Read N Characters Given Read4 - Practice Coding Problems

Read N Characters Given Read4 - Problem

Given a file and assume that you can only read the file using a given method read4, implement a method read to read n characters.

Method read4:

The API read4 reads four consecutive characters from file, then writes those characters into the buffer array buf4. The return value is the number of actual characters read.

Note that read4() has its own file pointer, much like FILE *fp in C.

Definition of read4:

Parameter: char[] buf4 Returns: int

buf4[] is a destination, not a source. The results from read4 will be copied to buf4[].

Example of how read4 works:

File file("abcde"); // File is "abcde", initially file pointer (fp) points to 'a' char[] buf4 = new char[4]; // Create buffer with enough space read4(buf4); // read4 returns 4. Now buf4 = "abcd", fp points to 'e' read4(buf4); // read4 returns 1. Now buf4 = "e", fp points to end of file read4(buf4); // read4 returns 0. Now buf4 = "", fp points to end of file

Method read:

By using the read4 method, implement the method read that reads n characters from file and store it in the buffer array buf. Consider that you cannot manipulate the file directly.

Definition of read:

Parameters: char[] buf, int n Returns: int

buf[] is a destination, not a source. You will need to write the results to buf[].

Note: Consider that you cannot manipulate the file directly. The file is only accessible for read4 but not for read. The read function will only be called once for each test case.

Input & Output

Example 1 — Basic Case

$ Input: file = "abcdefgh", n = 4

› Output: 4

💡 Note: First call to read4 returns "abcd" (4 characters), exactly what we need, so return 4

Example 2 — Need Multiple Calls

$ Input: file = "abcdefgh", n = 6

› Output: 6

💡 Note: First read4 gets "abcd" (4 chars), second read4 gets "ef" (2 chars), total 6 characters read

Example 3 — File Shorter Than n

$ Input: file = "abc", n = 5

› Output: 3

💡 Note: File only has 3 characters, so read4 returns 3, we can only read 3 characters total

Constraints

1 ≤ file.length ≤ 500
1 ≤ n ≤ 1000
file consists of English letters (a-z, A-Z), digits (0-9), and whitespace characters (' ', '\t', '\n')

Visualization

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Understanding the Visualization

Input

File content and number n of characters to read

Process

Use read4 API repeatedly, managing buffer transfers

Output

Return total characters actually read (min of n and file length)

Key Takeaway

🎯 Key Insight: Use read4 repeatedly and carefully manage the buffer to handle partial reads and exact character counts

Asked in

F Facebook 45 G Google 30 A Amazon 25

The key insight is to use read4 repeatedly and manage a buffer to extract exactly n characters. The main challenge is handling cases where n is not a multiple of 4 or when the file has fewer characters than requested. Best approach uses buffer management with Time: O(n), Space: O(1).

Common Approaches

Approach	Time	Space	Notes
✓ Direct Buffer Management	O(n)	O(1)	Use read4 repeatedly and manage the buffer to extract exactly n characters

Direct Buffer Management — Algorithm Steps

Step 1: Initialize variables to track total characters read
Step 2: Loop calling read4 until we have n characters or reach end of file
Step 3: Copy characters from read4 buffer to output buffer
Step 4: Return total characters actually read

Visualization

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Step-by-Step Walkthrough

Initial State

File pointer at start, need to read n characters

Call read4

Read up to 4 characters into temporary buffer

Copy to Output

Copy needed characters from temp buffer to output

Repeat

Continue until n characters read or file ends

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

static int filePointer = 0;

int read4(char buf4[], const char* fileContent) {
    int charsRead = 0;
    int fileLen = strlen(fileContent);
    
    for (int i = 0; i < 4; i++) {
        if (filePointer < fileLen) {
            buf4[i] = fileContent[filePointer];
            filePointer++;
            charsRead++;
        } else {
            break;
        }
    }
    return charsRead;
}

int solution(const char* fileContent, int n) {
    filePointer = 0;
    char* buf = (char*)malloc(n * sizeof(char));
    int totalRead = 0;
    
    while (totalRead < n) {
        char buf4[4];
        int charsRead = read4(buf4, fileContent);
        
        if (charsRead == 0) {
            break;
        }
        
        int charsToCopy = (charsRead < n - totalRead) ? charsRead : n - totalRead;
        for (int i = 0; i < charsToCopy; i++) {
            buf[totalRead + i] = buf4[i];
        }
        
        totalRead += charsToCopy;
    }
    
    free(buf);
    return totalRead;
}

int main() {
    char fileContent[1000];
    int n;
    
    fgets(fileContent, sizeof(fileContent), stdin);
    fileContent[strcspn(fileContent, "\n")] = 0;
    
    scanf("%d", &n);
    
    int result = solution(fileContent, n);
    printf("%d\n", result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

We read each character at most once, calling read4 approximately n/4 times

✓ Linear Growth

Space Complexity

O(1)

Only use a fixed-size buffer of 4 characters for read4 calls

✓ Linear Space

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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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