Read N Characters Given read4 II - Call Multiple Times

Hard

Asked by:

10 views

The API: int read4(char *buf) reads at most 4 characters from a file.

The API: int read4(char *buf) reads at most 4 characters from a file. The return value is the actual number of characters read. For example, it returns 3 if there is only 3 characters left in the file.

By using the read4 API, implement the function int read(char *buf, int n) that reads exactly n characters from the file.

Note:
The read function may be called multiple times.

Example 1:

Input: read(buf, 1)
Output: 1
Explanation: The first time the function is called, the API reads 4 characters from the file, which are "a, b, c, d". Then the function only returns the first character to the buffer buf

Example 2:

Input: read(buf, 5)
Output: 5
Explanation: The first time the function is called, the API reads 4 characters from the file, which are "a, b, c, d". Then the function returns the first 4 characters to the buffer buf. 
The second time the function is called, the API reads one character from the file, which is "e". Then the function returns the first character to the buffer buf.

Constraints:

1 <= n <= 1000

Read N Characters Given read4 II - Call Multiple Times

Hard

Asked by:

10 views

The API: int read4(char *buf) reads at most 4 characters from a file.

The API: int read4(char *buf) reads at most 4 characters from a file. The return value is the actual number of characters read. For example, it returns 3 if there is only 3 characters left in the file.

By using the read4 API, implement the function int read(char *buf, int n) that reads exactly n characters from the file.

Note:
The read function may be called multiple times.

Example 1:

Input: read(buf, 1)
Output: 1
Explanation: The first time the function is called, the API reads 4 characters from the file, which are "a, b, c, d". Then the function only returns the first character to the buffer buf

Example 2:

Input: read(buf, 5)
Output: 5
Explanation: The first time the function is called, the API reads 4 characters from the file, which are "a, b, c, d". Then the function returns the first 4 characters to the buffer buf. 
The second time the function is called, the API reads one character from the file, which is "e". Then the function returns the first character to the buffer buf.

Constraints:

1 <= n <= 1000

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 size limit for `n`, the number of characters to read, and for the file being read from?
If `read4` returns less than 4 characters (e.g., at the end of the file), how should `read` handle that situation across multiple calls?
What should `read` return if `n` is zero or negative?
Is the buffer `buf` guaranteed to be large enough to hold `n` characters, or do I need to perform any size checks?
Is the file path provided to `read4` constant across multiple calls to `read`, or might it change?

Brute Force Solution

Approach

We have a source that gives us characters in chunks of four, and we need to read a specific number of characters from it. The brute force approach simulates reading from the source character by character until we have read the desired amount.

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

Imagine an empty box where we'll put the characters we read.
Ask the source for the next four characters.
Take one character from those four and put it in our box.
Check if our box now has the number of characters we need.
If it does, then we're done.
If not, take another character from those four and add it to the box.
Again, check if we have enough characters now.
Keep doing this until we've used all four characters from the source.
If we still don't have enough characters, ask the source for another four characters and repeat the process of taking them one by one and putting them in our box until we have read the desired number of characters.

Code Implementation

def read_n_characters(read4, buffer, number_of_characters_to_read):
    characters_read = 0
    temporary_buffer = [''] * 4
    temporary_buffer_index = 0
    temporary_buffer_size = 0

    while characters_read < number_of_characters_to_read:
        # Check if temporary_buffer is empty and needs refilling
        if temporary_buffer_index == temporary_buffer_size:
            temporary_buffer_size = read4(temporary_buffer)
            temporary_buffer_index = 0

        # If read4 returns 0, then it is the end of the file
        if temporary_buffer_size == 0:
            break

        # Read character from temporary buffer to the main buffer.
        while (characters_read < number_of_characters_to_read and
               temporary_buffer_index < temporary_buffer_size):

            # Add the next character to the main buffer
            buffer[characters_read] = temporary_buffer[temporary_buffer_index]
            characters_read += 1
            temporary_buffer_index += 1

    return characters_read

Try it yourself on LeetCode

Big(O) Analysis

Time Complexity

O(n) – The read function reads at most n characters. The inner loop iterates at most 4 times for each call to read4. The number of calls to read4 is bounded by n/4 (we will never read more than n characters), and each read4 call takes constant time. Therefore, the overall time complexity is directly proportional to the number of characters read (n), resulting in O(n).

