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:
-
Initialization: Create instance variables within the
Solution
class to keep track of the bufferself.buf4
fromread4
, the current indexself.i
inbuf4
, and the number of charactersself.size
read in the last call toread4
. -
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. -
Filling buf4: Inside the loop, check if
self.i
equalsself.size
, which means we have processed all characters inbuf4
from the previousread4
call, or it's the first iteration. If true, callread4(self.buf4)
to read the next chunk of characters from the file intobuf4
. The number of characters actually read is stored inself.size
, andself.i
is reset to 0. -
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 inbuf4
(self.i < self.size
). During this loop, copy characters one by one fromself.buf4
to the target bufferbuf
usingbuf[j] = self.buf4[self.i]
. Incrementself.i
andj
with each character copied. -
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. -
Return Value: After filling
buf
or when the file end is reached, exit the loop and returnj
, the number of characters actually read and written intobuf
.
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.
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Start EvaluatorExample 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:
-
First
read
call to read 8 characters:- We initialize
self.buf4
to[]
, and set bothself.i
andself.size
to0
. - Enter the while loop because the target
n
is 8, and our read countj
is currently0
. - Since
self.i == self.size
, we callread4
and it fillsself.buf4
with the first 4 characters, soself.buf4 = ['H', 'e', 'l', 'l']
andself.size = 4
. - We now copy from
self.buf4
tobuf
. Since our targetn
is 8, we continue to copy untilself.buf4
is exhausted or we reach the targetn
. - By the end of this iteration,
buf = ['H', 'e', 'l', 'l']
,self.i = 4
, andj = 4
. We still need to read 4 more characters to meet our target. - We go through the loop again, call
read4
which now fillsself.buf4
with the next 4 characters:['o', 'W', 'o', 'r']
, andself.size = 4
. - We transfer these 4 characters to
buf
, resulting inbuf = ['H', 'e', 'l', 'l', 'o', 'W', 'o', 'r']
and updatingj
to8
, which meets our target. We exit the loop and return8
.
- We initialize
-
Second
read
call to read 5 characters:- We start again with
self.buf4
still having['o', 'W', 'o', 'r']
but this time our initialself.i
is4
andself.size
is4
since we didn't reset them after the last call (as they're being used to maintain state). - As
self.i
is equal toself.size
, we callread4
again. It reads the final 2 characters of the file:['l', 'd']
, and setsself.size = 2
. - We copy
['l', 'd']
intobuf
to getbuf = ['l', 'd']
. Nowself.i
is2
,self.size
is2
, and our read countj
is2
. - 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 return2
.
- We start again with
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
1class Solution:
2 def __init__(self):
3 self.buffer_4 = [''] * 4 # Buffer to store read characters from read4
4 self.buffer_index = 0 # The next read position in buffer_4
5 self.buffer_size = 0 # The number of characters read from read4
6
7 def read(self, buf: List[str], n: int) -> int:
8 """
9 Reads n characters from the file using read4.
10
11 :param buf: Destination buffer to store characters
12 :param n: Number of characters to read
13 :return: The number of characters actually read
14 """
15
16 # The total number of characters read so far
17 total_read = 0
18 # Loop until we read n characters or reach the end of the file
19 while total_read < n:
20 if self.buffer_index == self.buffer_size: # All characters in buffer are read
21 self.buffer_size = read4(self.buffer_4) # Read next 4 characters
22 self.buffer_index = 0 # Reset buffer index
23 if self.buffer_size == 0: # End of file reached
24 break
25
26 # Transfer characters from buffer_4 to buf while we haven't read n characters
27 while total_read < n and self.buffer_index < self.buffer_size:
28 buf[total_read] = self.buffer_4[self.buffer_index]
29 self.buffer_index += 1
30 total_read += 1
31
32 # Return the total number of characters read
33 return total_read
34
1// Extends the functionality of Reader4 class by implementing a custom reader method.
2public class Solution extends Reader4 {
3 private char[] internalBuffer = new char[4]; // Buffer to hold read characters from read4.
4 private int bufferIndex = 0; // Pointer to track the current position within internalBuffer.
5 private int contentSize = 0; // Amount of valid characters in internalBuffer after a call to read4.
6
7 /**
8 * Reads n characters from the file into buf.
