Problem Description

You start with a binary string s of length n that contains all zeros (like "0000..."). Your goal is to transform it into a given target binary string target of the same length.

You can perform a special flip operation: choose any index i (where 0 <= i < n) and flip all bits from position i to the end of the string (position n-1). Flipping means changing all '0' characters to '1' and all '1' characters to '0'.

For example, if you have string "00000" and flip from index 2, you get "00111". If you then flip from index 1, you get "01000".

The problem asks you to find the minimum number of flip operations needed to transform the initial all-zeros string into the target string.

The solution uses a clever observation: as you scan the target string from left to right, you need to perform a flip operation whenever the current bit differs from what it should be after considering all previous flips. The variable ans tracks both the count of operations and whether an odd or even number of flips have been performed (using ans & 1). When (ans & 1) ^ int(v) equals 1, it means the current position needs a flip to match the target, so we increment the operation count.

Intuition

The key insight is to think about the problem from left to right. When we scan the target string from the beginning, we need to understand what state each position is in after all the flips we've done so far.

Initially, all bits are 0. As we move through the target string:

• If we encounter a 1 in the target and our current bit is 0, we need to flip from this position
• If we encounter a 0 in the target and our current bit is 1, we also need to flip from this position

But here's the clever part: we don't need to actually maintain the entire string and perform the flips. Instead, we can track the cumulative effect of all flips using a parity counter.

Every flip operation from position i affects all positions from i to the end. So for any position j, the actual value at that position depends on how many flips have started at or before position j. If an odd number of flips have occurred, the bit is inverted from its original state; if even, it remains the same as the original.

Since we start with all zeros:

• After an even number of flips (0, 2, 4...), a position that started as 0 remains 0
• After an odd number of flips (1, 3, 5...), a position that started as 0 becomes 1

Therefore, at each position, we check: does the current state (determined by the parity of flips so far) match what we need in the target? If not, we perform a flip starting from this position. The expression (ans & 1) ^ int(v) checks exactly this - it returns 1 when there's a mismatch between the current state and the target, indicating we need another flip operation.

This greedy approach works because once we fix a position by flipping, all previous positions remain unchanged, so we never need to revisit them.

Learn more about Greedy patterns.

Solution Approach

The solution implements a greedy algorithm that processes the target string from left to right, maintaining a counter that serves dual purposes: counting operations and tracking parity.

Here's how the implementation works:

1. Initialize counter: Start with ans = 0, which will count the number of flip operations.

2. Iterate through target: Process each character v in the target string sequentially.

3. Check current state: For each position, we need to determine:

   • What the current bit value should be after all previous flips
   • Whether it matches the target bit at this position

4. Parity tracking: The expression ans & 1 gives us the parity of flips so far:

   • If ans & 1 = 0 (even number of flips), the current position is still 0
   • If ans & 1 = 1 (odd number of flips), the current position is 1

5. XOR comparison: The expression (ans & 1) ^ int(v) performs an XOR between:

   • The current state (ans & 1)
   • The target value (int(v))

   This XOR returns 1 (true) when they differ, meaning we need a flip operation.

6. Perform flip: When (ans & 1) ^ int(v) evaluates to true, increment ans += 1. This both:

   • Counts the new flip operation
   • Updates the parity for subsequent positions

The algorithm's efficiency comes from not actually performing the flips on a string. Instead, it uses the mathematical property that the effect of multiple flips can be represented by their parity. Each position only needs to be considered once, making this an O(n) time complexity solution with O(1) space complexity.

For example, with target = "10111":

• Position 0: Need 1, have 0 → flip (ans = 1)
• Position 1: Need 0, have 1 (due to previous flip) → flip (ans = 2)
• Position 2: Need 1, have 0 (even flips) → flip (ans = 3)
• Position 3: Need 1, have 1 (odd flips) → no flip
• Position 4: Need 1, have 1 (odd flips) → no flip
• Result: 3 operations

Example Walkthrough

Let's walk through the algorithm with target = "10111":

Initial state: Start with string "00000" and ans = 0

Position 0 (target = '1'):

