Given two integer vectors A and B, we have to pick one element from each vector such that their xor is maximum and we need to return this maximum xor value from the function int Solution::solve(vector &A, vector &B).
I found out that the code below is not passing all the test cases when I'm declaring and initializing the head pointer globally right beneath the class Trienode. Why is that?
Code
class Trienode
{
public:
Trienode* left;
Trienode* right;
Trienode()
{
left=0;
right=0;
}
};
// Trienode* head = new Trienode;
int Max_Xor_Pair(Trienode* head, vector<int> B)
{
int n=B.size();
int max_xor=INT_MIN;
for(int i=0; i<n; i++)
{
int pair1 = B[i];
int pair2 = 0;
Trienode* curr=head;
for(int j=31; j>=0; j--)
{
int bit = (pair1>>j)&1;
if(bit)
{
if(curr->left)
curr=curr->left;
else
{
curr=curr->right;
pair2 += pow(2,j);
}
}
else
{
if(curr->right)
{
curr=curr->right;
pair2 += pow(2,j);
}
else
curr=curr->left;
}
}
int curr_xor = pair1 ^ pair2;
max_xor = max(max_xor, curr_xor);
}
return max_xor;
}
void Insert(Trienode* head, int num)
{
Trienode* curr=head;
for(int i=31; i>=0; i--)
{
int x = num;
int bit= (x>>i)&1;
if(bit)
{
if(!curr->right)
{
Trienode* temp = new Trienode;
curr->right=temp;
}
curr=curr->right;
}
else
{
if(!curr->left)
{
Trienode* temp = new Trienode;
curr->left=temp;
}
curr=curr->left;
}
}
}
int Solution::solve(vector<int> &A, vector<int> &B) {
Trienode* head = new Trienode;
for(int x:A)
Insert(head,x);
return Max_Xor_Pair(head,B);
}
Sample Input
A : [ 15891, 6730, 24371, 15351, 15007, 31102, 24394, 3549, 19630, 12624, 24085, 19955, 18757, 11841, 4967, 7377, 13932, 26309, 16945, 32440, 24627, 11324, 5538, 21539, 16119, 2083, 22930, 16542, 4834, 31116, 4640, 29659, 22705, 9931, 13978, 2307, 31674, 22387, 5022, 28746, 26925, 19073, 6271, 5830, 26778, 15574, 5098, 16513, 23987, 13291, 9162 ]
B : [ 18637, 22356, 24768, 23656, 15575, 4032, 12053, 27351, 1151, 16942 ]
When head is a global variable, and you don't have this line in Solution::solve:
Trienode* head = new Trienode;
...then head will retain its value after the first test case has finished, and so the second test case will not start with an empty tree. Each test case will add more nodes to the one tree. Of course this means that, except for the first test case, the tree rooted by head is not the intended tree.
To make the version with the global variable work, reset it in Solution::solve:
head->left = head->right = nullptr;
BTW, you should also initialise these members with nullptr (instead of 0) in your TrieNode constructor. This better reflects the intent.
You can also go with this approach:
Code:
struct Node {
Node* left;
Node* right;
};
class MaxXorHelper{
private : Node* root;
public :
MaxXorHelper() {
root = new Node();
}
void addElements(vector<int> &arr) {
for(int i=0; i<arr.size(); i++) {
Node* node = root;
int val = arr[i];
for(int j=31; j>=0; j--) {
int bit = (val >> j) & 1;
if(bit == 0) {
if(!node->left) {
node->left = new Node();
}
node = node->left;
}
else {
if(!node->right) {
node->right = new Node();
}
node = node->right;
}
}
}
}
int findMaxXor(vector<int> &arr) {
int maxXor = INT_MIN;
for(int i=0; i<arr.size(); i++) {
Node* node = root;
int val2 = 0;
int val1 = arr[i];
for(int j=31; j>=0; j--) {
int bit = (val1 >> j) & 1;
if(bit == 0) {
if(node->right) {
val2 |= (1 << j);
node = node->right;
} else{
node = node->left;
}
}
else {
if(node->left) {
node = node->left;
} else{
val2 |= (1 << j);
node = node->right;
}
}
}
int curXor = val1 ^ val2;
maxXor = max(maxXor, curXor);
}
return maxXor;
}
};
int Solution::solve(vector<int> &A, vector<int> &B) {
MaxXorHelper helper;
helper.addElements(A);
return helper.findMaxXor(B);
}
Related
I use nearly the same method as in the discussion. But mine reaches the time limitation but his passes all cases. I want to know how to improve my code and why there is difference?
