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Home Code Solutions Hackerrank Algorithms

Brick Tiling – HackerRank Solution

Brick Tiling - HackerRank Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.

BhautikBhalala by BhautikBhalala
May 28, 2022
Reading Time: 1 min read
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Table of Contents

  • Brick Tiling – HackerRank Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.
  • Solutions of Algorithms Data Structures Hard HackerRank:
    • Here are all the Solutions of Hard , Advanced , Expert Algorithms of Data Structure of Hacker Rank , Leave a comment for similar posts
  • C++ Brick Tiling HackerRank Solution
  • Java Brick Tiling HackerRank Solution
  • Python 3 Brick Tiling HackerRank Solution
  • Python 2 Brick Tiling HackerRank Solution
  • C Brick Tiling HackerRank Solution
    • Warmup Implementation Strings Sorting Search Graph Theory Greedy Dynamic Programming Constructive Algorithms Bit Manipulation Recursion Game Theory NP Complete Debugging
    • Leave a comment below
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Brick Tiling – HackerRank Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.

Solutions of Algorithms Data Structures Hard HackerRank:

Here are all the Solutions of Hard , Advanced , Expert Algorithms of Data Structure of Hacker Rank , Leave a comment for similar posts

C++ Brick Tiling HackerRank Solution


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#include <map>
#include <set>
#include <list>
#include <cmath>
#include <queue>
#include <stack>
#include <bitset>
#include <vector>
#include <cstdio>
#include <string>
#include <sstream>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define PB push_back
#define MP make_pair
#define SZ(v) ((int)(v).size())
#define abs(x) ((x) > 0 ? (x) : -(x))
#define FOREACH(e,x) for(__typeof(x.begin()) e=x.begin();e!=x.end();++e)
typedef long long LL;

const int MOD = 1000000007;
const int inf = 0x3f3f3f3f;
const int maxn = 25;
const int maxs = 1 << 8;
LL dp[maxn][maxs][maxs];
int state[maxn];
int cur, nxt;
int n, m, initJ, initK;

bool valid(int mask, int pos) {
	return (mask & (1 << pos)) == 0;
}

void dfs(int dep, int s1, int s2, int s3, int current) {
	if (current >= (1 << m)) return;
	if (current == (1 << m) - 1) {
		dp[nxt][s2][s3] = (dp[nxt][s2][s3] + dp[cur][initJ][initK]) % MOD;
		return;
	}
	if (s1 & (1 << dep)) {
		dfs(dep + 1, s1, s2, s3, current);
	} else {
		// ### <- dep
		//   #
		if (dep + 1 < m && valid(s2, dep) && valid(s3, dep) && valid(s3, dep + 1)) {
			dfs(dep + 1, s1, s2 | (1 << dep), s3 | (1 << dep) | (1 << (dep + 1)), current | (1 << dep));
		}
		//   # 
		// ### <- dep
		if (dep >= 1 && valid(s2, dep) && valid(s3, dep) && valid(s3, dep - 1)) {
			dfs(dep + 1, s1, s2 | (1 << dep), s3 | (1 << dep) | (1 << (dep - 1)), current | (1 << dep));
		}
		// #   <- dep
		// ###
		if (dep + 1 < m && valid(s1, dep + 1) && valid(s2, dep + 1) && valid(s3, dep + 1)) {
			dfs(dep + 2, s1, s2 | (1 << (dep + 1)), s3 | (1 << (dep + 1)), current | (1 << dep) | (1 << (dep + 1)));
		}
		// ### <- dep
		// #  
		if (dep + 1 < m && valid(s1, dep + 1) && valid(s2, dep) && valid(s3, dep)) {
			dfs(dep + 2, s1, s2 | (1 << dep), s3 | (1 << dep), current | (1 << dep) | (1 << (dep + 1)));
		}
		// ## <- dep
		//  #
		//  #
		if (dep + 2 < m && valid(s2, dep) && valid(s2, dep + 1) && valid(s2, dep + 2)) {
			dfs(dep + 1, s1, s2 | (1 << dep) | (1 << (dep + 1)) | (1 << (dep + 2)), s3, current | (1 << dep));
		}
		//  #
		//  #
		// ## <- dep
		if (dep >= 2 && valid(s2, dep) && valid(s2, dep - 1) && valid(s2, dep - 2)) {
			dfs(dep + 1, s1, s2 | (1 << (dep - 2)) | (1 << (dep - 1)) | (1 << dep), s3, current | (1 << dep));
		}
		// ## <- dep
		// #
		// #
		if (dep + 2 < m && valid(s2, dep) && valid(s1, dep + 1) && valid(s1, dep + 2)) {
			dfs(dep + 3, s1, s2 | (1 << dep), s3, current | (1 << dep) | (1 << (dep + 1)) | (1 << (dep + 2)));
		}
		// #  <- dep
		// #
		// ##
		if (dep + 2 < m && valid(s1, dep + 1) && valid(s1, dep + 2) && valid(s2, dep + 2)) {
			dfs(dep + 3, s1, s2 | (1 << (dep + 2)), s3, current | (1 << dep) | (1 << (dep + 1)) | (1 << (dep + 2)));
		}
	}
}

