LeetCode 158: Read N Characters Given Read4 II - Call Multiple Times

i recently solved the read n characters given read4 ii problem on leetcode, and it’s a great example of system design and file i/o operations. this hard problem tests your understanding of buffering, state management, and efficient file reading.

Problem Statement

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 from the file using read4.

the read4 method has the following signature:

int read4(char[] buf4)

the read4 method reads 4 consecutive characters from the file, writes those characters into the buffer array buf4, and returns the number of characters read.

note that read4() has its own file pointer, much like file.readline() in python.

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

the read method may be called multiple times.

example:

file file("abcde"); // file is "abcde"
solution.read(buf, 5); // after calling your read method, buf should contain "abcde". we read a total of 5 characters from the file, so return 5.
solution.read(buf, 1); // we have reached the end of file, no more characters can be read. so return 0.

My Approach

when i first saw this problem, i immediately thought about using a buffer-based approach. the key insight is that we need to maintain state between multiple calls to the read method, efficiently managing a buffer to minimize calls to read4.

Initial Thoughts

i started by thinking about different approaches:

naive approach - call read4 for each character needed (inefficient)
buffer-based approach - maintain a buffer and reuse characters (optimal)
alternative approach - using a queue or circular buffer

Solution Strategy

i decided to use a buffer-based approach with the following strategy:

maintain buffer state - keep track of buffered characters and position
two-phase reading - first use buffered data, then call read4 if needed
efficient state management - preserve buffer state between calls
minimize read4 calls - reuse buffered data whenever possible

My Solution

/**
 * the read4 api is defined in the parent class reader4.
 *     read4(buf4: string[]): number
 */

class solution extends reader4 {
    private buffer: string[] = [];
    private bufferptr: number = 0;
    private buffercount: number = 0;
    
    /**
     * @param {string[]} buf destination buffer
     * @param {number} n number of characters to read
     * @return {number} the actual number of characters read
     */
    read(buf: string[], n: number): number {
        let copiedchars = 0;
        
        // first, copy any remaining characters from our buffer
        while (copiedchars < n && this.bufferptr < this.buffercount) {
            buf[copiedchars] = this.buffer[this.bufferptr];
            copiedchars++;
            this.bufferptr++;
        }
        
        // if we still need more characters, call read4
        while (copiedchars < n) {
            // reset buffer pointer since we're getting new data
            this.bufferptr = 0;
            
            // read new characters into our buffer
            this.buffercount = this.read4(this.buffer);
            
            // if no more characters available, break
            if (this.buffercount === 0) {
                break;
            }
            
            // copy characters from buffer to destination
            while (copiedchars < n && this.bufferptr < this.buffercount) {
                buf[copiedchars] = this.buffer[this.bufferptr];
                copiedchars++;
                this.bufferptr++;
            }
        }
        
        return copiedchars;
    }
}

Code Breakdown

let me walk through how this solution works:

1. Buffer State Management

private buffer: string[] = [];
private bufferptr: number = 0;
private buffercount: number = 0;

we maintain three key pieces of state:

buffer: stores characters from read4 calls
bufferptr: current position within the buffer
buffercount: total number of characters in the buffer

2. Two-Phase Reading Process

// phase 1: use existing buffer
while (copiedchars < n && this.bufferptr < this.buffercount) {
    buf[copiedchars] = this.buffer[this.bufferptr];
    copiedchars++;
    this.bufferptr++;
}

// phase 2: call read4 if needed
while (copiedchars < n) {
    // ... read4 logic
}

we first use any remaining characters from previous calls, then call read4 to get new data if needed.

3. Efficient Buffer Reuse

// reset buffer pointer since we're getting new data
this.bufferptr = 0;
this.buffercount = this.read4(this.buffer);

when we need new data, we reset the buffer pointer and get fresh characters from read4.

Example Walkthrough

let’s trace through multiple read calls: read(buf, 1), read(buf, 2), read(buf, 1)

initial state: buffer = [], bufferptr = 0, buffercount = 0

call 1: read(buf, 1)
- buffer is empty, need to call read4
- read4 returns ["a","b","c","d"], buffercount = 4
- copy "a" to buf, copiedchars = 1
- bufferptr = 1, remaining: ["b","c","d"]

call 2: read(buf, 2)
- copy "b","c" from buffer, copiedchars = 2
- bufferptr = 3, remaining: ["d"]

call 3: read(buf, 1)
- copy "d" from buffer, copiedchars = 1
- bufferptr = 4, buffer exhausted
- next read will call read4 again

Time and Space Complexity

time complexity: O(n) where n is the number of characters requested
space complexity: O(1) since we use a fixed-size buffer of 4 characters

the algorithm is optimal because:

we only call read4 when necessary
we reuse buffered data efficiently
we maintain constant space usage

Key Insights

buffer management - crucial for efficiency between multiple calls
state preservation - buffer position must be maintained between calls
two-phase reading - first use buffer, then call read4 if needed
minimize api calls - reuse buffered data to reduce read4 calls

Alternative Approaches

i also considered:

naive approach - call read4 for each character needed
- time complexity: O(n) but with many unnecessary read4 calls
- space complexity: O(1)
- much less efficient due to excessive api calls
queue-based approach - use a queue to store characters
- similar to buffer approach
- slightly more complex implementation
- same time/space complexity
circular buffer - implement a circular buffer for better memory usage
- more complex implementation
- same asymptotic complexity
- useful for very large files

Edge Cases to Consider

multiple small reads - handle consecutive small read requests efficiently
read more than available - handle requests larger than remaining file content
empty file - handle case where file has no content
buffer exhaustion - properly handle when buffer is completely used
end of file - detect when no more characters are available

Lessons Learned

this problem taught me:

system design principles - designing efficient file i/o operations
state management - maintaining state between function calls
buffer optimization - using buffering to improve performance
api design - designing interfaces that work well with external apis

Real-World Applications

this problem has direct applications in:

file system implementations - efficient file reading operations
stream processing - handling data streams with buffering
database systems - reading data from storage efficiently
network protocols - buffering network data for processing

Conclusion

the read n characters given read4 ii problem is an excellent exercise in system design and efficient file i/o operations. the key insight is using buffering to minimize expensive api calls while maintaining state between multiple read operations.

you can find my complete solution on leetcode.

this is part of my leetcode problem-solving series. i’m documenting my solutions to help others learn and to track my own progress.