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DrawLine.java
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DrawLine.java
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package com.ctci.bitmanipulation;
import java.util.Arrays;
/**
* @author rampatra
* @since 2019-03-21
*/
public class DrawLine {
/**
* A monochrome screen is stored as a single array of bytes, allowing eight consecutive pixels to be stored
* in one byte. The screen has width w, where w is divisible by 8 (that is, no byte will be split across rows).
* The height of the screen, of course, can be derived from the length of the array and the width. Implement a
* function that draws a horizontal line from (xl, y) to ( x2, y).
* <p>
* The method signature should look something like:
* {@code drawline(byte[] screen, int width, int xl, int x2, int y)}
* <p>
* Approach:
* First, find the numbers in which all bits has to be set. Next, find the starting number and apply the mask
* created from the starting offset. Do the same with the ending number.
*
* @param screen
* @param width
* @param x1
* @param x2
* @param y
*/
private static void drawLine(byte[] screen, int width, int x1, int x2, int y) {
int startOffset = x1 % 8;
int startFullByte = x1 / 8;
if (startOffset != 0) {
startFullByte++;
}
int endOffset = x2 % 8;
int endFullByte = x2 / 8;
if (endOffset != 7) {
endFullByte--;
}
// all bits have to be set in in-between numbers
for (int i = startFullByte; i <= endFullByte; i++) {
screen[width / 8 * y + i] |= (byte) 0xff;
}
/* 0xff is an integer literal which is like 000...11111111 (32 bits) but when we
cast it to a byte, we get rid of the initial 24 bits */
byte startByteMask = (byte) (0xff >> startOffset);
byte endByteMask = (byte) ~(0xff >> endOffset + 1);
if (x1 / 8 == x2 / 8) { // if starting and ending both lie in the same byte
screen[width / 8 * y + (x1 / 8)] |= (startByteMask & endByteMask);
} else {
screen[width / 8 * y + (startFullByte - 1)] |= startByteMask; // only specific bits set in the starting number
screen[width / 8 * y + (endFullByte + 1)] |= endByteMask; // only specific bits set in the ending number
}
}
public static void main(String[] args) {
/*
Consider the below screen with width 32 as an example:
byte[] screen = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12};
This screen has a width of 32 so you can assume the screen would be looking like:
9 10 11 12
5 6 7 8
1 2 3 4
x-axis is 5-20 (5th position to 20th position)
y-axis is 1
which means our line would lie in numbers 5, 6, and 7
so if you visualize these numbers in bits, it would be like:
00000101 00000110 00000111
^ ^
and after drawing the line, the bits would become:
00000111 11111111 11111111
and in the output we would see:
7, -1, -1 instead of 5, 6, 7
*/
byte[] screen = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12};
System.out.println("Input: " + Arrays.toString(screen));
drawLine(screen, 32, 5, 20, 1);
System.out.println("Output: " + Arrays.toString(screen));
System.out.println("---");
screen = new byte[]{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12};
System.out.println("Input: " + Arrays.toString(screen));
drawLine(screen, 32, 0, 5, 1);
System.out.println("Output: " + Arrays.toString(screen));
System.out.println("---");
screen = new byte[]{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12};
System.out.println("Input: " + Arrays.toString(screen));
drawLine(screen, 32, 3, 7, 1);
System.out.println("Output: " + Arrays.toString(screen));
System.out.println("---");
screen = new byte[]{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12};
System.out.println("Input: " + Arrays.toString(screen));
drawLine(screen, 16, 0, 7, 0);
System.out.println("Output: " + Arrays.toString(screen));
}
}