char str[10];
int main() {
	int ts;
	scanf("%d", &ts);
	while (ts--) {
		memset(state, 0, sizeof(state));
		scanf("%d%d", &n, &m);
		for (int i = 1; i <= n; i++) {
			scanf("%s", str);
			for (int j = 0; j < m; j++) {
				if (str[j] == '#') {
					state[i] |= (1 << j);
				}
			}
		}
		if (n == 1) {
			if (state[1] == (1 << m) - 1) puts("1");
			else puts("0");
			continue;
		}

		memset(dp, 0, sizeof(dp));
		dp[2][state[1]][state[2]] = 1;
		for (int i = 2; i <= n; i++) {
			cur = i, nxt = i + 1;
			for (int j = 0; j < (1 << m); j++) {
				for (int k = 0; k < (1 << m); k++) {
					if (dp[cur][j][k] == 0) continue;
					initJ = j; initK = k;
					dfs(0, j, k, state[i + 1], j);
				}
			}
		}
		cout << dp[n + 1][(1 << m) - 1][0] << endl;
	}
	return 0;
}

Java Brick Tiling HackerRank Solution


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import java.io.ByteArrayInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.InputMismatchException;

public class Solution {
	static InputStream is;
	static PrintWriter out;
//	static String INPUT = "1 2 2 ## ##";
	static String INPUT = "";
	
	static int[] makeMaskArray(int mask, int h, int amask)
	{
		int[] ret = new int[h];
		int p = 0;
		for(int i = 0;i < h;i++){
			if((i&mask) == 0 && (amask == -1 || (i&amask) != 0)){
				ret[p++] = i;
			}
		}
		return Arrays.copyOf(ret, p);
	}
	
	static void solve()
	{
		for(int T = ni();T >= 1;T--){
			int n = ni(), m = ni();
			char[][] map = nm(n,m);
			int[] pre = new int[1<<m*2+1];
			pre[(1<<m*2+1)-1] = 1;
			int[] cur = new int[1<<m*2+1];
			int mod = 1000000007;
			int omask = (1<<m*2+1)-1;
			int hmask = 1<<m*2;
			int mask1 = 1<<m-1|1<<2*m-1|1<<2*m;
			int mask2 = 1<<m-1|1<<2*m-1|1<<2*m-2;
			int mask3 = 1<<m-1|1<<m|1<<m+1;
			int mask4 = 1<<m-1|1<<m-2|1<<m-3;
			int mask5 = 1<<0|1<<m|1<<2*m;
			int mask6 = 1<<0|1<<m-1|1<<2*m-1;
			int mask7 = 1<<0|1<<1|1<<m+1;
			int mask8 = 1<<0|1<<1|1<<m-1;
			int[] val1 = makeMaskArray(mask1, 1<<2*m+1, -1);
			int[] val2 = makeMaskArray(mask2, 1<<2*m+1, hmask);
			int[] val3 = makeMaskArray(mask3, 1<<2*m+1, hmask);
			int[] val4 = makeMaskArray(mask4, 1<<2*m+1, hmask);
			int[] val5 = makeMaskArray(mask5, 1<<2*m+1, -1);
			int[] val6 = makeMaskArray(mask6, 1<<2*m+1, hmask);
			int[] val7 = makeMaskArray(mask7, 1<<2*m+1, hmask);
			int[] val8 = makeMaskArray(mask8, 1<<2*m+1, hmask);
			int[] valn = makeMaskArray(0, 1<<2*m+1, hmask);
			
