I was solving Dungeon Game on LeetCode. While I was able to come up with the recurrence relation (and determine that it was a DP question in the first place), what is the intuition behind starting from the bottom right, as shown in a highly upvoted solution:
class Solution {
public:
int calculateMinimumHP(vector<vector<int> > &dungeon) {
int M = dungeon.size();
int N = dungeon[0].size();
// hp[i][j] represents the min hp needed at position (i, j)
// Add dummy row and column at bottom and right side
vector<vector<int> > hp(M + 1, vector<int>(N + 1, INT_MAX));
hp[M][N - 1] = 1;
hp[M - 1][N] = 1;
for (int i = M - 1; i >= 0; i--) {
for (int j = N - 1; j >= 0; j--) {
int need = min(hp[i + 1][j], hp[i][j + 1]) - dungeon[i][j];
hp[i][j] = need <= 0 ? 1 : need;
}
}
return hp[0][0];
}
};
I thought it to be pretty similar to the Unique Paths question, and hence started solving from the top left (but it doesn't retrieve correct results). My solution (if at all required) is:
class Solution {
public int calculateMinimumHP(int[][] dungeon) {
if(dungeon==null || dungeon.length==0 || dungeon[0]==null || dungeon[0].length==0)
return -1;
int maxVal = 0;
int[][] dp = new int[dungeon.length][dungeon[0].length];
dp[0][0] = dungeon[0][0];
for(int i=1; i<dungeon.length; i++)
dp[i][0] = dp[i-1][0] + dungeon[i][0];
for(int i=1; i<dungeon[0].length; i++)
dp[0][i] = dp[0][i-1] + dungeon[0][i];
int chosenOne=0;
for(int i=1; i<dungeon.length; i++) {
for(int j=1; j<dungeon[0].length; j++) {
if(dp[i-1][j] < 0 && dp[i][j-1] < 0) {
chosenOne = Math.max(dp[i-1][j], dp[i][j-1]);
}
else {
if(dp[i-1][j] > 0 && dp[i][j-1] > 0) {
chosenOne = Math.min(dp[i-1][j], dp[i][j-1]);
}
else {
chosenOne = (dp[i-1][j]>0?dp[i-1][j]:dp[i][j-1]);
}
}
dp[i][j] = dungeon[i][j] + chosenOne;
maxVal = Math.min(maxVal, dp[i][j]);
}
}
return maxVal+1;
}
}
One of the comments here by 'JaiMataDi' mentions that it is so because we don't know the initial health point of the knight. But how about the approach that I used?
To be precise, could some one please point out the intuition behind why we start at the bottom right and not the top left?
Thanks!
So I've traced the segfault to the line, but I don't understand why this is a segfault. Can someone elaborate on the error of my ways?
Here are the variable declarations.
size_t i, j, n, m, chunk_size, pixel_size;
i = j = n = m = 0;
chunk_size = 256;
pixel_size = 4;
Here are the array declarations.
uint8_t** values = new uint8_t*[chunk_size];
for (i = 0; i < chunk_size; ++ i)
values[i] = new uint8_t[chunk_size];
float** a1 = new float*[chunk_size];
for (i = 0; i < chunk_size; ++i)
a1[i] = new float[chunk_size];
And here is where the segfault occurs.
float delta, d;
for (i = 0; i < 256; ++i) {
for (j = n = m = d = 0; j < 256; j = m) {
while (i == 0 || d != 0) {
d = a1[i][m]; <------SEGFAULT per GDB
++m;
}
delta = (d - a1[i][j]) / m;
n = j + 1;
while (n < j + m) {
a1[i][n] = a1[i][n - 1] + delta;
++n;
}
}
}
I'm fairly new to C++ and can't figure out why this would be a segfault. Is this not the proper way to set a variables value to a variable in an array? Is that the source of my segfault?
Note: The point of this whole thing is too expand a 4x4 array to a 256x256 array with my simpleton interpolation formula.
while (i == 0 || d != 0) {
d = a1[i][m]; <------SEGFAULT per GDB
++m;
}
This is an endless while loop, in some cases (e.g. in the first iteration of the outer loop).
