Read N Characters Given read4 II - Call Multiple Times

Hard

Read N Characters Given read4 II - Call Multiple Times

Hard

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 maximum number of characters that read4 can read at once?
If read4 returns less than 4 characters, does it mean we have reached the end of the file?
Can I assume the input buffer 'buf' is large enough to hold 'n' characters?
What should I return if the input 'n' is zero or negative?
Does read4 maintain its own internal buffer, or does each call to read4 start from the current file position?

Brute Force Solution

Approach

The goal is to read a specific number of characters, but the reading function only reads a limited amount at a time. The brute force strategy involves repeatedly calling the reading function and handling any leftover characters from previous calls until we have enough characters.

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

First, check if we still have any previously read characters left over from the last time we called the reading function.
If we do, use those characters first to fulfill the read request. Remove these characters from our saved leftover characters.
Keep track of how many characters we have already read to ensure we don't read more than required.
Call the provided reading function to get more characters if needed.
If the reading function returns less than the maximum, we've reached the end of the data.
Save any unneeded characters from the last read for the next time the function is called.
Repeat calling the reading function and using leftover characters until the requested number of characters have been read or until the end of the data.

Code Implementation

class Solution:
    def __init__(self):
        self.remaining_characters = ""

    def read(self, buffer, number_of_characters):
        characters_read = 0

        # Use leftover characters if available
        while self.remaining_characters and characters_read < number_of_characters:
            buffer[characters_read] = self.remaining_characters[0]
            self.remaining_characters = self.remaining_characters[1:]
            characters_read += 1

        # Keep reading until enough chars are read
        while characters_read < number_of_characters:
            read_buffer = [''] * 4
            chars_from_read4 = read4(read_buffer)

            # Reached end of file
            if chars_from_read4 == 0:
                break

            read_string = ''.join(read_buffer[:chars_from_read4])

            # Use chars to fill buffer
            characters_to_copy = min(number_of_characters - characters_read, chars_from_read4)
            for i in range(characters_to_copy):
                buffer[characters_read] = read_string[i]
                characters_read += 1

            # Store unused characters for next read
            if characters_to_copy < chars_from_read4:
                self.remaining_characters = read_string[characters_to_copy:]
                
        return characters_read

Big(O) Analysis

Time Complexity

O(n) – The algorithm reads at most n characters. While there's a loop involved in reading characters using read4, the total number of characters read will never exceed n. The read4 function is called repeatedly, but the total number of iterations of all loops in the function is proportional to n, since the function aims to read n characters. Therefore, the time complexity is O(n).

Space Complexity

O(4) – The algorithm stores leftover characters from the read4 function calls. This is implemented using a fixed-size buffer of 4 characters. It also stores an integer to track the number of valid characters in the buffer, and another integer to track the read offset within the buffer. Therefore, the extra space used is constant and independent of the number of characters to read (N).

Optimal Solution

Approach

This problem is about reading data in chunks, but the catch is that we might not always use up the whole chunk. We need to remember the leftover data from previous calls to make sure we don't lose any information when the function is called again.

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

First, we check if there's any leftover data from the last time we read. If so, we use that data first.
If we need more data than what's leftover, we call the provided 'read4' function to get the next chunk of data.
We then copy as much as we need from this new chunk into the output buffer.
If 'read4' returns less than 4 characters, it means we've reached the end of the file, so we need to stop reading further.
If we didn't use all the data from the 'read4' call, we save the remaining data for the next function call.
We keep track of how much data we've read and return that number.

Code Implementation

class Solution:

    def __init__(self):
        self.remaining_characters = []

    def read(self, buffer, number_of_characters):
        characters_read = 0

        # Use up remaining characters from the last read, if any.
        while self.remaining_characters and characters_read < number_of_characters:
            buffer[characters_read] = self.remaining_characters.pop(0)
            characters_read += 1

        # Read more characters until enough is read or end of file is reached.
        while characters_read < number_of_characters:
            read_buffer = [''] * 4
            characters_from_read4 = read4(read_buffer)

            # If read4 returns less than 4, it's the end of the file.
            if characters_from_read4 == 0:
                break

            # Copy characters into the output buffer and save any leftover.
            for i in range(characters_from_read4):
                if characters_read < number_of_characters:
                    buffer[characters_read] = read_buffer[i]
                    characters_read += 1
                else:
                    # Store the remaining characters for the next call
                    self.remaining_characters.append(read_buffer[i])

        return characters_read

Big(O) Analysis

Time Complexity

O(n) – The read function reads at most n characters. Inside the while loop, the operations like copying characters from the buffer or calling read4 take constant time for each character read. The while loop iterates until n characters are read or read4 returns less than 4, indicating the end of the file. Thus, the time complexity is linearly proportional to the number of characters read, resulting in O(n).

Space Complexity

O(1) – The algorithm utilizes a fixed-size buffer (typically an array of size 4) to store leftover characters read from the read4 function. It also uses a couple of integer variables to keep track of the read index and buffer size of the stored leftover characters. These data structures do not scale with the input size N, the number of characters to read. Therefore, the auxiliary space used remains constant, regardless of how large N is, and the space complexity is O(1).

Edge Cases

Case	How to Handle
n is 0	Return 0 immediately, as no characters need to be read.
read4 returns 0 on the first call	Return 0 immediately, as there are no characters to read from the source.
n is larger than total characters available in the file	Return the number of characters actually read, which may be less than n.
read4 returns less than 4 characters, but n is not yet reached	Return the number of characters read so far; do not keep trying to read more if read4 returns less than 4.
Internal buffer from previous calls still has data	Prioritize reading from the internal buffer before calling read4 again.
read4 returns 4 consistently, but the total number of characters is not a multiple of 4	Ensure the algorithm correctly handles the last chunk of characters when fewer than 4 are read.
Repeated calls with small n values, depleting the buffer slowly	Algorithm should efficiently manage the internal buffer and not unnecessarily call read4.
Integer overflow when tracking total characters read.	If applicable, prevent integer overflow by using a sufficiently large data type or adding check constraints if the number of chars to read is bounded.