Problem Description

In this LeetCode problem, we are given an initially zeroed binary string of length n, where the indexes are 1-indexed. Throughout a sequence of steps defined by an integer array flips, we flip the bits of this binary string from 0 to 1. The ith element of flips represents the bit position in the binary string that will be flipped during the ith step of the process.

A binary string is considered prefix-aligned after the ith step if all the bits from the beginning of the string to position i are set to 1, while the rest of the string remains at 0. The task is to calculate the total number of times the binary string is prefix-aligned during the entire flipping process.

For example:

- Initial binary string of length 5: 00000
- flips sequence: [3,2,4,1,5]

After each flip:
- Step 1, flip position 3: 00100 (Not prefix-aligned as the first bit is still 0)
- Step 2, flip position 2: 01100 (Not prefix-aligned as the first two bits are not all 1)
- Step 3, flip position 4: 01110 (Not prefix-aligned as the first four bits are not all 1)
- Step 4, flip position 1: 11110 (Not prefix-aligned as the first four bits are not all 1)
- Step 5, flip position 5: 11111 (Prefix-aligned as all the bits are 1)

Hence, the binary string is prefix-aligned 1 time during the flipping process.

Understanding the problem is crucial before attempting to create a solution.

Intuition

The intuition behind the solution is to keep track of the highest bit position (let's call it mx) that has been flipped at each step. For each step i, we compare mx with the current step index i: if mx is equal to i, it means that all bits up to the current step index have been flipped to 1, and hence the string is prefix-aligned.

Here is how this intuition is applied to the solution:

1. Initialize the counter ans to keep track of the number of times the string is prefix-aligned, and mx to record the highest bit position flipped so far.
2. Iterate through each flip in the sequence, keeping track of the current step number i (starting from 1).
3. Update mx to be the maximum between its current value and the bit position x of the current flip.
4. If mx is equal to i after flipping the bit at position x, increment ans by 1 since the string is currently prefix-aligned.
5. Continue the loop until all flips are processed.
6. Return ans, the total count of prefix-aligned occurrences.

Using this approach, we can efficiently determine the number of times the input binary string is prefix-aligned during the flipping process by only keeping track of the maximum flip position and the current step index.

Solution Approach

The solution provided follows an iterative approach with two primary variables in play: ans and mx. The variable ans serves as a counter for the instances when the binary string is prefix-aligned, and mx keeps track of the maximum index that has been flipped so far.

1. Initialize ans to 0. This variable will count the number of times the binary string is prefix-aligned during the flip sequence.
2. Initialize mx to 0. This variable represents the maximum position of the flipped bit encountered up to the current step.

We use a for loop to iterate through each flip in the provided flips list. The loop uses the built-in enumerate function in Python to loop over flips, starting from index 1 because the problem is 1-indexed.

3. During each iteration of the loop, we get two values: i, the step number starting from 1, and x, the current position to flip.
4. For each flip, update mx to be the maximum between the current mx and the flip position x. This is achieved by the expression mx = max(mx, x), which ensures mx always reflects the farthest position that was flipped till the current step.
5. The check mx == i will be True if all bits from the start to the current position i are 1 (thus, the string is prefix-aligned). If so, we increment ans by 1. This leverages the fact that a binary string is prefix-aligned if the maximum flipped position at step i is equal to i itself. Any flip sequence that has the largest flip within the range [1, i] at i ensures that all preceding bits are already flipped to 1.
6. Finally, the loop concludes once all the elements in flips have been iterated, and the ans value is returned. ans now contains the number of times the binary string was prefix-aligned during the process.

This algorithm doesn't use any complex data structures and requires no additional space besides the two variables ans and mx, making it space-efficient. The time complexity of the algorithm is O(n) where n is the number of flips since we are going through each flip exactly once. It efficiently solves the problem by keeping track of only the current state necessary to determine prefix alignment at each step without reconstructing the binary string at each instance.

Example Walkthrough

Let's walk through a small example to illustrate the solution approach.

Suppose we have an initially zeroed binary string of length 4 and the following flips sequence: [1,3,2,4].

• Initial binary string: 0000

We iterate through each flip, updating the maximum position mx flipped, and checking after each step if the string is prefix-aligned.

• Step 1, flip position 1: The binary string after the flip becomes 1000. mx is updated to 1. Since mx (1) equals the step number i (1), we increment ans by 1. The string is prefix-aligned.

  • New binary string: 1000
  • mx = 1
  • ans = 1

• Step 2, flip position 3: The binary string becomes 1010. The new mx is 3, which is now the highest position flipped so far. Since mx (3) does not equal the step number i (2), ans is not incremented. The string is not prefix-aligned.

  • New binary string: 1010
  • mx = 3
  • ans still = 1

• Step 3, flip position 2: After this flip, the binary string is 1110, and mx remains 3. As mx (3) is not equal to the step number i (3), ans remains the same. The string is not prefix-aligned.

  • New binary string: 1110
  • mx still = 3
  • ans still = 1

• Step 4, flip position 4: The final binary string is 1111. mx is updated to 4, which now equals the step number i (4). We increment ans by 1 again because the string is prefix-aligned.

  • New binary string: 1111
  • mx = 4
  • ans = 1 + 1 = 2

At the end of the flipping process, we have encountered 2 instances where the binary string is prefix-aligned, which means our answer ans is 2.

Using the solution approach, we keep track of the maximum flipped position and check for prefix alignment in each step efficiently. The variables ans and mx enable us to do this without needing additional data structures or performing complex operations. This makes the algorithm space-efficient and straightforward to implement with a time complexity of O(n), where n is the number of flips.

Solution Implementation

Time and Space Complexity

Time Complexity

The given algorithm consists of a single for loop that iterates through every element in the flips list exactly once. The enumeration of flips does not change the overall time complexity. Inside the loop, the algorithm performs a constant amount of work on each iteration, including a comparison and an assignment. Therefore, the time complexity of this function is O(n) where n is the number of elements in flips.

Space Complexity

The space complexity of the algorithm is defined by the amount of additional memory used by the algorithm as a function of the input size. Within the provided algorithm, ans and mx are the only variables that occupy extra space, and their memory footprint does not depend on the size of the input. As a result, the space complexity is O(1) because the amount of extra memory used is constant, irrespective of the input size.

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