Here is my entire code:
boolean[] visited;
public int dfs(int step, int[] edges, int node, Map<Integer, Integer> path) {
path.put(node, step);
visited[node] = true;
if (edges[node] == -1) {
return -1;
}
if (path.containsKey(edges[node])) {
return step - path.get(edges[node]) + 1;
}
return dfs(step + 1, edges, edges[node], path);
}
public int longestCycle(int[] edges) {
int res = -1;
visited = new boolean[edges.length];
for (int i = 0; i < edges.length; i++) {
if (visited[i]) {
continue;
}
int maxCircleLength = dfs(0, edges, i, new HashMap<Integer, Integer>());
res = Math.max(maxCircleLength, res);
}
return res;
}
This is his solution:
public int longestCycle(int[] edges) {
int longest = -1;
boolean visited[] = new boolean[edges.length]; // global visisted
HashMap<Integer, Integer> map;
for(int i=0; i<edges.length; i++){
if(visited[i]) continue;
int distance = 0, curr_node = i;
map = new HashMap<>(); // local visited
while(curr_node != -1){
if(map.containsKey(curr_node)){
longest = Math.max(longest, distance - map.get(curr_node));
break;
}
if(visited[curr_node]) break;
visited[curr_node] = true;
map.put(curr_node, distance);
curr_node = edges[curr_node];
distance++;
}
}
return longest;
}
I tried this problem by backtracking and did some optimization, though I am getting TLE. what further optimization can I do on this code?
Abridged problem statement - Task is to print all different r combinations of a string s (a r combination of a string s is a collection of exactly r letters from different positions in s).There may be different permutations of the same combination; consider only the one that has its r
characters in non-decreasing order. If s = "abaa" and s = 3.Then output should be (aaa,aab).
My code(in c)
int compare_chars(const void* a, const void* b);
char s[50];
int len;
int r ;
char combination[50];
void combinate(int index,int at)
{
if(at == r)
{
combination[at] = '\0';
puts(combination);
return ;
}
int i = index+1;
for ( ; i <= len-r+at ;)
{
char temp = s[I];
combination[at] = temp;
combinate(i,at+1);
while(s[i] == temp and i <= len-r+at)
i++;
}
return ;
}
int solve()
{
while ((scanf("%s %i",s,&r)) == 2)
{
len = strlen(s);
if(len == r)
{
printf("%s\n",s);
continue;
}
qsort(s,len,sizeof(char),compare_chars);
combinate(-1,0);
}
return 0;
}
int main()
{
int a = 1;
int t = 1;
while (a <= t)
{
int kk = solve();
}
return 0;
}
int compare_chars(const void* a, const void* b)
{
char arg1 = *(const char*)a;
char arg2 = *(const char*)b;
if (arg1 < arg2) return -1;
if (arg1 > arg2) return 1;
return 0;
}
I got a homework where I needed to wright a program to find MST of a graph. I tried running it on the school server but I get a run-time error. On big servers like HackerEarth or HackerRank, however, I got correct answers on all test cases. The boundary for the number of edges and vertices is 100000 and for the weight of an edge 10000. Vertices are labeled from 0 to n.
