CPMagical CP Library

Problem solution · Java

Read N Characters Given Read4 II - Call multiple times

Read N Characters Given Read4 II - Call multiple times: a Java solution using direct simulation. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Direct simulation
Source
walkccc LeetCode Solutions
Length
32 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Direct simulation

For Read N Characters Given Read4 II - Call multiple times, the implementation follows the problem’s operations directly while maintaining only the state needed for the next decision.

  1. Translate each rule into one explicit state update.
  2. Maintain the invariant after every processed item.
  3. Return the accumulated state once all relevant input has been handled.

Code notes

  • 32 lines of Java from the credited upstream file 158.java.
  • The implementation visibly relies on sequence storage.
  • 1 loop block detected.

Complexity

Count the number and nesting of passes over the input, then include the maintained containers in the memory estimate.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeRead N Characters Given Read4 II - Call multiple times · JavaJava
Use this to learn the idea, then write your own version.
/** * 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 the characters in the buf4 are consumed.        i4 = 0;           // Reset the buf4's index.        n4 = read4(buf4); // Read <= 4 characters from the file to the 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}

Keep practising

Did this explanation save you time? I'm a Grade 11 student building this free library to make difficult algorithms easier to understand.

Buy me a coffee ↗