Space Complexity

O(1) – The provided algorithm utilizes a fixed-size buffer of 4 characters from the read4 function. It also implicitly uses a few integer variables to track progress within this buffer and the overall desired character count. The space used by these variables and the fixed-size buffer is constant regardless of the value of N (the number of characters to read). Therefore, the space complexity is O(1).

Optimal Solution

Approach

This problem is about repeatedly reading characters from a source that provides them in chunks. We need to manage any leftover characters from previous reads so they aren't lost and can be used in subsequent requests.

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

First, remember that the 'read4' function gives you characters in blocks of four. Sometimes you'll get fewer than four, which means the source is running out of characters.
Keep track of any leftover characters you read in a previous 'read4' call but didn't need to return yet. These leftovers are like a little stash of characters waiting to be used.
When someone asks you to read characters, first use up any characters you have stored in your stash from previous calls. Only go to 'read4' if your stash is empty.
When you call 'read4', you might get fewer characters than you asked for. Also, 'read4' might give you more than the number of characters you need to return right now. If this happens, put the extra characters into your stash for later.
Make sure to stop reading when either 'read4' returns fewer than four characters (meaning the end of the source) or when you've read enough characters to fulfill the request.
The trick is to efficiently use your stash and only call 'read4' when absolutely necessary, while always keeping track of whether the source has more characters or not.

Code Implementation

class Solution:
    def __init__(self):
        self.previous_read_buffer = [''] * 4
        self.previous_read_pointer = 0
        self.previous_read_count = 0

    def read(self, buffer, number_of_chars):
        chars_read = 0
        
        while chars_read < number_of_chars:
            # First, use any characters left over from the previous read4 call
            if self.previous_read_pointer < self.previous_read_count:
                buffer[chars_read] = self.previous_read_buffer[self.previous_read_pointer]
                chars_read += 1
                self.previous_read_pointer += 1
            else:
                # Only call read4 if the previous buffer is empty.
                self.previous_read_pointer = 0
                self.previous_read_count = read4(self.previous_read_buffer)
                
                # Check for end of file condition
                if self.previous_read_count == 0:
                    break

                # If read4 returned characters, use them
                buffer[chars_read] = self.previous_read_buffer[self.previous_read_pointer]
                chars_read += 1
                self.previous_read_pointer += 1

        return chars_read

Try it yourself on LeetCode

Big(O) Analysis

Time Complexity

O(n) – The algorithm reads at most n characters from the input. While the internal read4 function might be called multiple times, the total number of characters read via read4 is bounded by n, the number of characters requested. The use of the buffer and pointers does not change the fundamental linear relationship between the input size n and the execution time. Therefore, the time complexity is O(n).

Space Complexity

O(1) – The solution maintains a buffer (stash) to store leftover characters from read4. The size of this buffer is fixed at 4 characters, as read4 returns a maximum of 4 characters at a time. Therefore, the space used for the buffer is constant and does not depend on N (the number of characters to read). No other data structures scale with the input size, resulting in constant auxiliary space.

Edge Cases

Case	How to Handle
n is zero	Return 0 immediately as no characters need to be read into the buffer.
buf is null or undefined	Throw an IllegalArgumentException or return 0 after logging an error, depending on the specific requirements, as there's no valid buffer to write to.
read4 returns 0 indicating end of file on the first call	Return 0, indicating no characters were read.
n is smaller than 4	Read up to 'n' characters from the internal buffer or call read4 once, copying only 'n' characters into buf and returning 'n'.
n is larger than the remaining characters in the file	Read as many characters as possible from the file, copying them into buf and return the actual number of characters read.
Multiple calls to read where the total n across calls exceeds the file size	Ensure the internal buffer and read4 calls are managed correctly to avoid reading past the end of the file in subsequent calls, returning only available characters.
Internal buffer from previous read calls contains characters, but n is smaller than the buffer size	First use the available characters from the internal buffer before calling read4, returning the number of characters actually read (which may be less than n).
Internal buffer is full and next read4 returns 0	Return 0, because no characters can be read from the internal buffer as well from the file.