Here is my code:
#include <stdio.h>
#include <algorithm>
using namespace std;
class Edge
{
public:
int x, y;
long long w;
bool operator<(const Edge& next) const;
};
bool Edge::operator<(const Edge& next) const
{
return this->w < next.w;
}
class DisjointSet
{
public:
int parent, rank;
};
DisjointSet set[100100];
Edge edges[100100];
int findWithPathCompression(int x)
{
if(set[x].parent != x)
set[x].parent = findWithPathCompression(set[x].parent);
return set[x].parent;
}
bool unionByRank(int x1, int y1)
{
int x = findWithPathCompression(x1);
int y = findWithPathCompression(y1);
if(x == y)
return false;
if(set[x].rank > set[y].rank)
set[y].parent = x;
else if(set[y].rank > set[x].rank)
set[x].parent = y;
else
{
set[y].parent = x;
set[x].rank++;
}
return true;
}
int main()
{
int n, m, e = 0, c = 0;
long long r = 0;
scanf("%d %d",&n,&m);
for(int i = 0; i <= n; i++)
{
set[i].parent = i;
set[i].rank = 1;
}
for(int i = 0; i < m; i++)
{
scanf("%d %d %lld",&edges[i].x,&edges[i].y,&edges[i].w);
edges[i].x;
edges[i].y;
}
sort(edges,edges + m);
while(e != n - 1)
{
if(unionByRank(edges[c].x,edges[c].y))
{
r += edges[c].w;
e++;
}
c++;
}
printf("%lld\n",r);
}
I have data from a camera in mono 8bit.
This is converted into an int vector using
std::vector<int> grayVector(size);
// convert / copy pointer data into vector: 8 bit
if (static_cast<XI_IMG_FORMAT>(format) == XI_MONO8)
{
quint8* imageIterator = reinterpret_cast<quint8*> (pMemVoid);
for (size_t count = 0; count < size; ++count)
{
grayVector[count] = static_cast<int>(*imageIterator);
imageIterator++;
}
}
Next, I need to convert this into a QImage. If I set the image format to QImage::Format_Mono the app crashes. With QImage::Format_RGB16 I get strippes, and with QImage::Format_RGB32 everything is black.
I would like to know how to do this the best, efficient and correct way?
// convert gray values into QImage data
QImage image = QImage(static_cast<int>(sizeX), static_cat<int>(sizeY), QImage::Format_RGB16);
for ( int y = 0; y < sizeY; ++y )
{
int yoffset = sizeY*y;
QRgb *line = reinterpret_cast<QRgb *>(image.scanLine(y)) ;
for ( int x = 0; x < sizeX ; ++x )
{
int pos = x + yoffset;
int color = grayVector[static_cast<size_t>(pos)];
*line++ = qRgb(color, color, color);
}
}
The conversion to int is unnecessary and you do it in a very inefficient way; all you need is to use the QImage::Format_Grayscale8 available since Qt 5.5 (mid-2015).
Anyway, what you really want is a way to go from XI_IMG to QImage. The default BP_UNSAFE buffering policy should be adequate - the QImage will do a format conversion, so taking the data from XiApi's internal buffer is OK. Thus the following - all of the conversions are implemented in Qt and are quite efficient - much better than most any naive code.
I didn't check whether some Xi formats may need a BGR swap. If so, then the swap can be set to true in the format selection code and the rest will happen automatically.
See also: xiAPI manual.
static QVector<QRgb> grayScaleColorTable() {
static QVector<QRgb> table;
if (table.isEmpty()) {
table.resize(256);
auto *data = table.data();
for (int i = 0; i < table.size(); ++i)
data[i] = qRgb(i, i, i);
}
return table;
}
constexpr QImage::Format grayScaleFormat() {
return (QT_VERSION >= QT_VERSION_CHECK(5,5,0))
? QImage::Format_Grayscale8
: QImage::Format_Indexed8;
}
QImage convertToImage(const XI_IMG *src, QImage::Format f) {
Q_ASSERT(src->fmt == XI_MONO16);
Q_ASSERT((src->padding_x % 2) == 0);
if (src->fmt != XI_MONO16) return {};
const quint16 *s = static_cast<const quint16*>(src->bp);
const int s_pad = src->padding_x/2;
