Question

Formatted question description: https://leetcode.ca/all/158.html

Given a file and assume that you can only read the file using a given method read4, implement a method read to read n characters. Your method read may be called multiple times.

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 store it 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 may be called multiple times.
Please remember to RESET your class variables declared in Solution, as static/class variables are persisted across multiple test cases. Please see here for more details.
You may assume the destination buffer array, buf, is guaranteed to have enough space for storing n characters.
It is guaranteed that in a given test case the same buffer buf is called by read.

Example 1:

Input: file = "abc", queries = [1,2,1]
Output: [1,2,0]
Explanation: The test case represents the following scenario:
File file("abc");
Solution sol;
sol.read(buf, 1); // After calling your read method, buf should contain "a". We read a total of 1 character from the file, so return 1.
sol.read(buf, 2); // Now buf should contain "bc". We read a total of 2 characters from the file, so return 2.
sol.read(buf, 1); // We have reached the end of file, no more characters can be read. So return 0.
Assume buf is allocated and guaranteed to have enough space for storing all characters from the file.

Example 2:

Input: file = "abc", queries = [4,1]
Output: [3,0]
Explanation: The test case represents the following scenario:
File file("abc");
Solution sol;
sol.read(buf, 4); // After calling your read method, buf should contain "abc". We read a total of 3 characters from the file, so return 3.
sol.read(buf, 1); // We have reached the end of file, no more characters can be read. So return 0.

Constraints:

1 <= file.length <= 500
file consist of English letters and digits.
1 <= queries.length <= 10
1 <= queries[i] <= 500

Algorithm

Two variables readPos and writePos are used to record the position of reading and writing, i starts looping from 0 to n,

If the read and write positions are the same at this time, call the read4 function, assign the result to writePos, and set readPos to zero,
If writePos is zero, indicating that there is nothing in buf, return the current coordinate i. Then use the readPos position of the built-in buff variable to overwrite the i position of the input string buf,
If the traversal is completed, return n

Code

public class Read_N_Characters_Given_Read4 {
    public class Solution extends Reader4 {
        /**
         * @param buf    Destination buffer
         * @param n   　　Maximum number of characters to read
         * @return　     The number of characters read
         */

        public int read(char[] buf, int n) {
            boolean isEof = false;
            int charsRead = 0;
            char[] buf4 = new char[4];

            while (!isEof && charsRead < n) {
                int size = read4(buf4);
                if (size < 4) {
                    isEof = true;
                }

                if (charsRead + size > n) {
                    size = n - charsRead;
                }

                /*
                        arraycopy(Object src,  int  srcPos,
                                  Object dest, int destPos,
                                  int length);
                 */
                // if last iteration and size==0, then copy length==0, not over-copy
                System.arraycopy(buf4, 0, buf, charsRead, size);
                charsRead += size;
            }

            return charsRead;
        }

    }

    private class Reader4 {
        public int read4(char[] buf4) {
            return 1; // stub
        }
    }
}


############

/**
 * The read4 API is defined in the parent class Reader4.
 *     int read4(char[] buf4);
 */

public class Solution extends Reader4 {
    private char[] buf4 = new char[4];
    private int i;
    private int size;

    /**
     * @param buf Destination buffer
     * @param n   Number of characters to read
     * @return    The number of actual characters read
     */
    public int read(char[] buf, int n) {
        int j = 0;
        while (j < n) {
            if (i == size) {
                size = read4(buf4);
                i = 0;
                if (size == 0) {
                    break;
                }
            }
            while (j < n && i < size) {
                buf[j++] = buf4[i++];
            }
        }
        return j;
    }
}

// OJ: https://leetcode.com/problems/read-n-characters-given-read4-ii-call-multiple-times/
// Time: O(N)
// Space: O(1)
// Forward declaration of the read4 API.
int read4(char *buf);
class Solution {
public:
    /**
     * @param buf Destination buffer
     * @param n   Number of characters to read
     * @return    The number of actual characters read
     */
    char leftover[4] = { '\0' };
    int read(char *buf, int n) {
        strcpy(buf, leftover);
        int cnt = strlen(leftover);
        while (cnt < n) {
            int x = read4(buf + cnt);
            if (!x) break;
            cnt += x;
        }
        if (cnt > n) {
            memcpy(leftover, buf + n, cnt - n);
            leftover[cnt - n] = '\0';
        } else leftover[0] = '\0';
        cnt = min(cnt, n);
        return cnt;
    }
};

# The read4 API is already defined for you.
# def read4(buf4: List[str]) -> int:


class Solution:
    def __init__(self):
        self.buf4 = [None] * 4
        self.i = self.size = 0

    def read(self, buf: List[str], n: int) -> int:
        j = 0
        while j < n:
            if self.i == self.size:
                self.size = read4(self.buf4)
                self.i = 0
                if self.size == 0:
                    break
            while j < n and self.i < self.size:
                buf[j] = self.buf4[self.i]
                self.i += 1
                j += 1
        return j

############

# The read4 API is already defined for you.
# @param buf, a list of characters
# @return an integer
# def read4(buf):
from collections import deque


class Solution(object):
  def __init__(self):
    self.rBuf = deque([])

  def read(self, buf, n):
    """
    :type buf: Destination buffer (List[str])
    :type n: Maximum number of characters to read (int)
    :rtype: The number of characters read (int)
    """
    cnt = 0
    tmp = [""] * 4
    while cnt < n:
      r = read4(tmp)
      for i in range(r):
        self.rBuf.append(tmp[i])
      for i in range(min(n - cnt, len(self.rBuf))):
        buf[cnt] = self.rBuf.popleft()
        cnt += 1
      if r == 0:
        break
    return cnt

/**
 * The read4 API is already defined for you.
 *
 *     read4 := func(buf4 []byte) int
 *
 * // Below is an example of how the read4 API can be called.
 * file := File("abcdefghijk") // File is "abcdefghijk", initially file pointer (fp) points to 'a'
 * buf4 := make([]byte, 4) // Create buffer with enough space to store characters
 * read4(buf4) // read4 returns 4. Now buf = ['a','b','c','d'], fp points to 'e'
 * read4(buf4) // read4 returns 4. Now buf = ['e','f','g','h'], fp points to 'i'
 * read4(buf4) // read4 returns 3. Now buf = ['i','j','k',...], fp points to end of file
 */

var solution = func(read4 func([]byte) int) func([]byte, int) int {
	buf4 := make([]byte, 4)
	i, size := 0, 0
	// implement read below.
	return func(buf []byte, n int) int {
		j := 0
		for j < n {
			if i == size {
				size = read4(buf4)
				i = 0
				if size == 0 {
					break
				}
			}
			for j < n && i < size {
				buf[j] = buf4[i]
				i, j = i+1, j+1
			}
		}
		return j
	}
}

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

Question

Algorithm

Code

All Problems

All Solutions

Welcome to Subscribe On Youtube

Question

Algorithm

Code

All Problems

All Solutions