9 *
10 * @param buf Destination buffer to store the characters read from the file.
11 * @param n Number of characters to read from the file.
12 * @return The number of actual characters read, which could be less than n if EOF is reached.
13 */
14 public int read(char[] buf, int n) {
15 int readCharsCount = 0; // Counter to keep track of the number of characters read into buf.
16
17 // Continue reading until the specified number of characters (n) is read or until the end of file is reached.
18 while (readCharsCount < n) {
19 // If internalBuffer has been fully read, reload it with new data from read4.
20 if (bufferIndex == contentSize) {
21 contentSize = read4(internalBuffer); // Read 4 characters and save the number of characters read.
22 bufferIndex = 0; // Reset buffer index.
23 // If no characters are read, end of file has been reached, so break out of the loop.
24 if (contentSize == 0) {
25 break;
26 }
27 }
28 // Transfer characters from the internal buffer to buf until we either fill buf or exhaust internalBuffer.
29 while (readCharsCount < n && bufferIndex < contentSize) {
30 buf[readCharsCount++] = internalBuffer[bufferIndex++];
31 }
32 }
33 // Return the total number of characters that were successfully read into buf.
34 return readCharsCount;
35 }
36}
37
1/**
2 * The read4 API is defined in the parent class Reader4.
3 * int read4(char *buf4);
4 */
5
6class Solution {
7public:
8 /**
9 * Reads n characters from the file and writes into buffer.
10 * @param buffer Destination buffer
11 * @param n Number of characters to read
12 * @return The number of actual characters read
13 */
14 int read(char* buffer, int n) {
15 int totalRead = 0; // Total number of characters read
16
17 // Continue reading until we have read n characters or there is no more to read.
18 while (totalRead < n) {
19 // Refill the tempBuffer if it's empty
20 if (bufferIndex == bufferSize) {
21 bufferSize = read4(tempBuffer);
22 bufferIndex = 0; // Reset buffer index
23 // If no characters were read, we've reached the end of the file
24 if (bufferSize == 0) break;
25 }
26
27 // Read from tempBuffer into buffer until we have read n characters
28 // or the tempBuffer is exhausted.
29 while (totalRead < n && bufferIndex < bufferSize) {
30 buffer[totalRead++] = tempBuffer[bufferIndex++];
31 }
32 }
33
34 return totalRead; // Return the total number of characters read
35 }
36
37private:
38 char tempBuffer[4]; // Temporary buffer to store read4 results
39 int bufferIndex = 0; // Index for the next read character in tempBuffer
40 int bufferSize = 0; // Represents how many characters read4 last read into tempBuffer
41};
42
1// Define global variables to keep track of the temporary buffer and indices.
2let tempBuffer: string[] = new Array(4); // Temporary buffer to store read4 results
3let bufferIndex: number = 0; // Index for the next read character in tempBuffer
4let bufferSize: number = 0; // Represents how many characters read4 last read into tempBuffer
5
6// Mocked read4 API function to match the context. This function should be replaced with the actual implementation.
7function read4(buf4: string[]): number {
8 // Implementation of read4 API should be provided.
9 return 0;
10}
11
12/**
13 * Reads 'n' characters from the file and writes into the buffer.
14 * @param buffer Destination buffer which is a string array
15 * @param n Number of characters to read
16 * @return The number of actual characters read
17 */
18function read(buffer: string[], n: number): number {
19 let totalRead: number = 0; // Total number of characters read
20
21 // Continue reading until we have read 'n' characters or there is no more content to read.
22 while (totalRead < n) {
23 // Refill the tempBuffer if it's empty.
24 if (bufferIndex === bufferSize) {
25 bufferSize = read4(tempBuffer);
26 bufferIndex = 0; // Reset buffer index
27
28 // If no characters were read, we've reached the end of the file.
29 if (bufferSize === 0) break;
30 }
31
32 // Read from tempBuffer into the buffer until we have read 'n' characters or tempBuffer is exhausted.
33 while (totalRead < n && bufferIndex < bufferSize) {
34 buffer[totalRead++] = tempBuffer[bufferIndex++];
35 }
36 }
37
38 // Return the total number of characters read.
39 return totalRead;
40}
41
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.
What are the two properties the problem needs to have for dynamic programming to be applicable? (Select 2)
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