if (f == QImage::Format_BGR30 ||
f == QImage::Format_A2BGR30_Premultiplied ||
f == QImage::Format_RGB30 ||
f == QImage::Format_A2RGB30_Premultiplied)
{
QImage ret{src->width, src->height, f};
Q_ASSERT((ret->bytesPerLine() % 4) == 0);
const int d_pad = ret->bytesPerLine()/4 - ret->width();
quint32 *d = (quint32*)ret.bits();
if (s_pad == d_pad) {
const int N = (src->width + s_pad) * src->height - s_pad;
for (int i = 0; i < N; ++i) {
quint32 const v = (*s++) >> (16-10);
*d++ = 0xC0000000 | v << 20 | v << 10 | v;
}
} else {
for (int j = 0; j < src->height; ++j) {
for (int i = 0; i < src->width; ++i) {
quint32 const v = (*s++) >> (16-10);
*d++ = 0xC0000000u | v << 20 | v << 10 | v;
}
s += s_pad;
d += d_pad;
}
}
return ret;
}
QImage ret{src->width, src->height, grayScaleFormat()};
const int d_pad = ret->bytesPerLine() - ret->width();
auto *d = ret.bits();
if (s_pad == d_pad) {
const int N = (src->width + s_pad) * src->height - s_pad;
for (int i = 0; i < N; ++i) {
*d++ = (*s++) >> 8;
} else {
for (int j = 0; j < src->height; ++j) {
for (int i = 0; i < src->width; ++i)
*d++ = (*s++) >> 8;
s += s_pad;
d += d_pad;
}
}
return ret;
}
QImage fromXiImg(const XI_IMG *src, QImage::Format dstFormat = QImage::Format_ARGB32Premultiplied) {
Q_ASSERT(src->width > 0 && src->height > 0 && src->padding_x >= 0 && src->bp_size > 0);
Q_ASSERT(dstFormat != QImage::Format_Invalid);
bool swap = false;
int srcPixelBytes = 0;
bool externalConvert = false;
QImage::Format srcFormat = QImage::Format_Invalid;
switch (src->fmt) {
case XI_MONO8:
srcPixelBytes = 1;
srcFormat = grayScaleFormat();
break;
case XI_MONO16:
srcPixelBytes = 2;
externalConvert = true;
break;
case XI_RGB24:
srcPixelBytes = 3;
srcFormat = QImage::Format_RGB888;
break;
case XI_RGB32:
srcPixelBytes = 4;
srcFormat = QImage::Format_RGB32;
break;
};
if (srcFormat == QImage::Format_Invalid && !externalConvert) {
qWarning("Unhandled XI_IMG image format");
return {};
}
Q_ASSERT(srcPixelBytes > 0 && srcPixelBytes <= 4);
int bytesPerLine = src->width * srcPixelBytes + src->padding_x;
if ((bytesPerLine * src->height - src->padding_x) > src->bp_size) {
qWarning("Inconsistent XI_IMG data");
return {};
}
QImage ret;
if (!externalConvert)
ret = QImage{static_cast<const uchar*>(src->bp), src->width, src->height,
bytesPerLine, srcFormat};
else
ret = convertToImage(src, dstFormat);
if (ret.format() == QImage::Format_Indexed8)
ret.setColorTable(grayScaleColorTable());
if (ret.format() != dstFormat)
ret = std::move(ret).convertToFormat(dstFormat);
if (swap)
ret = std::move(ret).rgbSwapped();
if (!ret.isDetached()) // ensure that we don't share XI_IMG's data buffer
ret.detach();
return ret;
}
I want to print value in an array (array name Array_index) which I write of variable name bit_1. But when I tried to enter inside this array, the value always show zero. Please advise me how to do a correct way in Arduino.
This is my code:
int test_number = 0;
unsigned int Array_index[] = {};
int bit_1 = 0;
int Andbit = 0;
int arrSize = 0;
void setup()
{
Serial.begin( 9600 );
}
void loop()
{
int count = 0;
test_number = random(10);
Serial.println(test_number);
for (bit_1 = 0; bit_1 <= 15; bit_1++)
{
Andbit = test_number & 1;
if (Andbit == 1)
{
Array_index[count] = bit_1;
//Serial.println(Array_index[count]);
count=count++;
}
else
{
}
test_number = test_number >> 1;
int arrSize = sizeof(Array_index) / sizeof( int );
Serial.println(arrSize);
for (int y = 0; y < arrSize; y++)
{
Serial.println(Array_index[y]);
}
}
while(1)
{
}
}
unsigned int Array_index[] = {}; is an array with 0 elements.
You should define it as unsigned int Array_index[16];.