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FAQ on the xz-utils backdoor

Background

On March 29th, 2024, a backdoor was discovered in xz-utils, a suite of software that gives developers lossless compression. This package is commonly used for compressing release tarballs, software packages, kernel images, and initramfs images. It is very widely distributed, statistically your average Linux or macOS system will have it installed for

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const char fragmentShader_yuv420p[] =
{
    "precision mediump float;\n"
    "uniform sampler2D Ytex;\n"
    "uniform sampler2D Utex,Vtex;\n"
    "varying vec2 vTextureCoord;\n"
    "void main(void) {\n"
    " float nx,ny,r,g,b,y,u,v;\n"
    " mediump vec4 txl,ux,vx;"
    " nx=vTextureCoord[0];\n"

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class EnumExample {
    private enum Status {
        Init, Success, Failed, Loading
    }
    
    public static enum ProcessStatus {
        LOAD_STAGE(Status.Init),
        
        END_STAGE(Status.Failed);
        

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public static int getMaxCodecInstanceByName(String name) {
    final Vector<MediaCodec> codecs = new Vector<>();
    final MediaFormat format = MediaFormat.createVideoFormat(MIME_TYPE, 1920, 1080);
    MediaCodec codec = null;

    for (int i = 0; i < max; i++) {
        try {
            codec = MediaCodec.createDecoderByType(MediaFormat.MIMETYPE_VIDEO_AVC);
            codec.configure(format, null, null, 0);
            codec.start();

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/*
You are climbing a stair case. It takes n steps to reach to the top.
Each time you can either climb 1 or 2 steps. In how many distinct ways can you climb to the top?
*/
public class Solution { 
    public int climbStairs(int n) {
        if(n == 0 || n == 1) return 1; 
        int finalWays = 0; 
        int f1 = 1;
        int f2 = 1;

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/*
Give a string, and a dictionary, find out if this string could be segmented by a space which's substring 
are in this dictionary. 

For example, String "leetcode", dictionary ["leet", "code"], 
then return true, cause this string could be segmented as "leet code";
*/

public class Solution {
    public boolean wordBreak(String str, Set<String> dict) {

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/*
  Implement an algorithm to detemine if a string has all unique characters. 
  What if you cannot use additional data structure.
*/

// solution 1, with an array, time complexity = O(n), space complexity = O(1)
public class  solution  {
    public boolean isUniqueChar(String str) { 
        if(str == null || str.length() == 0)
            return false;

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private fun hopAnimation() {
        val animation = bottomContainer
            .animate()
            .translationYBy(-40f)
            .alpha(0.9f)
            .setDuration(400)
        animation.withEndAction {
            bottomContainer
                .animate()
                .alpha(0.5f)

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// Generate data
int arraySize = 32768;
int data[] = new int[arraySize];

Random rnd = new Random(0);
for (int c = 0; c < arraySize; ++c)
    data[c] = rnd.nextInt() % 256;

// notice !!!
Arrays.sort(data);

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public class PhotoMultipartRequest<T> extends Request<T> {


private static final String FILE_PART_NAME = "file";

    private MultipartEntityBuilder mBuilder = MultipartEntityBuilder.create();
    private final Response.Listener<T> mListener;
    private final File mImageFile;
    protected Map<String, String> headers;