Your outer loop starts out with i = 0 and the inner loops starts with d = 0 and the logic controlling the while loop is not sufficient (see code comment).
for (i = 0; i < 256; ++i) {
for (j = n = m = d = 0; j < 256; j = m) {
// Here i == 0 is ALWAYS true (so d != 0 is ignored due to
// short-circuit evaluation) and then 'm' is continuously incremented
// until it goes out of bounds
while (i == 0 || d != 0) {
d = a1[i][m]; <------SEGFAULT per GDB
++m;
}
delta = (d - a1[i][j]) / m;
n = j + 1;
while (n < j + m) {
a1[i][n] = a1[i][n - 1] + delta;
++n;
}
}
The problem is in the following lines :
while (i == 0 || d != 0) {
d = a1[i][m]; <------SEGFAULT per GDB
++m;
}
Your while loop will keep on going while i equals 0. Since you never increment i in your while loop, m keeps on incrementing forever until arriving out of bounds, causing the segfault issue that you are having.
Make sure you check the values of i and m, so that they are in the allocated memory range and your code will work.
To represent all the cells I'm using a variable std::vector<bool> cells to represent the cells where true represents a live cell. To update the canvas, I have a variable std::vector<int> neighborCounts, where neighborCounts[i] represents how many live neighbors cell[i] has. In the update loop, I loop through all the cells, and if its a live cell, I add 1 to the neighborCounts of all the adjacent cells. Then after determining all of the neighbors, I once again loop through all the cells and perform the updates based on the number of neighbors it has.
std::vector<int> neighborCounts(NUM_ROWS * ROW_SIZE, 0);
for (int x = 0; x < ROW_SIZE; x++)
{
for (int y = 0; y < NUM_ROWS; y++)
{
if (cells[convertCoord(x, y)])
{
for (int m = x - 1; m <= x + 1; m++)
{
for (int n = y - 1; n <= y + 1; n++)
{
if (!(m == x && n == y) && m >= 0 && m < ROW_SIZE && n >= 0 && n < NUM_ROWS)
{
neighborCounts[convertCoord(m, n)] += 1;
}
}
}
}
}
}
for (int x = 0; x < ROW_SIZE; x++)
{
for (int y = 0; y < NUM_ROWS; y++)
{
int coord = convertCoord(x, y);
int numNeighbors = neighborCounts[coord];
if (cells[coord])
{
if (numNeighbors < 2)
{
cells[coord] = false;
}
else if (numNeighbors > 3)
{
cells[coord] = false;
}
}
else
{
if (numNeighbors == 3)
{
cells[coord] = true;
}
}
}
}
Then in the render function, I loop through all the cells and if its alive I draw it to the screen.
This seems to be the most straightforward way to do this, but there's definitely room for further optimization. How do powder toys and other apps manage a huge array of particles and yet still manage to run amazingly smooth? What is the most efficient way to manage all the cells in Conway's Game of Life?
Integer Range = 1;
for(Integer k = -Range; k <= Range; ++k)
{
for(Integer j = -Range; j <= Range; ++j)
{
for(Integer i = -Range; i <= Range; ++i)
{
Integer MCID = GetCellID(&CONSTANT_BOUNDINGBOX,CIDX +i, CIDY + j,CIDZ
+ k);
if(MCID < 0 || MCID >= c_CellNum)
{
continue;
}
unsigned int TriangleNum = c_daCell[MCID].m_TriangleNum;
for(unsigned int l = 0; l < TriangleNum; ++l)
{
TriangleID=c_daCell[MCID].m_TriangleID[l];
if( TriangleID >= 0 && TriangleID < c_TriangleNum && TriangleID
!= NearestID)// No need to calculate again for the same triangle
{
CDistance Distance ;
Distance.Magnitude = CalcDistance(&c_daTriangles[TriangleID], &TargetPosition,
&Distance.Direction);
if(Distance.Magnitude < NearestDistance.Magnitude)
{
NearestDistance = Distance;
NearestID = TriangleID;
}
}
}
}
}
}
}
c_daSTLDistance[ID] = NearestDistance;
c_daSTLID[ID] = NearestID;
GetCellID is the function to return the cellid in the variable CID with CIDX,CIDY,CIDZ with its position in the 3 axes
here the above code is a function to calculate the distance ,actually STL distance between a point and the triangles of the stl. This code runs fine however the problem is it is too slow as it has large number of loops within the code. Now my concern is to optimize the loop. Is there any technique of optimizing the loops within the code?
