A clear explanation of implementing read using the given read4 API and copying only the needed characters.
Problem Restatement
We are given a file, but we cannot access the file directly.
The only way to read from the file is through a provided API:
read4(buf4)This function reads up to 4 characters from the file and writes them into buf4.
It returns the actual number of characters read.
We need to implement:
read(buf, n)This function should read up to n characters from the file, write them into buf, and return the number of characters actually read.
The file may contain fewer than n characters.
LeetCode states that read4 has its own file pointer, buf4 is a destination buffer, and read will be called only once for each test case. The destination buffer buf has enough space for n characters.
Input and Output
| Item | Meaning |
|---|---|
| Input | Destination buffer buf and integer n |
| Output | Number of characters actually read |
| Given API | read4(buf4) |
read4 return value | Number of characters read, from 0 to 4 |
| File access | Only through read4 |
| Important detail | read is called once per test case |
Example method shape:
def read(self, buf: List[str], n: int) -> int:
...Examples
Example 1:
file = "abc"
n = 4We ask for 4 characters, but the file contains only 3.
So buf should contain:
"abc"The function returns:
3Example 2:
file = "abcde"
n = 5We need all 5 characters.
One call to read4 reads:
"abcd"The second call reads:
"e"The function returns:
5Example 3:
file = "abcde"
n = 3One call to read4 may read 4 characters into the temporary buffer:
"abcd"But we only copy the first 3 characters into buf.
The function returns:
3First Thought: Keep Calling read4
Since read4 gives us at most 4 characters at a time, we can repeatedly call it until one of two things happens:
- We have copied
ncharacters intobuf. read4returns fewer than 4 characters, meaning the file has ended.
The important point is that read4 writes into a temporary buffer, not directly into the final answer buffer.
So we need:
buf4 = [""] * 4Then after each call:
count = read4(buf4)we copy only the characters we still need.
Key Insight
There are two limits at the same time.
The first limit is how many characters read4 actually returned:
countThe second limit is how many characters the user still wants:
n - totalSo after one read4 call, the number of characters we should copy is:
min(count, n - total)This avoids copying too many characters when n is not a multiple of 4.
Algorithm
Create a temporary buffer of size 4:
buf4 = [""] * 4Set:
total = 0While total < n:
- Call
read4(buf4). - Let
countbe the number of characters read. - Copy at most
min(count, n - total)characters frombuf4tobuf. - Increase
total. - If
count < 4, stop because the file has ended.
Return total.
Walkthrough
Use:
file = "abcde"
n = 5Start:
total = 0First call:
count = read4(buf4)buf4 contains:
"abcd"and:
count = 4We still need 5 characters, so we copy 4.
Now:
buf = ["a", "b", "c", "d"]
total = 4Second call:
count = read4(buf4)buf4 contains:
"e"and:
count = 1We still need:
5 - 4 = 1So we copy 1 character.
Now:
buf = ["a", "b", "c", "d", "e"]
total = 5Return:
5Correctness
The algorithm copies characters into buf in the same order that read4 returns them.
Each call to read4 advances the file pointer, so repeated calls read consecutive parts of the file.
After each call, the algorithm copies only:
min(count, n - total)characters.
This means it never writes more than n characters into buf.
If read4 returns fewer than 4 characters, the file has no more characters after that call, so stopping is correct.
If the algorithm reaches total == n, it has read the requested number of characters, so stopping is correct.
Therefore, the returned value is exactly the number of characters copied into buf, which is the number of characters actually read.
Complexity
Let k be the number of characters actually copied, where k <= n.
| Metric | Value | Why |
|---|---|---|
| Time | O(k) | Each copied character is written once |
| Space | O(1) | The temporary buffer always has size 4 |
Implementation
# The read4 API is already defined for you.
# def read4(buf4: List[str]) -> int:
class Solution:
def read(self, buf: List[str], n: int) -> int:
buf4 = [""] * 4
total = 0
while total < n:
count = read4(buf4)
chars_to_copy = min(count, n - total)
for i in range(chars_to_copy):
buf[total] = buf4[i]
total += 1
if count < 4:
break
return totalCode Explanation
We create a temporary buffer:
buf4 = [""] * 4This is where read4 writes its result.
We track how many characters have been copied into buf:
total = 0The loop runs until we have copied n characters:
while total < n:Each iteration reads up to 4 characters:
count = read4(buf4)Then we compute how many of those characters should be copied:
chars_to_copy = min(count, n - total)This handles cases like n = 3, where read4 might return 4 but we only need 3.
The copy loop writes characters into the destination buffer:
for i in range(chars_to_copy):
buf[total] = buf4[i]
total += 1If read4 returns fewer than 4 characters:
if count < 4:
breakthen we reached the end of the file.
Finally:
return totalreturns the number of characters written into buf.
Testing
LeetCode provides read4, but we can simulate it locally.
from typing import List
class Reader4:
def __init__(self, file: str):
self.file = file
self.ptr = 0
def read4(self, buf4: List[str]) -> int:
count = 0
while count < 4 and self.ptr < len(self.file):
buf4[count] = self.file[self.ptr]
self.ptr += 1
count += 1
return count
class Solution(Reader4):
def read(self, buf: List[str], n: int) -> int:
buf4 = [""] * 4
total = 0
while total < n:
count = self.read4(buf4)
chars_to_copy = min(count, n - total)
for i in range(chars_to_copy):
buf[total] = buf4[i]
total += 1
if count < 4:
break
return total
def run_tests():
sol = Solution("abc")
buf = [""] * 4
assert sol.read(buf, 4) == 3
assert "".join(buf[:3]) == "abc"
sol = Solution("abcde")
buf = [""] * 5
assert sol.read(buf, 5) == 5
assert "".join(buf[:5]) == "abcde"
sol = Solution("abcde")
buf = [""] * 3
assert sol.read(buf, 3) == 3
assert "".join(buf[:3]) == "abc"
sol = Solution("")
buf = [""] * 2
assert sol.read(buf, 2) == 0
sol = Solution("abcdefg")
buf = [""] * 4
assert sol.read(buf, 4) == 4
assert "".join(buf[:4]) == "abcd"
print("all tests passed")
run_tests()| Test | Why |
|---|---|
file = "abc", n = 4 | File shorter than requested |
file = "abcde", n = 5 | Needs multiple read4 calls |
file = "abcde", n = 3 | Must copy less than read4 returns |
file = "", n = 2 | Empty file |
file = "abcdefg", n = 4 | Exact one full read4 call |