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Link for the Problem – Read N Characters Given Read4– LeetCode Problem
Read N Characters Given Read4– LeetCode Problem
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:
![Read N Characters Given Read4 LeetCode Programming Solutions | LeetCode Problem Solutions in C++, Java, & Python [💯Correct] 2 157 example](https://www.goodtecher.com/wp-content/uploads/2021/08/157_example-1024x687.png)
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 forread
. - 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 storingn
characters.
Example 1:
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.
Example 2:
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.
Example 3:
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.
Example 4:
Input: file = "leetcode", n = 5 Output: 5 Explanation: After calling your read method, buf should contain "leetc". We read a total of 5 characters from the file, so return 5.
Constraints:
1 <= file.length <= 500
file
consist of English letters and digits.1 <= n <= 1000
Read N Characters Given Read4– LeetCode Solutions
Read N Characters Given Read4 Solution in C++:
/** * The read4 API is defined in the parent class Reader4. * int read4(char *buf4); */ class Solution { public: /** * @param buf Destination buffer * @param n Number of characters to read * @return The number of actual characters read */ int read(char* buf, int n) { char* buf4 = new char[4]; int i4 = 0; // buf4's index int n4 = 0; // buf4's size int i = 0; // buf's index while (i < n) { if (i4 == n4) { // all characters in buf4 are consumed i4 = 0; // reset buf4's index n4 = read4(buf4); // read 4 (or less) chars from file to buf4 if (n4 == 0) // reach the EOF return i; } buf[i++] = buf4[i4++]; } return i; } };
Read N Characters Given Read4 Solution in Java:
/** * The read4 API is defined in the parent class Reader4. * int read4(char[] buf4); */ public class Solution extends Reader4 { /** * @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) { char[] buf4 = new char[4]; int i4 = 0; // buf4's index int n4 = 0; // buf4's size int i = 0; // buf's index while (i < n) { if (i4 == n4) { // all characters in buf4 are consumed i4 = 0; // reset buf4's index n4 = read4(buf4); // read 4 (or less) chars from file to buf4 if (n4 == 0) // reach the EOF return i; } buf[i++] = buf4[i4++]; } return i; } }
Read N Characters Given Read4 Solution in Python:
""" The read4 API is already defined for you. def read4(buf4: List[chr]) -> 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 = [' '] * 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 """ class Solution: def read(self, buf: List[str], n: int) -> int: buf4 = [' '] * 4 i4 = 0 # buf4's index n4 = 0 # buf4's size i = 0 # buf's index while i < n: if i4 == n4: # all characters in buf4 are consumed i4 = 0 # reset buf4's index n4 = read4(buf4) # read 4 (or less) chars from file to buf4 if n4 == 0: # reach the EOF return i buf[i] = buf4[i4] i += 1 i4 += 1 return i