157. Read N Characters Given Read4

EasyArrayInteractiveSimulation
Leetcode Link

Problem Description

The problem is to simulate the behavior of reading characters from a file using a predefined API read4. The read4 API reads up to 4 characters from a file and stores them into a buffer. It returns the number of actual characters read, which could be fewer than 4 if the end of the file is reached.

You need to implement a read function that reads n characters into a given buffer buf. The function should only use the read4 method to interact with the file, meaning it cannot access the file content directly. The goal is to fill the buffer buf with n characters if they are available in the file, and return the count of characters actually read. If there are fewer than n characters left in the file when read is called, the function should read the remaining characters, write them to buf, and return the actual count.

Additionally, the function may be called multiple times with different values of n. The read function has to maintain the reading position in the file across multiple calls.

In summary, there are three key requirements to the problem:

  1. Read characters from a file and store them in the provided buffer buf.
  2. Use the given read4 API function to read from the file.
  3. Return the actual number of characters read, which should not exceed n.

Intuition

To solve this problem, we need to understand that the read function needs to interact with the read4 function to access the file data. Since we can only read 4 characters at a time, we might not always read exactly n characters in one go. As a result, we'll generally need to call read4 multiple times, accumulating characters until we've read as many as n or have reached the end of the file.

The solution should execute in a loop where each iteration attempts to read up to 4 new characters into a temporary buffer buf4, and then copy these characters to the final destination buffer buf. The loop needs to keep track of two main variables:

  1. The number of characters that have been read and added to the destination buffer buf.
  2. The number of characters read from the file in the current iteration using read4.

The loop must continue until we have either read n characters or there are no more characters to read from the file (i.e., read4 returns fewer than 4 characters).

Each time read4 is called, we iterate through the characters in buf4 and add them to buf. If the destination buffer buf is filled with n characters, the function should return immediately, indicating that the requested number of characters has been read.

The variable i in the provided code snippet keeps track of the number of characters copied to buf, and it's incremented after each character is copied. When i equals n, or when read4 does not return any new characters (indicating the end of the file), the function should return i as the total count of characters read.

Solution Approach

The solution provided follows an iterative approach, which effectively combines the usage of the buffer from read4 and the main buffer buf where the final output will be stored. Here's a step-by-step explanation of how the solution works:

  1. Initialization: Before entering the loop, an index i is initialized to keep track of the number of characters copied to the destination buffer buf. This i will be incremented after copying each character. A temporary buffer buf4 with space for 4 characters is also created to hold the characters read from the file by read4.

  2. Reading from File: Inside the while loop, read4 is called, and its return value is stored in v. The return value represents the number of characters actually read from the file, which can be anything from 0 to 4.

  3. Copying to Destination Buffer: After reading characters into buf4, a for loop iterates over the characters in buf4. For each character, the following is done:

    • Copy from buf4 to buf at the current index i.
    • Increment i.
    • Check if i is equal to or greater than n, which means we've read the required number of characters. If true, we return n immediately, as we've fulfilled the request.
  4. Ending the Loop: The while loop checks if v is at least 4, indicating that there might be more characters to read. If v is less than 4, it means the end of the file has been reached, or there are not enough characters left to fill buf4 completely, and the loop ends.

  5. Returning Read Characters: Outside the loop, after reading all necessary characters or reaching the end of the file, i (which represents the actual count of characters copied to buf) is returned.

The algorithm used here is straightforward and leverages both the read4 API constraint (i.e., reading up to only 4 characters at a time) and the requirement to copy a precise number of characters to the destination buffer buf. No complex data structures are required, and the pattern is to iterate and copy until the condition is satisfied.

This solution is robust and easy to understand. It calculates the right amount of characters needed and uses minimal extra space, only requiring an additional small buffer buf4 to bridge between read4 and the final buf.

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Example Walkthrough

Let's go through an example to see how the solution approach works. Imagine we have a file with the content "LeetCode" and we want to read 6 characters from it. We will use the read method to accomplish this task.

