LeetCode 157 - Read N Characters Given Read4

This problem asks us to implement a read function using a restricted API called read4. The file itself is hidden from us, and the only way to access its contents is by repeatedly calling read4.

leetcodeeasyarraysimulationinteractive

LeetCode Problem 157

Difficulty: 🟢 Easy
Topics: Array, Simulation, Interactive

Solution

Problem Understanding

This problem asks us to implement a read function using a restricted API called read4. The file itself is hidden from us, and the only way to access its contents is by repeatedly calling read4.

The read4 API reads up to 4 consecutive characters from the file into a temporary buffer called buf4. It returns the number of characters actually read. If the file still has at least 4 unread characters, the return value will be 4. If fewer than 4 characters remain, it returns the remaining count. If the file pointer has already reached the end of the file, it returns 0.

Our goal is to implement:

read(buf, n)

This function should read at most n characters from the file into the destination buffer buf, then return the number of characters actually read.

The important detail is that we are not allowed to access the file directly. We must repeatedly call read4 until either:

  1. We have read exactly n characters, or
  2. We reach the end of the file.

The constraints are relatively small:

  • file.length <= 500
  • n <= 1000

This means efficiency is not extremely critical from a performance perspective, but the problem is designed to test careful simulation and buffer handling.

Several edge cases are important:

  • The file may contain fewer characters than n
  • n may not be a multiple of 4
  • The last read4 call may return fewer than 4 characters
  • We must avoid copying extra characters into buf
  • We must stop immediately after reading n characters, even if read4 returned more usable data

A naive implementation can easily introduce off-by-one errors or accidentally copy too many characters.

Approaches

Brute Force Approach

The brute-force approach is to repeatedly call read4, append everything returned into an intermediate list or string, and then truncate the result to the first n characters.

Conceptually, this works because read4 eventually exhausts the file. Once all characters are collected, we can simply copy the first n characters into the destination buffer.

However, this approach is wasteful because it may read and store more data than necessary. For example, if we only need 1 more character but read4 gives us 4, we unnecessarily keep the extra data even though the problem guarantees that read is only called once.

It also introduces unnecessary auxiliary storage.

Optimal Approach

The key observation is that we can process the data incrementally.

Each call to read4 gives us up to 4 characters. We immediately copy only the required amount into the destination buffer. We maintain a counter tracking how many characters have already been written into buf.

The algorithm stops when:

  • We have copied exactly n characters, or
  • read4 returns fewer than 4 characters, meaning we reached the end of the file

This approach avoids unnecessary storage and processes characters exactly once.

Approach Time Complexity Space Complexity Notes
Brute Force O(file.length) O(file.length) Reads everything into temporary storage before truncating
Optimal O(n) O(1) Copies characters directly into destination buffer

Algorithm Walkthrough

  1. Create a temporary buffer buf4 of size 4.

This buffer is required because read4 writes its results into it. We cannot pass the destination buffer directly because read4 always expects a 4-character temporary array. 2. Maintain a variable total_read.

This variable tracks how many characters have already been copied into the destination buffer buf. 3. Enter a loop that continues while total_read < n.

We stop once we have filled the requested number of characters. 4. Call read4(buf4).

The function returns the number of characters actually read from the file. Store this value in count. 5. If count == 0, stop immediately.

This means the file pointer has reached the end of the file and there are no more characters available. 6. Copy characters from buf4 into buf.

We copy one character at a time while:

  • We still have unread characters in buf4
  • We have not yet copied n characters overall
  1. Increment total_read after each copied character.

This ensures we always know how many characters have been placed into the destination buffer. 8. Continue until either:

  • total_read == n
  • read4 returns 0
  1. Return total_read.

This is the actual number of characters successfully read.

Why it works

The algorithm maintains the invariant that buf[0:total_read] always contains exactly the characters read so far from the file, in correct order.

Each call to read4 advances the file pointer by the number of characters returned. Since we copy characters sequentially and stop exactly at n, we never skip characters or overwrite previously copied data.

Because the loop terminates either after reading n characters or after reaching end-of-file, the returned value is always correct.

Python Solution

"""
The read4 API is already defined for you.

    @param buf4, a list of characters
    @return an integer
    def read4(buf4):
"""

from typing import List

class Solution:
    def read(self, buf: List[str], n: int) -> int:
        """
        :type buf: Destination buffer (List[str])
        :type n: Number of characters to read (int)
        :rtype: The number of actual characters read (int)
        """
        
        buf4 = [''] * 4
        total_read = 0
        
        while total_read < n:
            count = read4(buf4)
            
            if count == 0:
                break
            
            for i in range(count):
                if total_read == n:
                    break
                
                buf[total_read] = buf4[i]
                total_read += 1
        
        return total_read

The implementation begins by allocating a temporary buffer buf4 with capacity 4. This is necessary because read4 always writes into a separate 4-character array.

The variable total_read tracks how many characters have already been copied into the destination buffer.

Inside the main loop, we repeatedly call read4(buf4). The returned value tells us how many valid characters were placed into buf4.

If read4 returns 0, we know we have reached the end of the file and must stop.

Otherwise, we iterate through the returned characters and copy them into the destination buffer one at a time. Before each copy, we check whether we have already reached n characters. This prevents us from copying excess characters when the final read4 call returns more data than we still need.

Finally, we return total_read, which represents the exact number of characters successfully read.

Go Solution

/**
 * The read4 API is already defined for you.
 *
 *     read4 := func(buf4 []byte) int
 */

var solution = func(read4 func([]byte) int) func([]byte, int) int {

    return func(buf []byte, n int) int {

        buf4 := make([]byte, 4)
        totalRead := 0

        for totalRead < n {

            count := read4(buf4)

            if count == 0 {
                break
            }

            for i := 0; i < count && totalRead < n; i++ {
                buf[totalRead] = buf4[i]
                totalRead++
            }
        }

        return totalRead
    }
}

The Go implementation follows the same logic as the Python version, but there are some language-specific differences.

