Solution Explanation for LeetCode 157: Read N Characters Given Read4

This problem involves reading n characters from a file using a helper function read4 that reads only 4 characters at a time. The challenge is to efficiently use read4 to fulfill the request for n characters.

Approach:

The solution uses a simple iterative approach. It repeatedly calls read4 to read chunks of 4 characters until either n characters are read or the end of the file is reached. The characters read from read4 are then copied to the destination buffer buf.

Algorithm:

1. Initialization: Create a buffer buf4 (size 4) to store characters read by read4. Initialize a counter i to track the number of characters written to the destination buffer buf, and v to indicate the number of characters read by read4 in each iteration (initially 5 to ensure the loop starts).

2. Iteration: The while loop continues as long as v (characters read from read4) is greater than or equal to 4, meaning there are potentially more characters to read.

3. Read4 call: Inside the loop, call read4 to read characters into buf4. The return value v is the number of characters actually read.

4. Copy characters: Iterate through the characters in buf4 (up to v characters). Copy each character to the destination buffer buf, incrementing i. If i reaches n, then n characters have been read, so return n.

5. Return: After the loop, i holds the total number of characters read. Return i.

Time Complexity Analysis:

• The read4 function is called at most ⌈n/4⌉ times (ceiling of n divided by 4), as it reads 4 characters at a time.
• Copying characters from buf4 to buf takes O(n) time in the worst case (when all n characters are read).
• Therefore, the overall time complexity is O(n), dominated by the copying step.

Space Complexity Analysis:

The space complexity is O(1) because we use a fixed-size buffer buf4 regardless of the input size n.

Code Implementation (Python):

class Solution:
    def read(self, buf, n):
        i = 0
        buf4 = [0] * 4 #buffer for read4
        v = 5 #initial value to enter the while loop
        while v >= 4:
            v = read4(buf4)
            for j in range(v):
                buf[i] = buf4[j]
                i += 1
                if i >= n:
                    return n
        return i

Code Implementation (Java):

class Solution extends Reader4 {
    public int read(char[] buf, int n) {
        char[] buf4 = new char[4];
        int i = 0, v = 5;
        while (v >= 4) {
            v = read4(buf4);
            for (int j = 0; j < v; ++j) {
                buf[i++] = buf4[j];
                if (i >= n) {
                    return n;
                }
            }
        }
        return i;
    }
}

Code Implementation (C++):

class Solution {
public:
    int read(char *buf, int n) {
        char buf4[4];
        int i = 0, v = 5;
        while (v >= 4) {
            v = read4(buf4);
            for (int j = 0; j < v; ++j) {
                buf[i++] = buf4[j];
                if (i >= n) {
                    return n;
                }
            }
        }
        return i;
    }
};

Code Implementation (Go):

func read(buf []byte, n int) int {
	buf4 := make([]byte, 4)
	i, v := 0, 5
	for v >= 4 {
		v = read4(buf4)
		for j := 0; j < v; j++ {
			buf[i] = buf4[j]
			i++
			if i >= n {
				return n
			}
		}
	}
	return i
}

The implementations in other languages are very similar, differing only in syntax. The core algorithm remains the same.