.
Approach 1: Greedy Counting (O(n) time, O(1) space)
The key observation: every time a 1 appears before a 0, that pair will eventually require a move. This turns the problem into counting inversions of the form (1,0). Scan the string left to right while tracking how many 1s have been seen. When a 0 appears, all previously seen 1s could move past it through operations, so add that count to the result. Continue until the end of the string. This greedy scan works because each 1 can cross each 0 at most once, and counting these pairs directly gives the maximum number of operations.
This technique is common in problems involving binary ordering and swap-style operations, especially in greedy and string processing tasks. The algorithm performs a single pass and uses only a couple of counters.
Approach 2: Scan and Balance Counting (O(n) time, O(1) space)
Another way to view the problem is by balancing the number of 1s that still need to move. Iterate through the string while maintaining a running count of active 1s. Each time a 0 appears and there are pending 1s to the left, those 1s will eventually move across this position. Add the active 1 count to the total operations. Conceptually, the algorithm treats each 0 as a checkpoint that all previous 1s must cross.
This formulation highlights the counting aspect of the problem and mirrors how many swap-like transformations occur when pushing elements to the end. Implementation remains a single linear scan with constant memory.
Recommended for interviews: The greedy counting approach. It reduces the problem to counting 1-before-0 pairs in one pass, which demonstrates strong pattern recognition. Explaining the inversion-style insight shows the interviewer you can convert a simulation problem into a simple counting strategy.
In this approach, we will iterate through the string and count the number of operations we can perform. Each valid operation will involve swapping a pair of '10' to '01' by tracking the zeros followed by ones.
The function maximumOperations iterates through the string, counting the number of zeros we encounter. When a '1' follows the counted zeros, we can perform operations to push this '1' through all zero positions, incrementing our operations counter. This allows all '1's to continually shift through to the end.
Time Complexity: O(n), as we iterate through the string once.
Space Complexity: O(1), using constant extra space.
In this approach, we track the imbalance between '1's and '0's to calculate possible operations. The idea is to find how many '1's can become "stuck" behind '0's and count how many operations it would take to move each '1' to the end of the '0s'.
Here, the string is scanned to calculate an imbalance, which increases when a '1' is encountered and contributes to potential operations whenever a '0' follows. This approach tallies up by simulating moving '1's over '0's, counting each necessary move.
Time Complexity: O(n), going through the string once.
Space Complexity: O(1), no extra space beyond constant counters.
| Approach | Complexity |
|---|---|
| Greedy Approach | Time Complexity: O(n), as we iterate through the string once. |
| Scan and Balance Counting | Time Complexity: O(n), going through the string once. |
| Approach | Time | Space | When to Use |
|---|---|---|---|
| Greedy Counting | O(n) | O(1) | Best general solution. Single pass counting of 1-before-0 pairs. |
| Scan and Balance Counting | O(n) | O(1) | Useful when reasoning about active 1s moving across zeros during the scan. |
Maximum Number of Operations to Move Ones to the End | Interview Style | Leetcode 3228 | MIK • codestorywithMIK • 6,127 views views
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