I need to place numbers within a grid such that it doesn't collide with each other. This number placement should be random and can be horizontal or vertical. The numbers basically indicate the locations of the ships. So the points for the ships should be together and need to be random and should not collide.
I have tried it:
int main()
{
srand(time(NULL));
int Grid[64];
int battleShips;
bool battleShipFilled;
for(int i = 0; i < 64; i++)
Grid[i]=0;
for(int i = 1; i <= 5; i++)
{
battleShips = 1;
while(battleShips != 5)
{
int horizontal = rand()%2;
if(horizontal == 0)
{
battleShipFilled = false;
while(!battleShipFilled)
{
int row = rand()%8;
int column = rand()%8;
while(Grid[(row)*8+(column)] == 1)
{
row = rand()%8;
column = rand()%8;
}
int j = 0;
if(i == 1) j= (i+1);
else j= i;
for(int k = -j/2; k <= j/2; k++)
{
int numberOfCorrectLocation = 0;
while(numberOfCorrectLocation != j)
{
if(row+k> 0 && row+k<8)
{
if(Grid[(row+k)*8+(column)] == 1) break;
numberOfCorrectLocation++;
}
}
if(numberOfCorrectLocation !=i) break;
}
for(int k = -j/2; k <= j/2; k++)
Grid[(row+k)*8+(column)] = 1;
battleShipFilled = true;
}
battleShips++;
}
else
{
battleShipFilled = false;
while(!battleShipFilled)
{
int row = rand()%8;
int column = rand()%8;
while(Grid[(row)*8+(column)] == 1)
{
row = rand()%8;
column = rand()%8;
}
int j = 0;
if(i == 1) j= (i+1);
else j= i;
for(int k = -j/2; k <= j/2; k++)
{
int numberOfCorrectLocation = 0;
while(numberOfCorrectLocation != i)
{
if(row+k> 0 && row+k<8)
{
if(Grid[(row)*8+(column+k)] == 1) break;
numberOfCorrectLocation++;
}
}
if(numberOfCorrectLocation !=i) break;
}
for(int k = -j/2; k <= j/2; k++)
Grid[(row)*8+(column+k)] = 1;
battleShipFilled = true;
}
battleShips++;
}
}
}
}
But the code i have written is not able to generate the numbers randomly in the 8x8 grid.
Need some guidance on how to solve this. If there is any better way of doing it, please tell me...
How it should look:
What My code is doing:
Basically, I am placing 5 ships, each of different size on a grid. For each, I check whether I want to place it horizontally or vertically randomly. After that, I check whether the surrounding is filled up or not. If not, I place them there. Or I repeat the process.
Important Point: I need to use just while, for loops..
You are much better of using recursion for that problem. This will give your algorithm unwind possibility. What I mean is that you can deploy each ship and place next part at random end of the ship, then check the new placed ship part has adjacent tiles empty and progress to the next one. if it happens that its touches another ship it will due to recursive nature it will remove the placed tile and try on the other end. If the position of the ship is not valid it should place the ship in different place and start over.
I have used this solution in a word search game, where the board had to be populated with words to look for. Worked perfect.
