158. Read N Characters Given read4 II - Call Multiple Times

Problem Description

The problem provides a hypothetical file system and a function read4 which can read 4 characters from a file at a time and store them into an array buf4. The main task is to implement a new method read which can read n characters from the file and store them in an array buf. This read function can be called multiple times, which adds complexity because it needs to handle subsequent reads correctly, taking into consideration what was read in previous calls.

Constraints are that you can't access the file directly, and your class should be stateful to remember its state between multiple calls to read. Moreover, the buf array provided as an argument to the read method has sufficient space to store n characters.

The description also underscores that the static/class variables maintained in your Solution class must be reset as they are persistent across multiple test cases. The function read returns the number of characters actually read and stored in buf.

Intuition

To solve this problem, we build around the functionality of the read4 method, which is provided to us. Since we can only read the file by using read4, which reads up to 4 characters at a time, we have to call read4 multiple times until we have read n characters or until we reach the end of the file.

However, since the read method can be called multiple times and should continue reading from where it left off, we need some stateful variables to store information between calls. This is necessary to keep track of any leftover characters from the last call to read4 that were not used in the previous calls to read.

Here's where the variables self.buf4, self.i, and self.size come into play. The variable self.buf4 is an array of 4 characters to store data from the read4 calls. self.i keeps track of the current position in buf4, and self.size stores the number of characters read in the last call to read4.

The intuition behind the solution is to use a loop to fill buf with characters until we've read n characters or we've read all available characters in the file. Inside the loop, we check if we need to call read4 to fill our buffer buf4 (when self.i == self.size). If read4 returns 0, it means we've reached the end of the file, and we can break out of the loop.

After filling buf4, we transfer characters from buf4 to buf while incrementing both self.i and our counter j up to n, ensuring we don't read beyond the requested number of characters or beyond what's available in the buffer.

The method returns the number of characters written to buf, indicated by j.

Solution Approach

The solution involves calling the API read4 and handling the buffer management manually. The two important aspects here are filling the buffer buf appropriately, and maintaining the state between multiple calls to read. The implementation can be described in the following steps:

1. Initialization: Create instance variables within the Solution class to keep track of the buffer self.buf4 from read4, the current index self.i in buf4, and the number of characters self.size read in the last call to read4.

2. Iteration: Use a while-loop to continue reading characters until we have read n characters (the target amount) or until there are no more characters to read from the file.

3. Filling buf4: Inside the loop, check if self.i equals self.size, which means we have processed all characters in buf4 from the previous read4 call, or it's the first iteration. If true, call read4(self.buf4) to read the next chunk of characters from the file into buf4. The number of characters actually read is stored in self.size, and self.i is reset to 0.

4. Transferring to buf: After ensuring buf4 is filled, enter another loop which runs as long as there are characters left to read (j < n) and there are characters available in buf4 (self.i < self.size). During this loop, copy characters one by one from self.buf4 to the target buffer buf using buf[j] = self.buf4[self.i]. Increment self.i and j with each character copied.

5. End of File Handling: If read4 returns 0, it indicates the end of the file. At this point, break out of the loop since no more characters can be read.

6. Return Value: After filling buf or when the file end is reached, exit the loop and return j, the number of characters actually read and written into buf.

Using these steps, the algorithm ensures that the reading process can be paused and resumed across multiple calls to read. It correctly handles buffering of characters and maintains state between calls. Moreover, this approach does not depend on how many times read is called or the number of characters requested in each call; it always returns the correct number of characters read from the file.

Example Walkthrough

Let's consider a file containing the text "HelloWorld" and assume we want to read its contents using our method read. For demonstration, let's say we're making two calls to read: the first call to read 8 characters, and the second call to read 5 characters. Here's how our solution handles it:

1. First read call to read 8 characters:

   • We initialize self.buf4 to [], and set both self.i and self.size to 0.
   • Enter the while loop because the target n is 8, and our read count j is currently 0.
   • Since self.i == self.size, we call read4 and it fills self.buf4 with the first 4 characters, so self.buf4 = ['H', 'e', 'l', 'l'] and self.size = 4.
   • We now copy from self.buf4 to buf. Since our target n is 8, we continue to copy until self.buf4 is exhausted or we reach the target n.
   • By the end of this iteration, buf = ['H', 'e', 'l', 'l'], self.i = 4, and j = 4. We still need to read 4 more characters to meet our target.
   • We go through the loop again, call read4 which now fills self.buf4 with the next 4 characters: ['o', 'W', 'o', 'r'], and self.size = 4.
   • We transfer these 4 characters to buf, resulting in buf = ['H', 'e', 'l', 'l', 'o', 'W', 'o', 'r'] and updating j to 8, which meets our target. We exit the loop and return 8.

2. Second read call to read 5 characters:

   • We start again with self.buf4 still having ['o', 'W', 'o', 'r'] but this time our initial self.i is 4 and self.size is 4 since we didn't reset them after the last call (as they're being used to maintain state).
   • As self.i is equal to self.size, we call read4 again. It reads the final 2 characters of the file: ['l', 'd'], and sets self.size = 2.
   • We copy ['l', 'd'] into buf to get buf = ['l', 'd']. Now self.i is 2, self.size is 2, and our read count j is 2.
   • Since we have reached the end of the file and read4 cannot read more characters, the loop ends. We can't read 5 characters as requested because only 2 are left in the file. Thus we return 2.

So, over the course of two read calls asking for 8 and then 5 characters, we returned 8 characters the first time and 2 the second time, accurately reflecting the contents of "HelloWorld" and the stateful nature of consecutive reads.

Solution Implementation

Time and Space Complexity

The time complexity of this read function is O(n). Each call to read4 reads at most 4 characters until n characters are read or there is no more content to read from the file. In the worst case, the function will call read4 ceil(n/4) times to read n characters.

The space complexity of the solution is O(1). The solution uses a constant amount of extra space, buf4 of size 4, to store the read characters temporarily and a few integer variables (self.i, self.size, and j) to keep track of positions, which does not depend on the size of the input n.

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