Read N Characters Given read4 II – Call Multiple Times LeetCode Programming Solutions | LeetCode Problem Solutions in C++, Java, & Python [💯Correct]

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In this post, you will find the solution for the Read N Characters Given read4 II – Call Multiple Times in C++, Java & Python-LeetCode problem. We are providing the correct and tested solutions to coding problems present on LeetCode. If you are not able to solve any problem, then you can take help from our Blog/website.

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Link for the ProblemRead N Characters Given read4 II – Call Multiple Times– LeetCode Problem

Read N Characters Given read4 II - Call Multiple Times– LeetCode Problem

Problem:

The API:int read4(char *buf)reads 4 characters at a time from a file.

The return value is the actual number of characters read. For example, it returns 3 if there is only 3 characters left in the file.

By using theread4API, implement the functionint read(char *buf, int n)that reads n characters from the file.

Note:
Thereadfunction may be called multiple times.

Example 1:

Given buf = "abc"
read("abc", 1) // returns "a"
read("abc", 2); // returns "bc"
read("abc", 1); // returns ""

Example 2:

Given buf = "abc"
read("abc", 4) // returns "abc"
read("abc", 1); // returns ""
Read N Characters Given read4 II - Call Multiple Times– LeetCode Solutions
Read N Characters Given read4 II - Call Multiple Times 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) {
    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;
  }

 private:
  char* buf4 = new char[4];
  int i4 = 0;  // buf4's index
  int n4 = 0;  // buf4's size
};
Read N Characters Given read4 II - Call Multiple Times 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) {
    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;
  }

  private char[] buf4 = new char[4];
  private int i4 = 0; // buf4's index
  private int n4 = 0; // buf4's size
}
Read N Characters Given read4 II - Call Multiple Times Solution in Python:
# The read4 API is already defined for you.
# def read4(buf4: List[str]) -> int:

class Solution:
  def read(self, buf: List[str], n: int) -> int:
    i = 0  # buf's index

    while i < n:
      if self.i4 == self.n4:  # all characters in buf4 are consumed
        self.i4 = 0  # reset buf4's index
        self.n4 = read4(self.buf4)  # read 4 (or less) chars from file to buf4
        if self.n4 == 0:  # reach the EOF
          return i
      buf[i] = self.buf4[self.i4]
      i += 1
      self.i4 += 1

    return i

  buf4 = [' '] * 4
  i4 = 0  # buf4's index
  n4 = 0  # buf4's size

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