This problem involves implementing a read
function that reads n
characters from a file using a given read4
function, which reads only four characters at a time. The challenge is that read
can be called multiple times, and we need to efficiently manage the buffer between calls.
The solution uses a buffer (buf4
) to store the characters read by read4
. The read
function keeps track of the current index (i
) within buf4
and the number of characters read in the current read4
call (size
).
Algorithm:
Initialization: The solution initializes buf4
, i
(current index in buf4
), and size
(number of characters read by the last read4
call).
Main Loop: The while j < n
loop continues until n
characters have been read or the end of the file is reached.
Check Buffer: Inside the loop, if i == size
checks if the internal buffer buf4
is empty.
Read from File: If buf4
is empty, read4(buf4)
is called to read characters from the file. The return value (size
) indicates the number of characters read. If size
is 0, the end of the file is reached, and the loop breaks.
Copy to Output Buffer: The while j < n and i < size
loop copies characters from buf4
to the output buffer buf
until either n
characters have been copied or buf4
is exhausted.
Return: The function returns j
, which represents the total number of characters read and copied to buf
.
Time Complexity:
The time complexity is O(n), where n is the number of characters to read. The algorithm iterates through the input at most once. The read4
function is called at most n/4 times in the worst case.
Space Complexity:
The space complexity is O(1). The solution uses a fixed-size buffer (buf4
) of size 4, irrespective of the input size n
.
Code Explanation (Python):
class Solution:
def __init__(self):
self.buf4 = [None] * 4 # Internal buffer for read4
self.i = 0 # Index into buf4
self.size = 0 # Number of chars in buf4
def read(self, buf, n):
j = 0 # Index into output buffer buf
while j < n: # Loop until n chars read or EOF
if self.i == self.size: # If internal buffer is empty
self.size = read4(self.buf4) # Read from file
self.i = 0 # Reset index
if self.size == 0: # EOF
break
while j < n and self.i < self.size: # Copy from internal to output buffer
buf[j] = self.buf4[self.i]
self.i += 1
j += 1
return j
The Java, C++, and Go code implementations follow the same logic and algorithm, differing only in syntax and data structure specifics. They all achieve the same O(n) time and O(1) space complexity.