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Introduction

LeetCode 157 asks us to implement a method for reading n characters from a file using a helper function Read4, which reads up to 4 characters at a time. The challenge is to return exactly n characters while handling the edge cases of reading from the file, such as less than 4 characters remaining or the end of the file.

The Read4 function is predefined and reads up to 4 characters from the file into a buffer. The goal is to implement the read() method that reads n characters using Read4.

Problem Statement

We are given a file with a read4 function, which reads at most 4 bytes of the file at a time. The task is to implement a function read that reads exactly n characters from the file. You need to make use of the read4 API to achieve this.

Here is the API signature:

go

func read(buf []byte, n int) int

• buf: The buffer to store the result.
• n: The number of characters to read.

The function should return the number of characters actually read.

Helper Function:

The read4 function reads 4 characters at a time from the file into a buffer. Its signature is:

go

func read4(buf []byte) int

It reads up to 4 bytes from the file and returns the number of bytes actually read. The number of bytes can be less than 4 if it reaches the end of the file.

Approach

Key Idea:

The read method needs to repeatedly call read4 until it has read the desired n characters or the end of the file is reached. Here’s the approach to solve this:

1. Iterate until we read n characters:

• Call read4 and store the result in a temporary buffer.
• Copy the characters into the buf array until we have collected n characters.

1. Handle the number of characters read:

• If fewer than 4 characters are returned by read4 (indicating the end of the file), we stop.
• Keep track of the total number of characters read and stop when n characters have been read.

Steps:

1. Initialize a pointer totalRead to 0 to track how many characters have been read.
2. Use a loop to call read4 until n characters are read or there are no more characters to read.
3. In each loop iteration, check how many characters read4 returns.
4. Add these characters to the result buffer and update the totalRead.

Go Implementation

go

// The read4 function is predefined for us to use.
// It reads up to 4 bytes from the file into the provided buffer and returns the number of bytes read.
func read4(buf []byte) int {
    // This is just a mock of the read4 function, in real usage it reads from file.
    // The implementation is not needed here, as it's predefined.
    return 0
}

// Read N characters from the file using read4.
func read(buf []byte, n int) int {
    totalRead := 0
    tempBuf := make([]byte, 4)  // Temporary buffer to hold 4 characters

    // Keep reading until we've read the required number of characters or reached the end
    for totalRead < n {
        // Call read4 to read up to 4 characters
        count := read4(tempBuf)
        if count == 0 { // End of file reached
            break
        }

        // Copy the characters from tempBuf into the output buffer
        for i := 0; i < count && totalRead < n; i++ {
            buf[totalRead] = tempBuf[i]
            totalRead++
        }
    }

    return totalRead
}

Explanation of the Code:

1. The read4 Function:

• This is a predefined function that simulates reading 4 characters at a time from the file. In a real-world scenario, it would read from an actual file or stream.

2. The read Function:

• totalRead: Tracks how many characters have been successfully read into the buffer.
• tempBuf: A temporary buffer that holds the characters read from the file (at most 4 characters).
• We repeatedly call read4 to read chunks of up to 4 characters.
• In each loop iteration, we copy the characters from tempBuf to buf until we've read n characters or reached the end of the file.
• If fewer than 4 characters are read, we check for the end of the file and stop reading.

3. Copying Characters:

• We copy characters from tempBuf to buf one by one until we've read n characters or we exhaust the characters returned by read4.

Time and Space Complexity

OperationComplexityreadO(n)read4O(1)

• Time Complexity:
• The read function calls read4 multiple times, but each call to read4 is O(1). The number of read4 calls is proportional to n (the number of characters to read). Thus, the time complexity is O(n).
• Space Complexity:
• We use a temporary buffer tempBuf of size 4, so the space complexity is O(1). The space for the buf array is given as input, so it does not contribute to the overall space complexity.

Example

Example 1:

go

buf := make([]byte, 5)
n := 5
result := read(buf, n)
fmt.Println(result) // Output: 5
fmt.Println(string(buf)) // Output: "Hello" (Assuming 'read4' returns "Hello" in 4-byte chunks)

Explanation:

• We want to read 5 characters.
• If read4 returns "Hell" in one call and "o" in another, we will store "Hell" in the buffer first, then "o" to make the total 5 characters.

Conclusion

LeetCode 157 tasks us with reading n characters using a helper function read4. By using a temporary buffer to store characters returned by read4, we can efficiently accumulate the required characters into the provided buffer. This solution handles edge cases, such as reading fewer than 4 characters at the end of the file, while ensuring that the time complexity remains O(n).