This is a code from my word search game:
bool generate ( std::string word, BuzzLevel &level, CCPoint position, std::vector<CCPoint> &placed, CCSize lSize )
{
std::string cPiece;
if ( word.size() == 0 ) return true;
if ( !level.inBounds ( position ) ) return false;
cPiece += level.getPiece(position)->getLetter();
int l = cPiece.size();
if ( (cPiece != " ") && (word[0] != cPiece[0]) ) return false;
if ( pointInVec (position, placed) ) return false;
if ( position.x >= lSize.width || position.y >= lSize.height || position.x < 0 || position.y < 0 ) return false;
placed.push_back(position);
bool used[6];
for ( int t = 0; t < 6; t++ ) used[t] = false;
int adj;
while ( (adj = HexCoord::getRandomAdjacentUnique(used)) != -1 )
{
CCPoint nextPosition = HexCoord::getAdjacentGridPositionInDirection((eDirection) adj, position);
if ( generate ( word.substr(1, word.size()), level, nextPosition, placed, lSize ) ) return true;
}
placed.pop_back();
return false;
}
CCPoint getRandPoint ( CCSize size )
{
return CCPoint ( rand() % (int)size.width, rand() % (int)size.height);
}
void generateWholeLevel ( BuzzLevel &level,
blockInfo* info,
const CCSize &levelSize,
vector<CCLabelBMFont*> wordList
)
{
for ( vector<CCLabelBMFont*>::iterator iter = wordList.begin();
iter != wordList.end(); iter++ )
{
std::string cWord = (*iter)->getString();
// CCLog("Curront word %s", cWord.c_str() );
vector<CCPoint> wordPositions;
int iterations = 0;
while ( true )
{
iterations++;
//CCLog("iteration %i", iterations );
CCPoint cPoint = getRandPoint(levelSize);
if ( generate (cWord, level, cPoint, wordPositions, levelSize ) )
{
//Place pieces here
for ( int t = 0; t < cWord.size(); t++ )
{
level.getPiece(wordPositions[t])->addLetter(cWord[t]);
}
break;
}
if ( iterations > 1500 )
{
level.clear();
generateWholeLevel(level, info, levelSize, wordList);
return;
}
}
}
}
I might add that shaped used in the game was a honeycomb. Letter could wind in any direction, so the code above is way more complex then what you are looking for I guess, but will provide a starting point.
I will provide something more suitable when I get back home as I don't have enough time now.
I can see a potential infinite loop in your code
int j = 0;
if(i == 1) j= (i+1);
else j= i;
for(int k = -j/2; k <= j/2; k++)
{
int numberOfCorrectLocation = 0;
while(numberOfCorrectLocation != i)
{
if(row+k> 0 && row+k<8)
{
if(Grid[(row)*8+(column+k)] == 1) break;
numberOfCorrectLocation++;
}
}
if(numberOfCorrectLocation !=i) break;
}
Here, nothing prevents row from being 0, as it was assignd rand%8 earlier, and k can be assigned a negative value (since j can be positive). Once that happens nothing will end the while loop.
Also, I would recommend re-approaching this problem in a more object oriented way (or at the very least breaking up the code in main() into multiple, shorter functions). Personally I found the code a little difficult to follow.
A very quick and probably buggy example of how you could really clean your solution up and make it more flexible by using some OOP:
enum Orientation {
Horizontal,
Vertical
};
struct Ship {
Ship(unsigned l = 1, bool o = Horizontal) : length(l), orientation(o) {}
unsigned char length;
bool orientation;
};
class Grid {
public:
Grid(const unsigned w = 8, const unsigned h = 8) : _w(w), _h(h) {
grid.resize(w * h);
foreach (Ship * sp, grid) {
sp = nullptr;
}
}
bool addShip(Ship * s, unsigned x, unsigned y) {
if ((x <= _w) && (y <= _h)) { // if in valid range
if (s->orientation == Horizontal) {
if ((x + s->length) <= _w) { // if not too big
int p = 0; //check if occupied
for (int c1 = 0; c1 < s->length; ++c1) if (grid[y * _w + x + p++]) return false;
p = 0; // occupy if not
for (int c1 = 0; c1 < s->length; ++c1) grid[y * _w + x + p++] = s;
return true;
} else return false;
} else {
if ((y + s->length) <= _h) {
int p = 0; // check
for (int c1 = 0; c1 < s->length; ++c1) {
if (grid[y * _w + x + p]) return false;
p += _w;
}
p = 0; // occupy
for (int c1 = 0; c1 < s->length; ++c1) {
grid[y * _w + x + p] = s;
p += _w;
}
return true;
} else return false;
}
} else return false;
}
void drawGrid() {
for (int y = 0; y < _h; ++y) {
for (int x = 0; x < _w; ++x) {
if (grid.at(y * w + x)) cout << "|S";
else cout << "|_";
}
cout << "|" << endl;
}
cout << endl;
}
void hitXY(unsigned x, unsigned y) {
if ((x <= _w) && (y <= _h)) {
if (grid[y * _w + x]) cout << "You sunk my battleship" << endl;
else cout << "Nothing..." << endl;
}
}
private:
QVector<Ship *> grid;
unsigned _w, _h;
};
The basic idea is create a grid of arbitrary size and give it the ability to "load" ships of arbitrary length at arbitrary coordinates. You need to check if the size is not too much and if the tiles aren't already occupied, that's pretty much it, the other thing is orientation - if horizontal then increment is +1, if vertical increment is + width.
