[CSES][Introductory Problems] Grid Paths 

you done a great work man really your explanation make me feel what is the issue in the problem

Great work, man!! Loved the animation. It made understanding the solution a lot easier. And this definitely isn't an introductory problem.

Great work sir. You are the BEST!

How does the pruning decreases the time complexity? I think it's over complicating to handle all these cases.

Awesome work!

Could you give some tip on how did you make that animation? I think it could be very helpful in coming up with issues in our solution, than trying to figure out issues through Debug logs. Any link or tip on visualising the algorithm will be very helpful. Anyways, great explanation, thanks a lot, and keep up the good work.

really nice explanation, understood it but I don't think I'll be able to solve it in an interview

Can you explain why 88418 path ? I just want to know how to calculate that number using math.

can you please make a video on prime tuples(from mathematics section of cses) i tried using bitmasking it give tle in 2 cases

Amazing... Learnt alot ❤

Thanks for the explanation

#include #define int long long int #define F first #define S second #define pb push_back #define RIGHT 0 #define LEFT 1 #define DOWN 2 #define UP 3 #define isValid(a,b,c) (a>=b && a < c ? 1: 0) using namespace std; int dx[4] = {0,0,1,-1}; int dy[4] = {1,-1,0,0}; string str; int vis[7][7]; void solve() { } int countPaths(int x,int y, int pos) { if(pos == (int)str.length()) return (x==6 && y==0); if(x == 6 && y == 0) return 0; if(vis[x][y]) return 0; vector visited(4,-1); for(int k = 0; k < 4; k++) { if(isValid(x + dx[k],0,7) && isValid(y + dy[k],0,7)) visited[k] = vis[x + dx[k]][y+dy[k]]; } if(!visited[DOWN] && !visited[UP] && visited[RIGHT] && visited[LEFT]) return 0; if(!visited[RIGHT] && !visited[LEFT] && visited[DOWN] && visited[UP]) return 0; if(isValid(x-1,0,7) && isValid(y+1,0,7) && vis[x-1][y+1]==1) { if(!visited[RIGHT] && !visited[UP]) return 0; } if(isValid(x+1,0,7) && isValid(y+1,0,7) && vis[x+1][y+1]==1) { if(!visited[RIGHT] && !visited[DOWN]) return 0; } if(isValid(x-1,0,7) && isValid(y-1,0,7) && vis[x-1][y-1]==1) { if(!visited[LEFT] && !visited[UP]) return 0; } if(isValid(x+1,0,7) && isValid(y-1,0,7) && vis[x+1][y-1]==1) { if(!visited[LEFT] && !visited[DOWN]) return 0; } vis[x][y] = 1; int numOfPaths = 0; if(str[pos] == '?') { for(int k = 0; k < 4; k++) { if(isValid(x+dx[k],0,7) && isValid(y+dy[k],0,7)) numOfPaths += countPaths(x+dx[k], y+dy[k], pos+1); } } else if(str[pos] == 'R' && y + 1 < 7) numOfPaths+= countPaths(x,y+1,pos+1); else if(str[pos] == 'L' && y -1 >= 0) numOfPaths+= countPaths(x,y-1,pos+1); else if(str[pos] == 'U' && x -1 >= 0) numOfPaths+= countPaths(x-1,y,pos+1); else if(str[pos] == 'D' && x + 1 < 7) numOfPaths+= countPaths(x+1,y,pos+1); vis[x][y] = 0; return numOfPaths; } int32_t main() { #ifndef ONLINE_JUDGE freopen("input.txt","r",stdin); freopen("output.txt","w",stdout); #endif cin >> str; cout

Beautiful, thanks!

great video, nice explaination

Very cool cases thank you

thank you very much, sir

It would be good if you could attach your solution too. Anyway thanks for the video

Bro, Thanks a lot, ur too awesome.

Thanks !!

Awesome!

very nice

Tks bro

thansk dud

Gud one

👌👌👌

damn. DAMN. damn.

can any1 share the AC code plz

Still giving TLE