Solution Explanation

This problem can be solved efficiently using a greedy approach. The core idea is to track the current state (0 or 1) and flip the state only when necessary to match the target string.

Algorithm:

1. Initialization: We start with a count (or ans in the code examples) initialized to 0. This variable keeps track of the number of flips performed. The initial state of the binary string s is all zeros.

2. Iteration: We iterate through the characters of the target string.

3. Flip Check: For each character in target:

   • We check if the current state (count % 2) is different from the character's value (0 or 1). The XOR operation (^) helps us concisely perform this comparison. If they are different, it means a flip is required.
   • If a flip is needed, we increment the count (representing a flip operation).

4. Return: After iterating through the entire target string, the count variable will hold the minimum number of flip operations needed.

Time Complexity Analysis:

The algorithm iterates through the target string once. Therefore, the time complexity is O(n), where n is the length of the target string.

Space Complexity Analysis:

The algorithm uses a constant amount of extra space to store the count variable. Therefore, the space complexity is O(1).

Code Explanation (Python):

class Solution:
    def minFlips(self, target: str) -> int:
        ans = 0  # Initialize the flip count
        for v in target:  # Iterate through the target string
            v = int(v) #Convert the character to integer
            if (ans & 1) ^ v:  #Check if flip is required using XOR
                ans += 1  # Increment the flip count if needed
        return ans

The code in other languages (Java, C++, Go) follows the same logic with minor syntax differences to accommodate the respective language features. The core algorithmic approach remains identical, providing an efficient solution to the problem.