This gives flexibility to use the methods to quickly populate the grid with random data:
int main() {
Grid g(20, 20);
g.drawGrid();
unsigned shipCount = 20;
while (shipCount) {
Ship * s = new Ship(qrand() % 8 + 2, qrand() %2);
if (g.addShip(s, qrand() % 20, qrand() % 20)) --shipCount;
else delete s;
}
cout << endl;
g.drawGrid();
for (int i = 0; i < 20; ++i) g.hitXY(qrand() % 20, qrand() % 20);
}
Naturally, you can extend it further, make hit ships sink and disappear from the grid, make it possible to move ships around and flip their orientation. You can even use diagonal orientation. A lot of flexibility and potential to harness by refining an OOP based solution.
Obviously, you will put some limits in production code, as currently you can create grids of 0x0 and ships of length 0. It's just a quick example anyway. I am using Qt and therefore Qt containers, but its just the same with std containers.
I tried to rewrite your program in Java, it works as required. Feel free to ask anything that is not clearly coded. I didn't rechecked it so it may have errors of its own. It can be further optimized and cleaned but as it is past midnight around here, I would rather not do that at the moment :)
public static void main(String[] args) {
Random generator = new Random();
int Grid[][] = new int[8][8];
for (int battleShips = 0; battleShips < 5; battleShips++) {
boolean isHorizontal = generator.nextInt(2) == 0 ? true : false;
boolean battleShipFilled = false;
while (!battleShipFilled) {
// Select a random row and column for trial
int row = generator.nextInt(8);
int column = generator.nextInt(8);
while (Grid[row][column] == 1) {
row = generator.nextInt(8);
column = generator.nextInt(8);
}
int lengthOfBattleship = 0;
if (battleShips == 0) // Smallest ship should be of length 2
lengthOfBattleship = (battleShips + 2);
else // Other 4 ships has the length of 2, 3, 4 & 5
lengthOfBattleship = battleShips + 1;
int numberOfCorrectLocation = 0;
for (int k = 0; k < lengthOfBattleship; k++) {
if (isHorizontal && row + k > 0 && row + k < 8) {
if (Grid[row + k][column] == 1)
break;
} else if (!isHorizontal && column + k > 0 && column + k < 8) {
if (Grid[row][column + k] == 1)
break;
} else {
break;
}
numberOfCorrectLocation++;
}
if (numberOfCorrectLocation == lengthOfBattleship) {
for (int k = 0; k < lengthOfBattleship; k++) {
if (isHorizontal)
Grid[row + k][column] = 1;
else
Grid[row][column + k] = 1;
}
battleShipFilled = true;
}
}
}
}
Some important points.
As #Kindread said in an another answer, the code has an infinite loop condition which must be eliminated.
This algorithm will use too much resources to find a solution, it should be optimized.
Code duplications should be avoided as it will result in more maintenance cost (which might not be a problem for this specific case), and possible bugs.
Hope this answer helps...