• Current parity: ans & 1 = 0 & 1 = 0 (even flips, so position is still '0')
• Target needs: '1'
• Check mismatch: (0) ^ (1) = 1 (true, they differ!)
• Action: Perform flip from position 0 → ans = 1
• String becomes: "11111"

Position 1 (target = '0'):

• Current parity: ans & 1 = 1 & 1 = 1 (odd flips, so position is '1')
• Target needs: '0'
• Check mismatch: (1) ^ (0) = 1 (true, they differ!)
• Action: Perform flip from position 1 → ans = 2
• String becomes: "10000"

Position 2 (target = '1'):

• Current parity: ans & 1 = 2 & 1 = 0 (even flips, so position is '0')
• Target needs: '1'
• Check mismatch: (0) ^ (1) = 1 (true, they differ!)
• Action: Perform flip from position 2 → ans = 3
• String becomes: "10111"

Position 3 (target = '1'):

• Current parity: ans & 1 = 3 & 1 = 1 (odd flips, so position is '1')
• Target needs: '1'
• Check mismatch: (1) ^ (1) = 0 (false, they match!)
• Action: No flip needed

Position 4 (target = '1'):

• Current parity: ans & 1 = 3 & 1 = 1 (odd flips, so position is '1')
• Target needs: '1'
• Check mismatch: (1) ^ (1) = 0 (false, they match!)
• Action: No flip needed

Result: Minimum operations = 3

The key insight is that we don't actually maintain the string - we just track whether each position has been flipped an odd or even number of times using the parity of our counter.

Solution Implementation

Time and Space Complexity

Time Complexity: O(n) where n is the length of the target string. The algorithm iterates through each character in the target string exactly once, performing constant-time operations (bitwise AND, XOR, integer conversion, and conditional increment) for each character.

Space Complexity: O(1). The algorithm uses only a constant amount of extra space - a single integer variable ans to track the number of flips. The space usage does not depend on the input size, as we're not creating any additional data structures that scale with the input.

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

Common Pitfalls

1. Misunderstanding the Flip Operation Scope

A common mistake is thinking that each flip operation only affects a single bit or a limited range. The flip operation actually affects ALL bits from index i to the end of the string. This misunderstanding can lead to incorrect counting of operations.

Wrong approach:

def minFlips(self, target: str) -> int:
    # WRONG: Counting each '1' as needing a flip
    return sum(1 for bit in target if bit == '1')

Correct understanding: Each flip from position i affects all positions from i to n-1, creating a cascading effect that must be tracked.

2. Attempting to Simulate Actual Flips

Some may try to maintain an actual binary string and perform flips on it, which is inefficient and unnecessary.

Inefficient approach:

def minFlips(self, target: str) -> int:
    current = ['0'] * len(target)
    operations = 0
  
    for i in range(len(target)):
        if current[i] != target[i]:
            # Actually flipping all bits from i to end
            for j in range(i, len(target)):
                current[j] = '1' if current[j] == '0' else '0'
            operations += 1
  
    return operations

Better solution: Use parity tracking instead of actual string manipulation, as shown in the optimal solution.

3. Incorrect Parity Logic

Confusing what the parity represents or using wrong logical operations can lead to incorrect results.

Common mistakes:

# WRONG: Using AND instead of XOR
if (flip_count & 1) & int(char):  # This is incorrect logic
    flip_count += 1

# WRONG: Not considering the current state
if int(char) == 1:  # Ignores the effect of previous flips
    flip_count += 1

Correct approach: Use XOR to compare current state with target: (flip_count & 1) ^ int(char)

4. Off-by-One Errors with Bit Conversion

Forgetting to convert the character to integer or using incorrect conversion can cause issues.

Potential error:

# WRONG: Comparing character with integer directly
if (flip_count & 1) ^ char:  # 'char' is a string, not int
    flip_count += 1

Fix: Always convert the character to integer: int(char)

5. Not Recognizing the Greedy Nature

Some might think this problem requires dynamic programming or backtracking to find the optimal solution, leading to overcomplicated approaches.

Key insight: The greedy approach (flipping whenever current state doesn't match target) is always optimal because:

• Once we encounter a mismatch at position i, we must flip at that position
• Delaying the flip won't reduce the total number of operations
• The leftmost unmatched bit always determines the next necessary flip