Leetcode 157. Read N Characters Given Read4

Problem Explanation: The problem is to read n characters from a file using a given method read4. read4 reads 4 consecutive characters from the file and writes those characters in a buffer array buf. read4 has its own file pointer and it reads 4 characters at a time.

Then we need to implement method 'read' which uses 'read4' to read n characters from the file and stores them in a buffer array. We are not allowed to manipulate the file directly so the file is only read by 'read4' but not by 'read'. The read function is called once for each test case.

Walk Through an Example: For example, If we have a file "leetcode", and n = 5, after calling the read method, buf should contain "leetc". We read a total of 5 characters from 'read4' so it returns 5.

Approach of the Solution: The solution is simple, we will initialize an array 'buf4' of size 4 and two variables i4 and i to track the indices of array 'buf4' and 'buf' respectively. Then we will loop until we reach n and in the loop, we first check if i4 equals to the size of 'buf4'. If it is, It means we have consumed all characters in buf4, then we reset i4 and read next 4 characters from the file to 'buf4' using 'read4'. If we reach EOF (end of file) then we would return the index i.

In the end, we copy characters from 'buf4' to 'buf' and increment both the indices. Also, if i becomes equal to n, it signifies we have filled the buf array with n characters and break the while loop. Then we return i which is the total number of actual characters read.

Python Solution

1
2python
3class Solution:
4  def read(self, buf, n):
5    buf4 = [''] * 4  # Buffer to store characters read from read4
6    i4 = 0  # Index in buf4
7    n4 = 0  # Number of characters available in buf4
8    i = 0  # Total characters read
9
10    while i < n:
11      if i4 == n4:  # All characters in buf4 are consumed
12        i4 = 0  # Reset buf4's index
13        n4 = read4(buf4)  # Read 4 (or less) chars from file to buf4
14        if n4 == 0:       # Reach the end of file
15          return i  # Return the total characters read
16      buf[i] = buf4[i4]
17      i += 1
18      i4 += 1
19    return i

Java Solution

1
2java
3public class Solution {
4    private int i4 = 0;  // Index in buf4
5    private int n4 = 0;  // Number of characters available in buf4
6    private char[] buf4 = new char[4];  // Buffer to store characters read from read4
7
8    public int read(char[] buf, int n) {
9        int i = 0;  // Total characters read
10
11        while (i < n) {
12            if (i4 == n4) {  // All characters in buf4 are consumed
13                n4 = read4(buf4);  // Read 4 (or less) chars from file to buf4
14                i4 = 0;  // Reset buf4's index
15                if (n4 == 0)  // Reach the end of file
16                    return i;  // Return the total characters read
17            }
18            buf[i++] = buf4[i4++];
19        }
20        return i;
21    }
22}

JavaScript Solution

1
2javascript
3var Solution = function() {
4    this.buf4 = Array(4);  // Buffer to store characters read from read4
5    this.i4 = 0;  // Index in buf4
6    this.n4 = 0;  // Number of characters available in buf4
7}
8
9Solution.prototype.read = function(buf, n) {
10    let i = 0;  // Total characters read
11    while (i < n) {
12        if (this.i4 === this.n4) {  // All characters in buf4 are consumed
13            this.n4 = read4(this.buf4);  // Read 4 (or less) chars from file to buf4
14            this.i4 = 0;  // Reset buf4's index
15            if (this.n4 === 0)  // Reach the end of file
16                return i;  // Return the total characters read
17        }
18        buf[i++] = this.buf4[this.i4++];
19    }
20    return i;
21}
22
23### C# Solution
24

csharp public class Solution { private char[] buf4 = new char[4]; // Buffer to store characters read from read4 private int i4 = 0; // Index in buf4 private int n4 = 0; // Number of characters available in buf4

1public int Read(char[] buf, int n) {
2    int i = 0;  // Total characters read
3    while (i < n) {
4        if (i4 == n4) {  // All characters in buf4 are consumed
5            n4 = read4(buf4);  // Read 4 (or less) chars from file to buf4
6            i4 = 0;  // Reset buf4's index
7            if (n4 == 0)  // Reach the end of file
8                return i;  // Return the total characters read
9        }
10        buf[i++] = buf4[i4++];
11    }
12    return i;
13}

}

1
2
3
4Basic idea for this problem is pretty simple: we maintain a buffer `buf4` to hold characters read from `read4` and copy those characters to `buf` upon request. We also need to keep track of how many characters we have in `buf4` and how many characters we already used. If we exhaust `buf4`, we read more characters and reset the counters. Otherwise, we just keep copying characters to `buf`.Conclusion:
5The problem is not difficult after understanding the implications of read, read4, and the nature of the buffer. The solution we've explored uses a constant space, and does repetitive reads and writes, focusing only on ‘buf4’ and 'buf'. It's also worth noting that this solution guarantees that no reading occurs beyond the requested amount, even if the file contains more characters. 
6
7The solution was implemented in Python, Javascript, Java and C# by using a simple while loop to keep adding characters to the array until 'n' characters have been read or the file ends. We also ensured to reset the indices correctly each time we read with 'read4'. This problem is a good example of how certain problems are modeled and designed in software engineering using buffers and streams.