Read N Characters Given Read4 Python Solution

Tested Python solution for LeetCode 157 with 17 pytest cases. Generate a practice environment with lcpy.

LeetCode 157, Easy. Topics: Array, Interactive, Simulation. View on LeetCode.

Generate this problem as a practice environment: tested reference solution, 17 parametrized pytest cases, and a playground notebook:

lcpy gen -n 157   # by problem number
lcpy gen -s read_n_characters_given_read4   # by problem name

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Problem

Given a file and assume that you can only read the file using a given method read4, implement a method 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[].

Below is a high-level 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 to store characters
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 stores them in the buffer array buf. Consider that you cannot manipulate file directly.

The return value is the number of actual characters read.

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.
• You may assume the destination buffer array, buf, is guaranteed to have enough space for storing n characters.

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Examples

Input: file = "abc", n = 4
Output: 3
Explanation: After calling your read method, buf should contain "abc". We read a total of 3 characters from the file, so return 3.
Note that "abc" is the file's content, not buf. buf is the destination buffer that you will have to write the results to.

Input: file = "abcde", n = 5
Output: 5
Explanation: After calling your read method, buf should contain "abcde". We read a total of 5 characters from the file, so return 5.

Input: file = "abcdABCD1234", n = 12
Output: 12
Explanation: After calling your read method, buf should contain "abcdABCD1234". We read a total of 12 characters from the file, so return 12.

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Constraints

• 1 <= file.length <= 500
• file consists of English letters and digits.
• 1 <= n <= 1000

Note: In this repository the read4 API is provided as the read4(buf4) method of the File class in solution.py, and read receives that File instance as its third argument.

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Solution

Reference implementation from solution.py on GitHub, full suite in test_solution.py:

class File:
    # Test-harness API: backs the read4 interface with the file content
    def __init__(self, content: str) -> None:
        self.content = content
        self.pos = 0

    def read4(self, buf4: list[str]) -> int:
        # Reads up to 4 consecutive characters into buf4 and returns the
        # number of characters actually read
        chunk = self.content[self.pos : self.pos + 4]
        self.pos += len(chunk)
        for i, char in enumerate(chunk):
            buf4[i] = char
        return len(chunk)


class Solution:
    # Time: O(n)
    # Space: O(1)
    def read(self, buf: list[str], n: int, file: File) -> int:
        i = 0
        buf4 = [""] * 4
        while i < n:
            count = file.read4(buf4)
            if count == 0:
                break
            for j in range(min(count, n - i)):
                buf[i] = buf4[j]
                i += 1
        return i

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Complexity

Time	Space
O(n)	O(1)

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