Problem 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 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 = "0<strong><u>001</u></strong>10".

Return the maximum number of operations that you can perform.

Intuition

The problem revolves around shifting 1s past 0s efficiently in the binary string to maximize the number of such shifts. To find a solution, we can utilize a greedy approach:

1. Track 1s with Counting: We maintain a counter cnt to record the number of 1s encountered so far. This is useful because any contiguous set of 1s can potentially be shifted past a 0 that follows these 1s.

2. Iterate Through the String: As we iterate through the string, each time we encounter a 1, we increment our cnt. Whenever we hit a 0 and the previous character is a 1, it indicates a possible shift operation for all 1s counted.

3. Count the Operations: For each possible move operation, add the current cnt to the total number of operations since these many 1s can effectively be shifted past the 0.

In essence, the logic is about counting how many 1s can be shifted past every 0 encountered right after a block of 1s, leveraging these counts for the maximum number of operations possible.

Learn more about Greedy patterns.

Solution Approach

To solve this problem, we employ a greedy approach with two main components: counting and iterating. Here's a structured breakdown of the implementation:

1. Initialize Variables:

   • We initiate two variables: ans to store the final count of maximum operations and cnt to count the current number of 1s encountered in the string.

2. Iterate Through the String:

   • We loop through each character (c) in the string s using an index i.

3. Count the 1s:

   • If the current character c is 1, we increment cnt by one. This counting tracks how many 1s can potentially be moved.

4. Check and Calculate Operations at 0s:

   • If the current character is 0, and the previous character (s[i-1]) is 1, it indicates the end of a segment that can be shifted.
   • Increase ans by the number cnt, since all counted 1s can slide past this 0.

5. Return the Result:

   • After completing the iteration over s, ans will hold the total number of shift operations that were possible, which we return as the result.

In code, this approach is implemented as follows:

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

In summary, this solution effectively calculates the maximum number of operations by leveraging the properties of binary strings and using a count to track movable 1s as it progresses through each character.

Example Walkthrough

Let's consider the binary string s = "1101000" to illustrate the solution approach.

1. Initialization:

   • Set ans = 0 to store the total operations.
   • Set cnt = 0 to count the number of 1s seen so far.

2. Iterate Through the String:

   • Index 0: c = '1'
     • Increment cnt to 1.
   • Index 1: c = '1'
     • Increment cnt to 2.
   • Index 2: c = '0'
     • The previous character was a 1, which means we can shift 1s past this 0.
     • Add cnt (which is 2) to ans. Now, ans = 2.
   • Index 3: c = '1'
     • Increment cnt to 3.
   • Index 4: c = '0'
     • The previous character was a 1, again allowing a shift.
     • Add cnt (which is 3) to ans. Now, ans = 5.
   • Index 5: c = '0'
     • No shift possible since the previous character was not a 1.
   • Index 6: c = '0'
     • Similarly, no shift is possible here.

3. Return the Result:

   • The total number of shift operations is ans = 5.

Thus, the maximum number of operations that can be performed is 5, where we shift 1s past 0s at indices 2 and 4.

Solution Implementation

Time and Space Complexity

The time complexity of the code is O(n), where n is the length of the string s. This is because the code iterates through the string s once, performing constant-time operations for each character.

The space complexity is O(1) because the code uses a constant amount of extra space regardless of the input size. The variables ans and cnt do not depend on the length of the input string.

Learn more about how to find time and space complexity quickly.