Read N Characters Given Read4 - Practice Coding Problems

Read N Characters Given Read4 - Problem

Imagine you're working with a file reading system that has a quirky limitation: you can only read the file in chunks of exactly 4 characters at a time using a built-in method called read4. Your task is to implement a more flexible read method that can read any number of characters n from the file.

The Challenge: You need to bridge the gap between the fixed 4-character chunks and the variable-length reads that users want.

How read4 works:

It reads up to 4 consecutive characters from the file
Stores them in a buffer array buf4
Returns the actual number of characters read (could be less than 4 at end of file)
Maintains its own file pointer that advances automatically

Your Goal: Implement read(buf, n) that reads exactly n characters (or fewer if end of file is reached) and stores them in the destination buffer buf.

Example: If the file contains "abcdefgh" and you want to read 6 characters, your method should return 6 and fill buf with "abcdef".

Input & Output

Basic Reading Example

$ Input: File = "abc", n = 4

› Output: 3

💡 Note: We want to read 4 characters, but the file only contains 3 characters "abc". The function returns 3, and buf contains "abc".

Exact Multiple of 4

$ Input: File = "abcdefgh", n = 8

› Output: 8

💡 Note: The file has exactly 8 characters and we want to read 8. This requires exactly 2 calls to read4: first gets "abcd", second gets "efgh".

Partial Read

$ Input: File = "abcdefghijk", n = 5

› Output: 5

💡 Note: We only want 5 characters from an 11-character file. First read4 gets "abcd" (4 chars), second read4 gets "efgh" but we only copy "e" to reach our target of 5.

Constraints

0 ≤ n ≤ 10⁸
The file contains at most 10⁸ characters
read4 is guaranteed to work correctly and read up to 4 characters
You cannot manipulate the file directly - only through read4
The read function will only be called once per test case

Visualization

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Understanding the Visualization

Setup

Initialize counters and prepare a 4-character temporary buffer

Read Chunk

Call read4 to get up to 4 characters from the file

Smart Copy

Copy only the characters we need: min(available, remaining needed)

Continue or Stop

Repeat until we have n characters or reach end of file

Key Takeaway

🎯 Key Insight: The algorithm efficiently handles the mismatch between fixed-size file reads (4 chars) and variable user requests (n chars) by using smart copying logic: always copy min(available_chars, remaining_needed_chars).

Asked in

f Facebook 45 G Google 38 a Amazon 32 ⊞ Microsoft 28

The optimal approach uses a direct buffer management strategy with O(n) time complexity. The key insight is efficiently bridging the gap between fixed 4-character chunks from read4 and variable-length user requests by using a temporary buffer and careful copying logic.

Common Approaches

Approach	Time	Space	Notes
✓ Direct Buffer Management	O(n)	O(1)	Use read4 repeatedly and copy characters one by one until we have n characters

Direct Buffer Management — Algorithm Steps

Initialize variables to track total characters read and current position in destination buffer
Create a temporary 4-character buffer for read4
While we still need more characters and file has content:
Call read4 to fill temporary buffer with up to 4 characters
Copy the minimum of (remaining needed characters, characters actually read) to destination
Update counters and continue until done
Return total characters successfully read

Visualization

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Step-by-Step Walkthrough

Initialize

Set up variables and prepare temporary buffer

Call read4

Read up to 4 characters into temporary buffer

Copy characters

Transfer needed characters to destination buffer

Repeat

Continue until n characters read or end of file

Code -

solution.c — C

/**
 * The read4 API is defined in the parent class Reader4.
 *     int read4(char *buf4);
 */

int read(char* buf, int n) {
    int totalRead = 0;
    
    while (totalRead < n) {
        char buf4[4];
        int count = read4(buf4);
        
        if (count == 0) {
            break;
        }
        
        int toCopy = (count < n - totalRead) ? count : n - totalRead;
        for (int i = 0; i < toCopy; i++) {
            buf[totalRead++] = buf4[i];
        }
    }
    
    return totalRead;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

We need to read n characters, and each character is processed once. The number of read4 calls is O(n/4) which is still O(n)

✓ Linear Growth

Space Complexity

O(1)

Only uses a fixed-size temporary buffer of 4 characters regardless of input size

✓ Linear Space

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Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Direct Buffer Management — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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