#3228

Medium

auto_awesomeGreedy choice plus invariant validation

LeetCode Problem Workspace

Maximum Number of Operations to Move Ones to the End

This problem requires finding the maximum number of operations to move ones to the end of a binary string.

Problem Statement

You are given a binary string s consisting of characters '0' and '1'. You are allowed to perform an operation any number of times. In each operation, select the leftmost '1', and swap it with the nearest '0' to its right. Your goal is to return the maximum number of operations you can perform on the string.

The problem emphasizes using greedy choices and checking for invariant properties during the operation. The optimal strategy involves always performing the operation on the leftmost '1' to maximize the number of times the operation can be applied.

Examples

Example 1

Input: s = "1001101"

Output: 4

We can perform the following operations:

Example 2

Input: s = "00111"

Output: 0

Example details omitted.

Constraints

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

Solution Approach

Greedy Approach

The key insight is to always choose the leftmost '1' and swap it with the closest '0'. This ensures that you maximize the number of operations. As you process the string, count how many operations you can make until no further valid swaps are possible.

Invariant Validation

While performing swaps, maintain the invariant that the '1's are always moved to the right. You should check if the next '1' can still be swapped, and stop when no valid swaps can be made.

Time and Space Complexity

The time complexity of the solution depends on the implementation details, but it is generally linear. Space complexity can be kept constant if only a few variables are used for tracking the process.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity of the optimal solution is O(n), where n is the length of the binary string, as we traverse the string once. Space complexity is O(1) because we only need a few variables to track the current state of the string during the operation.

What Interviewers Usually Probe

The candidate demonstrates an understanding of greedy algorithms by explaining the choice of selecting the leftmost '1'.
The candidate should discuss the trade-offs of swapping earlier versus later '1's, as this maximizes operations.
The candidate should be able to articulate the time and space complexity for their solution.

Common Pitfalls or Variants

Common pitfalls

Misunderstanding the operation by performing swaps in non-greedy orders, reducing the number of valid operations.
Not considering edge cases where no operations are possible, such as when all '1's are already at the end of the string.
Failing to optimize for time complexity when implementing the solution, leading to inefficient algorithms.

Follow-up variants

Handling strings with only '0's or only '1's.
Allowing a fixed number of operations rather than maximizing the count.
Changing the operation such that you can only swap a fixed number of '1's.

FAQ

What is the optimal approach for solving Maximum Number of Operations to Move Ones to the End?

The optimal approach uses a greedy strategy where you perform the operation on the leftmost '1' each time to maximize the number of operations.

How do I ensure the greedy choice is valid in this problem?

By selecting the leftmost '1' and swapping it with the nearest '0' to its right, you maximize the number of valid operations. This greedy choice maintains the desired invariant.

What is the time complexity of the solution?

The time complexity is O(n), where n is the length of the binary string, as you traverse the string once to perform the swaps.

Are there any edge cases I should consider?

Yes, consider cases where the string has no '1's or all '1's are already at the end, in which no operations can be performed.

How can GhostInterview help with this problem?

GhostInterview provides detailed insights into using greedy algorithms and optimizing the solution for time and space complexity, ensuring an efficient solution.

terminal

Solution

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.

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

Continue Practicing

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