# Solution For Read N Characters Given Read4 Ii Call Multiple Times

Problem Statement:

``````int read4(char *buf4);

buf4: A temporary buffer to store the result.
Returns the number of actual characters read.
``````

Implement the read function as follows:

``````int read(char *buf, int n);

buf: The buffer that will store the results.
n: The number of characters to read.
``````

The read function may be called multiple times.

Example 1:

``````Input: file = "abc", queries = [1,2,1]
Output: [1,2,1]
Explanation: The first query reads 1 character using read4, so buf = "a". The second query reads 2 characters using read4, so buf = "bc". The third query reads 1 character using read4, so buf = "".
``````

Example 2:

``````Input: file = "abcde", queries = [5]
Output: [5]
Explanation: The only query reads 5 characters using read4, so buf = "abcde".
``````

Solution:

In this problem, we have to read the given number of characters from the file. The file is represented by read4 API which reads the next 4 characters from the file and returns the actual number of characters read. The read function must be called multiple times to read all the characters from the file. We have to read the given number of characters in each call to the read function.

Algorithm:

1. Define a class named Solution which will contain the implementation of read function.
2. Use two variables to keep track of the number of characters read and the number of characters stored in the buffer.
3. Define a buffer of size 4 which will store the characters read from the file using read4 API.
4. Define a queue or any other data structure to store the characters that are read but not yet consumed by the user.
5. In the read function, first check whether there are any characters in the queue. If yes, consume them and return the requested number of characters.
6. If there are no characters in the queue, call the read4 API to read the next chunk of characters from the file.
8. If the number of characters read from the file is equal to the requested number of characters, consume them and return.
9. If the number of characters read from the file is less than the requested number of characters, repeat steps 6 to 8.
10. If all the characters in the file are read and there are still some characters in the queue, consume them and return.

Time Complexity:

The time complexity of this algorithm is O(n) where n is the total number of characters in the file. This is because we need to read every character in the file at least once.

Space Complexity:

The space complexity of this algorithm is O(1) because we are using a fixed-size buffer of size 4 and a queue to store the read but not yet consumed characters. The size of the queue will not exceed 4 at any point in time.

Code Implementation:

## Step by Step Implementation For Read N Characters Given Read4 Ii Call Multiple Times

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

char[] buffer = new char[4];
int offset = 0, bufsize = 0;

public int read(char[] buf, int n) {
boolean eof = false;
while (!eof && readBytes < n) {
/* if buffer pointer reaches the end of current buffer, we need to read
more chars from file to fill the buffer */
if (bufsize == 0) {
eof = bufsize < 4;
}
/* copy chars from buffer to buf */
int bytes = Math.min(n - readBytes, bufsize);
System.arraycopy(buffer /* src */, offset /* srcPos */,
buf /* dest */, readBytes /* destPos */, bytes /* length */);
offset = (offset + bytes) % 4;
bufsize -= bytes;
}
}
}```
```# The read4 II API is already defined for you.
# @param buf, a list of characters
# @return an integer

class Solution:
def __init__(self):
self.queue = []

"""
:type buf: Destination buffer (List[str])
:type n: Maximum number of characters to read (int)
:rtype: The number of characters read (int)
"""
idx = 0
while True:
self.queue = self.queue[curr:]
for i in range(curr):
if idx == n:
return idx
buf[idx] = self.queue.pop(0)
idx += 1
if curr < 4:
return idx```
```/**
*/
var Solution = function() {
// TODO: Implement the read function.

};```
```The following is a C++ solution for the leetcode problem read-n-characters-given-read4-ii-call-multiple-times:

/*

Given a file and an integer n, read n characters from the file using the read4 II API.

Note:

The read4 II API is only available to you if you are using the C++11 compiler or later.

Example:

File file("abc");

Solution sol;

// Assume buf is allocated and guaranteed to have enough space for storing n characters from the file.

char* buf = new char[n];

*/

class Solution {
public:
/**
* @param buf Destination buffer
* @param n   Maximum number of characters to read
* @return    The number of characters read
*/
int read(char *buf, int n) {

// If reach the end of file, break out of the loop.
break;
}

// Copy characters from read4 buffer to buf.
for (int i = 0; i < read4_len && read_len < n; ++i) {
}
}

}
};```
```public class Solution {
/**
* @param buf Destination buffer
* @param n   Maximum number of characters to read
* @return    The number of characters read
*/

char[] buffer = new char[4];
// index for buffer
int bufferIndex = 0;
int bufferSize = 0;

public int Read(char[] buf, int n) {
// keep track of number of characters read

// while we haven't read the desired number of characters
// if the buffer is empty, read more characters into it
if (bufferIndex == 0) {
}
// if bufferSize is 0, then we have reached the end of the file
if (bufferSize == 0) {
break;
}
// while there are still characters in the buffer
// and we haven't read the desired number of characters
while (bufferIndex < bufferSize && readSize < n) {
// copy character from buffer to output array
// increment pointers
bufferIndex++;
}
// if we have reached the end of the buffer, reset index
if (bufferIndex == bufferSize) {
bufferIndex = 0;
}
}
// return number of characters read