Consider the case where the read function is used as follows: read(buf, 6). We start with buf being an empty reference to a character array.

  1. Initialization: A temporary buffer buf4 is created to store the characters read from the file by read4, and an index i is initialized to 0.

  2. First Call to read4:

    • read4 is called, returning a value of 4, since it reads "Leet" from the file.
    • Now, the buf4 contains "Leet".
  3. Copying to Destination Buffer (First Loop):

    • Characters from buf4 are copied to buf one by one.
    • i is incremented with each character copied. After copying "Leet", i is now 4.
  4. Check if More Characters are Needed: Since i (4) is less than n (6), the loop continues.

  5. Second Call to read4:

    • read4 is called again, now it reads "Code" from the file.
    • The buf4 contains "Code", and read4 returns 4, but we only need 2 more characters.
  6. Copying to Destination Buffer (Second Loop):

    • We only copy two characters because i needs to reach 6.
    • After copying "Co", i is now 6, which is equal to n.
  7. End Loop and Return: Since i has reached n, we stop and return i.

Therefore, after the read(buf, 6) call, buf contains "LeetCo", and the function returns 6 because that's the number of characters we wanted to read.

Remember, the actual buffer buf is intended to be large enough for the n characters, and the values are not shown as a string but copied into it like an array. The provided explanation simplifies the representation for clarity.

