Leetcode 158. Read N Characters Given Read4 II - Call multiple times

Problem Explanation:

This problem is asking to create a method which can read n characters from a file using the provided read4 method. The read4 method is able to read 4 characters at a time from a file and store those characters in the provided buffer. Our read method should be able to manage reading a variable amount of characters and may be called multiple times in succession.

We'll walk through an example where the file is represented as "abcdefghijk" and we want to read 7 characters.

• For the first call to read(buf, 7), the method will call read4(buf) until it has read 7 characters. After the first call to read4(buf), the buffer will store "abcd" and the file pointer will be at 'e'. After the second call to read4(buf), the buffer will store "efgh", but we only need "efg", so we keep "h" for the next read. The read function will return 7 as we've successfully read 7 characters ("abcdefg").
• For the next call to read(buf, 1), we don't need to call read4(buf) as we still have a character left over from last time. The buffer will contain "h" and the read function will return 1.

The Solution:

The solution uses the provided read4 method to solve this problem. It uses the index and size concept for buffer management.

Here's how each step of the algorithm work:

1. Before we start reading characters, initialize two pointers: i for buf's index and i4 for buf4's index, and a counter n4 for counting the total number of characters we have read from buf4.

2. We read characters into buf one by one.

3. For each character, if all characters in buf4 are consumed, reset buf4's index i4 and read another 4 (or less) characters from the file to buf4.

4. If the file reaches the EOF (indicated by read4 returns 0), return the number of characters we have already read.

Python Solution:

1
2python
3class Solution(object):
4    def __init__(self):
5        self.buf4_index = 0
6        self.buf4_size = 0
7        self.buf4 = ['']*4
8
9    def read(self, buf, n):
10        # buf's index
11        i = 0
12        while i < n:
13            if self.buf4_index < self.buf4_size:
14                buf[i] = self.buf4[self.buf4_index]
15                i += 1
16                self.buf4_index += 1
17            elif self.buf4_index == self.buf4_size:
18                self.buf4_size = read4(self.buf4)
19                self.buf4_index = 0
20                if self.buf4_size == 0:
21                    return i  # EOF
22        return i

Java Solution:

1
2java
3public class Solution extends Reader4 {
4    private int buf4_index = 0;
5    private int buf4_size = 0;
6    private char[] buf4 = new char[4];
7
8    public int read(char[] buf, int n) {
9        int i = 0;
10        while (i < n) {
11            if (buf4_index < buf4_size) {
12                buf[i] = buf4[buf4_index];
13                i++;
14                buf4_index++;
15            } else if (buf4_index == buf4_size){
16                buf4_size = read4(buf4);
17                buf4_index = 0;
18                if (buf4_size == 0) {
19                    return i;  // EOF
20                }
21            }
22        }
23        return i;
24    }
25}

JavaScript Solution:

1
2javascript
3var Solution = function() {
4    this.buf4_index = 0;
5    this.buf4_size = 0;
6    this.buf4 = [];
7};
8
9Solution.prototype.read = function(buf, n) {
10    let i = 0;
11    while (i < n) {
12        if (this.buf4_index < this.buf4_size) {
13            buf[i] = this.buf4[this.buf4_index];
14            i++;
15            this.buf4_index++;
16        } else if (this.buf4_index == this.buf4_size) {
17            this.buf4_size = read4(this.buf4);
18            this.buf4_index = 0;
19            if (this.buf4_size == 0) {
20                return i;  // EOF
21            }
22        }
23    }
24    return i;
25};

C++ Solution:

1
2c++
3class Solution {
4public:
5    Solution() : buf4_index(0), buf4_size(0) {}
6
7    int read(char *buf, int n) {
8        int i = 0;
9        while (i < n) {
10            if (buf4_index < buf4_size) {
11                buf[i++] = buf4[buf4_index++];
12            } else if (buf4_index == buf4_size) {
13                buf4_size = read4(buf4);
14                buf4_index = 0;
15                if (buf4_size == 0) {
16                    return i;  // EOF
17                }
18            }
19        }
20        return i;
21    }
22
23private:
24    char buf4[4];
25    int buf4_index, buf4_size;
26};

C# Solution:

1
2csharp
3public class Solution : Reader4 {
4    private int buf4_index = 0;
5    private int buf4_size = 0;
6    private char[] buf4 = new char[4];
7
8    public int Read(char[] buf, int n) {
9        int i = 0;
10        while (i < n) {
11            if (buf4_index < buf4_size) {
12                buf[i] = buf4[buf4_index];
13                i++;
14                buf4_index++;
15            } else if (buf4_index == buf4_size){
16                buf4_size = Read4(buf4);
17                buf4_index = 0;
18                if (buf4_size == 0) {
19                    return i;  // EOF
20                }
21            }
22        }
23        return i;
24    }
25}

For above solution, buf4, buf4_index, and buf4_size are declared as class variables, because they need to maintain their states between multiple calls of read(). That's the key to make the stuck data available for future readings.Time Complexity Analysis:

The time complexity of the code is O(n), where n is the number of characters we're aiming to read. This is because we're running a while loop, with the exit condition being when we've read n characters. In the worst-case scenario, we must go through this loop n times.

Space Complexity Analysis:

The space complexity of the code is O(1). Apart from the inputs, the only extra memory used is a 4-character buffer (buf4) that's reused in each function call, its associated index (buf4_index) and size (buf4_size). Since the sizes of these are constant and not increasing with the size of the function inputs, the space complexity is constant.

Tips for Interviews:

In interviews, it's important to clearly explain these complexities to your interviewer, as well as your understanding of the problem, your thought process in solving the problem and explaining the workings of your solution. It is equally important to write clean, organized, and well-commented code, as this demonstrates proper software engineering principles and makes your solution easier to understand. Interviewers would value your solution much more if you can prove that it will work on all possible edge cases and not just the provided or obvious ones. An efficient and elegant solution to a problem, that you can explain thoroughly, can impress your interviewers and increase your chances of success in your coding interviews.

Collaboration:

Working software is usually written by teams, so it's advantageous to be comfortable with pair programming. Even though your interviewer may not be writing code with you, behave as if you're pair programming. Describe what you're doing as you're doing it, ask for input when you're unsure, and actively involve the interviewer in your thinking and decision-making process.

Remember to stay upbeat and positive during the interview; show that you enjoy solving difficult problems and writing software. Even if you fail to solve the problem, the attitude you display can demonstrate your work ethic and positivity in the face of challenges, which are highly desirable qualities in a software developer.

Conclusion:

In conclusion, the problem we analyzed before where we're expected to implement a read method that can read n characters from a file using the provided read4 method which reads 4 characters at a time, is a complex problem that can evaluate the candidate's skills for their competence with multiple areas. It not only tests basic knowledge of arrays and buffers but also tests logic skills, coding practices, understanding of time complexity and object-oriented concepts. Hence, this or similar problems can be a valuable addition to your interview preparation material.