Go uses []byte instead of character arrays. The temporary buffer buf4 is allocated using make([]byte, 4).

The inner loop combines both stopping conditions directly in the loop header:

for i := 0; i < count && totalRead < n; i++

This avoids the explicit nested if statement used in Python.

There are no integer overflow concerns because the constraints are extremely small. Slice handling is straightforward because the destination buffer is guaranteed to have enough capacity.

Worked Examples

Example 1

file = "abc"
n = 4

Initial state:

Variable Value
total_read 0
file pointer 0
buf []

First read4 call

read4(buf4) -> 3
buf4 = ['a', 'b', 'c']

Copy process:

Step Character total_read buf
1 a 1 ['a']
2 b 2 ['a', 'b']
3 c 3 ['a', 'b', 'c']

Second read4 call

read4(buf4) -> 0

End of file reached.

Final result:

Return 3
buf = ['a', 'b', 'c']

Example 2

file = "abcde"
n = 5

First read4 call

read4(buf4) -> 4
buf4 = ['a', 'b', 'c', 'd']

After copying:

total_read buf
4 ['a', 'b', 'c', 'd']

Second read4 call

read4(buf4) -> 1
buf4 = ['e']

After copying:

total_read buf
5 ['a', 'b', 'c', 'd', 'e']

Stop because total_read == n.

Final result:

Return 5

Example 3

file = "abcdABCD1234"
n = 12

First read4 call

read4(buf4) -> 4
buf4 = ['a', 'b', 'c', 'd']

Second read4 call

read4(buf4) -> 4
buf4 = ['A', 'B', 'C', 'D']

Third read4 call

read4(buf4) -> 4
buf4 = ['1', '2', '3', '4']

Final state:

total_read buf
12 ['a','b','c','d','A','B','C','D','1','2','3','4']

Return:

12

Complexity Analysis

Measure Complexity Explanation
Time O(n) Each character is copied at most once
Space O(1) Only a fixed-size temporary buffer of length 4 is used

The time complexity is linear in the number of characters requested because every copied character requires constant work.

The space complexity is constant because the only extra memory used is the temporary buf4 array of size 4, regardless of input size.

Test Cases

# Helper simulation framework for local testing

class File:
    def __init__(self, content):
        self.content = content
        self.ptr = 0

    def read4(self, buf4):
        count = 0

        while count < 4 and self.ptr < len(self.content):
            buf4[count] = self.content[self.ptr]
            count += 1
            self.ptr += 1

        return count

def run_test(file_content, n):
    file_obj = File(file_content)

    def read4(buf4):
        return file_obj.read4(buf4)

    class Solution:
        def read(self, buf, n):
            buf4 = [''] * 4
            total_read = 0

            while total_read < n:
                count = read4(buf4)

                if count == 0:
                    break

                for i in range(count):
                    if total_read == n:
                        break

                    buf[total_read] = buf4[i]
                    total_read += 1

            return total_read

    solution = Solution()
    buf = [''] * n

    result = solution.read(buf, n)

    return result, ''.join(buf[:result])

# Provided examples
assert run_test("abc", 4) == (3, "abc")  # File shorter than n
assert run_test("abcde", 5) == (5, "abcde")  # Exact file length
assert run_test("abcdABCD1234", 12) == (12, "abcdABCD1234")  # Multiple full read4 calls

# Boundary cases
assert run_test("a", 1) == (1, "a")  # Smallest valid input
assert run_test("abcd", 4) == (4, "abcd")  # Exactly one read4 call
assert run_test("abcde", 1) == (1, "a")  # Stop before consuming full read4 result

# Additional edge cases
assert run_test("abcdef", 3) == (3, "abc")  # Partial read
assert run_test("abcdef", 6) == (6, "abcdef")  # Exact complete read
assert run_test("abcdef", 10) == (6, "abcdef")  # n larger than file length
assert run_test("123456789", 8) == (8, "12345678")  # Multiple read4 calls with leftover
Test Why
"abc", 4 Validates end-of-file before reaching n
"abcde", 5 Validates exact file-length read
"abcdABCD1234", 12 Validates multiple complete read4 operations
"a", 1 Smallest valid input
"abcd", 4 Exactly one full read4 call
"abcde", 1 Ensures extra read4 characters are not copied
"abcdef", 3 Tests stopping early
"abcdef", 6 Tests complete exact read
"abcdef", 10 Tests n larger than file length
"123456789", 8 Tests partial consumption from final read4 block

Edge Cases

One important edge case occurs when the file contains fewer characters than n. For example, the file may contain only "abc" while n = 10. A buggy implementation might continue looping forever or attempt to copy invalid characters after reaching the end of the file. Our implementation correctly handles this because read4 eventually returns 0, which immediately terminates the loop.

Another important case occurs when n is smaller than the number of characters returned by the current read4 call. For example, if we need only one more character but read4 returns four characters, we must copy only one character into the destination buffer. The inner condition:

if total_read == n:
    break

prevents accidental over-copying.

A third tricky case occurs when the file length is exactly a multiple of 4. In this situation, every read4 call before the end returns exactly 4. Some incorrect implementations mistakenly assume that a return value smaller than 4 is required to terminate. Our solution instead terminates when either total_read == n or read4 returns 0, so it handles exact multiples correctly.

A fourth subtle case involves the final partial block. Suppose the file length is 5. The first read4 returns 4, while the second returns 1. The implementation must copy exactly one valid character from the second buffer and ignore the remaining unused positions. Because we iterate only up to count, we never read garbage values from buf4.