Solution Implementation

1class Solution:
2    def read(self, buf, n):
3        """
4        Reads up to n characters from a file using read4() and stores them into buf.
5
6        :param buf: Destination buffer (List[str]) where characters are to be stored
7        :type buf: List[str]
8        :param n: Number of characters to read
9        :type n: int
10        :return: The actual number of characters read
11        :rtype: int
12        """
13
14        # Initialize the index pointer for the output buffer
15        index = 0
16
17        # Temporary buffer to store the output of read4
18        buf4 = [''] * 4
19
20        # Variable to store the number of characters read in last read4 operation
21        chars_read = 4  # Initialize to 4 to enter the while loop
22      
23        # Continue reading until we have read 4 or fewer chars (end of file)
24        while chars_read == 4:
25            # Read the next 4 (or fewer) chars from the file into buf4
26            chars_read = read4(buf4)
27          
28            # Iterate over the chars read into buf4
29            for i in range(chars_read):
30                # Copy the character from buf4 to the destination buffer
31                buf[index] = buf4[i]
32                index += 1  # Increment the index in the destination buffer
33              
34                # If we have read the required number of characters, return n
35                if index == n:
36                    return index
37      
38        # Return the actual number of characters read
39        return index
40
1public class Solution extends Reader4 {
2    /**
3     * Reads up to 'n' characters from the file and stores them in 'buf'.
4     * 
5     * @param buf Destination buffer to store read characters.
6     * @param n Number of characters to read.
7     * @return The number of actual characters read, which might be less than 'n' if the file has fewer characters.
8     */
9    public int read(char[] buf, int n) {
10        // Temporary buffer to hold chunks of read characters
11        char[] tempBuffer = new char[4];
12      
13        // Index for the destination buffer 'buf'
14        int bufIndex = 0;
15      
16        // Variable to hold the count of characters actually read in each read4 call
17        int charReadCount = 0;
18      
19        // Continue reading until there are fewer than 4 characters returned, which signifies end of file or buffer
20        do {
21            // Read up to 4 characters into tempBuffer
22            charReadCount = read4(tempBuffer);
23          
24            // Copy characters from tempBuffer to buf, up to the number of characters requested 'n'
25            for (int j = 0; j < charReadCount; ++j) {
26                buf[bufIndex] = tempBuffer[j];
27                bufIndex++;
28              
29                // If 'bufIndex' reaches 'n', we've read the required number of characters
30                if (bufIndex == n) {
31                    return n; // The requested number of characters have been read
32                }
33            }
34        } while (charReadCount == 4); // Continue if we read 4 characters, meaning there could be more to read
35      
36        // Return the number of characters actually stored in 'buf'
37        return bufIndex;
38    }
39}
40
1class Solution {
2public:
3    /**
4     * Reads characters into buf from a file and returns the actual number
5     * of characters read, which could be less than n if the end of file is reached.
6     * @param buf - Destination buffer to store read characters
7     * @param n   - Number of characters to be read 
8     * @return    - Actual number of characters read
9     */
10    int read(char* buf, int n) {
11        char tempBuffer[4]; // Temporary buffer to hold read chunks of 4 characters
12        int totalCharsRead = 0; // Total characters read
13
14        while (true) {
15            // Read up to 4 characters into tempBuffer from file
16            int charsRead = read4(tempBuffer);
17          
18            // Transfer characters from tempBuffer to destination buf
19            for (int j = 0; j < charsRead; ++j) {
20                buf[totalCharsRead++] = tempBuffer[j];
21
22                // If the number of characters requested (n) is reached,
23                // return the number of characters read so far.
24                if (totalCharsRead == n) {
25                    return n;
26                }
27            }
28
29            // Break the loop if we read less than 4 characters,
30            // which means end of file is reached
31            if (charsRead < 4) {
32                break;
33            }
34        }
35
36        // Return total number of characters actually read
37        return totalCharsRead;
38    }
39};
40
1// Assuming read4 is already defined elsewhere to read 4 characters at a time from the file
2declare function read4(tempBuffer: string[]): number;
3
4/**
5 * Reads characters into buf from a file and returns the actual number of characters read,
6 * which could be less than n if the end of file is reached.
7 * @param buf - Destination buffer to store read characters
8 * @param n - Number of characters to be read
9 * @return - Actual number of characters read
10 */
11let read = (buf: string[], n: number): number => {
12    // Temporary buffer to hold read chunks of 4 characters
13    let tempBuffer: string[] = [];
14    // Total characters read
15    let totalCharsRead = 0; 
16
17    while (true) {
18        // Read up to 4 characters into tempBuffer from file
19        let charsRead = read4(tempBuffer);
20
21        // Transfer characters from tempBuffer to destination buf
22        for (let j = 0; j < charsRead; ++j) {
23            // Store the character in the destination buffer
24            buf[totalCharsRead++] = tempBuffer[j];
25
26            // If the number of characters requested (n) is reached,
27            // return the total number of characters read so far.
28            if (totalCharsRead === n) {
29                return n;
30            }
31        }
32
33        // Break the loop if we read less than 4 characters, signaling end of file
34        if (charsRead < 4) {
35            break;
36        }
37    }
38
39    // Return total number of characters actually read
40    return totalCharsRead;
41};
42

Time and Space Complexity

Time Complexity

The time complexity of the code is determined by the number of times read4 is called and the number of times the inner loop runs. The read4 function is called until it returns less than 4 characters, which indicates the end of the file or the buffer is fully read.

The maximum number of times read4 can be called is ceil(n/4), since read4 reads 4 characters at a time and n is the total number of characters we want to read. In the worst case, the inner loop runs 4 times for each call to read4 (if read4 returns 4 characters each time). Therefore, the inner loop iteration count is at most 4 multiplied by the number of read4 calls, giving us a potential maximum of 4 * ceil(n/4) iterations.

However, due to the condition if i >= n inside the inner loop, the actual read process will stop as soon as n characters are read. Thus, the tight bound on the number of iterations of the inner loop is n. Therefore, the time complexity of the code is O(n).

Space Complexity

The space complexity of the algorithm is determined by the additional space used by the algorithm besides the input and output. The additional space in this algorithm is utilized for the buf4 array, which is a constant size of 4 characters.

No other additional space is growing with the input size n. Hence the space complexity is constant, or O(1).

Learn more about how to find time and space complexity quickly using problem constraints.


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