Number of Times Binary String Is Prefix-Aligned

Medium

Asked by:

6 views

Topics:

Arrays

You have a 1-indexed binary string of length n where all the bits are 0 initially. We will flip all the bits of this binary string (i.e., change them from 0 to 1) one by one. You are given a 1-indexed integer array flips where flips[i] indicates that the bit at index flips[i] will be flipped in the i^th step.

A binary string is prefix-aligned if, after the i^th step, all the bits in the inclusive range [1, i] are ones and all the other bits are zeros.

Return the number of times the binary string is prefix-aligned during the flipping process.

Example 1:

Input: flips = [3,2,4,1,5]
Output: 2
Explanation: The binary string is initially "00000".
After applying step 1: The string becomes "00100", which is not prefix-aligned.
After applying step 2: The string becomes "01100", which is not prefix-aligned.
After applying step 3: The string becomes "01110", which is not prefix-aligned.
After applying step 4: The string becomes "11110", which is prefix-aligned.
After applying step 5: The string becomes "11111", which is prefix-aligned.
We can see that the string was prefix-aligned 2 times, so we return 2.

Example 2:

Input: flips = [4,1,2,3]
Output: 1
Explanation: The binary string is initially "0000".
After applying step 1: The string becomes "0001", which is not prefix-aligned.
After applying step 2: The string becomes "1001", which is not prefix-aligned.
After applying step 3: The string becomes "1101", which is not prefix-aligned.
After applying step 4: The string becomes "1111", which is prefix-aligned.
We can see that the string was prefix-aligned 1 time, so we return 1.

Constraints:

n == flips.length
1 <= n <= 5 * 10⁴
flips is a permutation of the integers in the range [1, n].

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the range of values for `n` (the length of the binary string) and for the elements within the `flips` array?
If the `flips` array is empty, should I return 0?
Could you define more explicitly what constitutes a 'sorted prefix'? Is it sorted in ascending order?
Are the values in the `flips` array guaranteed to be within the range [1, n], inclusive?
If no prefix is ever prefix-aligned, what should the function return?

Brute Force Solution

Approach

The brute force approach involves examining every step of the binary string. We'll look at each prefix and check if it meets the alignment condition. Specifically, we will see if the maximum position of a one so far is equal to the count of ones up to that point.

Here's how the algorithm would work step-by-step:

Begin by checking the first digit of the binary string.
See if the largest location of a '1' from the start of the string up to this point is equal to the number of '1's from the start of the string up to this point.
Now, expand your view to the first two digits.
Again, check if the largest location of a '1' within these two digits equals the total number of '1's within these two digits.
Continue expanding your view, one digit at a time, always checking the condition about the largest location of a '1' and the number of '1's.
Every time the check passes for a prefix, increase a counter.
After checking all digits, the final count will be the number of times the condition was met.

Code Implementation

def prefix_aligned_count_brute_force(binary_string):
    prefix_aligned_count = 0
    max_position_so_far = 0
    ones_count = 0

    for index in range(len(binary_string)):
        if binary_string[index] == '1':
            # Update maximum position of '1' seen so far.
            max_position_so_far = max(max_position_so_far, index + 1)
            ones_count += 1

        # Crucial step: Checking the alignment condition.
        if max_position_so_far == ones_count:
            prefix_aligned_count += 1

    return prefix_aligned_count

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the binary string of length n once. In each iteration, it checks if the prefix is aligned based on the maximum index of '1' and the count of '1's. The operations inside the loop (finding max index and counting 1s up to the current position) can be done in constant time since we can maintain the max index and count of 1s so far with each iteration. Therefore, the time complexity is directly proportional to the length of the string n, resulting in a Big O complexity of O(n).

Space Complexity

O(1) – The algorithm iterates through the binary string, keeping track of the largest location of a '1' encountered so far and the number of '1's. It uses only a few integer variables to store these counts and the final result. The space used by these variables is constant and does not depend on the length of the input binary string, which we can denote as N. Therefore, the auxiliary space complexity is O(1).

Optimal Solution

Approach

The key insight is to realize we only care about the 'prefix-aligned' property. We can track the maximum value seen so far and the number of '1's we've encountered to efficiently determine when the string is prefix-aligned at each point.

Here's how the algorithm would work step-by-step:

Keep track of the largest number seen so far in the input string.
Also, keep track of how many '1's you have seen in the string.
For each position in the input string, check if the largest number seen up to that position is equal to the count of '1's.
If they are equal, it means that all the positions from 1 up to the current position are filled with '1's based on the value of the highest index seen so far, and so we increment our counter.
Finally, the total count of positions where the condition is met is the result.

Code Implementation

def prefix_aligned_ones(binary_string):
    max_index_seen = 0
    ones_count = 0
    aligned_count = 0

    for index, value in enumerate(binary_string):
        if int(value) == 1:
            ones_count += 1

        max_index_seen = max(max_index_seen, index + 1)

        # Crucial step: check for prefix-alignment.
        if max_index_seen == ones_count:
            aligned_count += 1

    return aligned_count

Big(O) Analysis

Time Complexity

O(n) – The provided solution iterates through the input string of size n exactly once. Inside the loop, it performs constant-time operations: updating the maximum value seen so far, counting the number of ones, and comparing these values. Since the number of operations grows linearly with the size of the input, the time complexity is O(n).

Space Complexity

O(1) – The algorithm uses a constant amount of extra space. It only stores two integer variables: one to track the largest number seen so far and another to count the number of 1s encountered. These variables occupy a fixed amount of memory, independent of the input string's length (N). Therefore, the auxiliary space complexity is O(1).

Edge Cases

Case	How to Handle
Null or empty input array	Return 0 immediately since no flips are performed.
Input array of size 1	Return 1 since the first element will always be aligned.
Input array with all elements equal to n	The prefix will be aligned after each flip, resulting in a count of n.
Input array with elements in strictly increasing order (1, 2, 3, ..., n)	The prefix will always be aligned after each flip, leading to a count of n.
Input array with elements in strictly decreasing order (n, n-1, n-2, ..., 1)	The prefix will be aligned only after all elements are flipped; thus, the count is 1.
Input array contains duplicate values	The latest occurrence of each index being flipped should be considered, so subsequent flips overwrite previous ones.
Integer overflow if 'n' is very large	Use a data type capable of storing large numbers (e.g., long in Java) for calculations involving 'n'.
Large input array causing time limit exceeded	Optimize the prefix check to O(1) by maintaining the maximum flipped index so far and comparing it with the current index.

Null or empty input array

How to Handle:

Return 0 immediately since no flips are performed.

Input array of size 1

How to Handle:

Return 1 since the first element will always be aligned.

Input array with all elements equal to n

How to Handle:

The prefix will be aligned after each flip, resulting in a count of n.

Input array with elements in strictly increasing order (1, 2, 3, ..., n)

How to Handle:

The prefix will always be aligned after each flip, leading to a count of n.

Input array with elements in strictly decreasing order (n, n-1, n-2, ..., 1)

How to Handle:

The prefix will be aligned only after all elements are flipped; thus, the count is 1.

Input array contains duplicate values

How to Handle:

The latest occurrence of each index being flipped should be considered, so subsequent flips overwrite previous ones.

Integer overflow if 'n' is very large

How to Handle:

Use a data type capable of storing large numbers (e.g., long in Java) for calculations involving 'n'.

Large input array causing time limit exceeded

How to Handle:

Optimize the prefix check to O(1) by maintaining the maximum flipped index so far and comparing it with the current index.