			for(int i = 0;i < n;i++){
				for(int j = 0;j < m;j++){
					Arrays.fill(cur, 0);
					if(map[i][j] == '.'){
						// xx
						// .x
						// .x
						if(i >= 2 && j >= 1){
							for(int k : val1){
								int nk = ((k|mask1)<<1|1)&omask;
								cur[nk] += pre[k];
								if(cur[nk] >= mod)cur[nk] -= mod;
							}
						}
						// x.
						// x.
						// xx
						if(i >= 2 && j >= 1){
							for(int k : val5){
								int nk = ((k|mask5)<<1|1)&omask;
								cur[nk] += pre[k];
								if(cur[nk] >= mod)cur[nk] -= mod;
							}
						}
						// xx
						// x.
						// x.
						if(i >= 2 && j+1 < m){
							for(int k : val2){
								int nk = ((k|mask2)<<1|1)&omask;
								cur[nk] += pre[k];
								if(cur[nk] >= mod)cur[nk] -= mod;
							}
						}
						// .x
						// .x
						// xx
						if(i >= 2 && j >= 1){
							for(int k : val6){
								int nk = ((k|mask6)<<1|1)&omask;
								cur[nk] += pre[k];
								if(cur[nk] >= mod)cur[nk] -= mod;
							}
						}
						// xxx
						// ..x
						if(i >= 1 && j >= 2){
							for(int k : val3){
								int nk = ((k|mask3)<<1|1)&omask;
								cur[nk] += pre[k];
								if(cur[nk] >= mod)cur[nk] -= mod;
							}
						}
						// x..
						// xxx
						if(i >= 1 && j >= 2){
							for(int k : val7){
								int nk = ((k|mask7)<<1|1)&omask;
								cur[nk] += pre[k];
								if(cur[nk] >= mod)cur[nk] -= mod;
							}
						}
						// xxx
						// x..
						if(i >= 1 && j+2 < m){
							for(int k : val4){
								int nk = ((k|mask4)<<1|1)&omask;
								cur[nk] += pre[k];
								if(cur[nk] >= mod)cur[nk] -= mod;
							}
						}
						// ..x
						// xxx
						if(i >= 1 && j >= 2){
							for(int k : val8){
								int nk = ((k|mask8)<<1|1)&omask;
								cur[nk] += pre[k];
								if(cur[nk] >= mod)cur[nk] -= mod;
							}
						}
						// none
						for(int k : valn){
							int nk = (k<<1)&omask;
							cur[nk] += pre[k];
							if(cur[nk] >= mod)cur[nk] -= mod;
						}
					}else{
						for(int k : valn){
							int nk = (k<<1|1)&omask;
							cur[nk] += pre[k];
							if(cur[nk] >= mod)cur[nk] -= mod;
						}
					}
					int[] dum = cur; cur = pre; pre = dum;
				}
			}
			out.println(pre[(1<<m*2+1)-1]);
		}
	}
	
	public static void main(String[] args) throws Exception
	{
		long S = System.currentTimeMillis();
		is = INPUT.isEmpty() ? System.in : new ByteArrayInputStream(INPUT.getBytes());
		out = new PrintWriter(System.out);
		
		solve();
		out.flush();
		long G = System.currentTimeMillis();
		tr(G-S+"ms");
	}
	
	private static boolean eof()
	{
		if(lenbuf == -1)return true;
		int lptr = ptrbuf;
		while(lptr < lenbuf)if(!isSpaceChar(inbuf[lptr++]))return false;
		
		try {
			is.mark(1000);
			while(true){
				int b = is.read();
				if(b == -1){
					is.reset();
					return true;
				}else if(!isSpaceChar(b)){
					is.reset();
					return false;
				}
			}
		} catch (IOException e) {
			return true;
		}
	}
	
	private static byte[] inbuf = new byte[1024];
	static int lenbuf = 0, ptrbuf = 0;
	
	private static int readByte()
	{
		if(lenbuf == -1)throw new InputMismatchException();
		if(ptrbuf >= lenbuf){
			ptrbuf = 0;
			try { lenbuf = is.read(inbuf); } catch (IOException e) { throw new InputMismatchException(); }
			if(lenbuf <= 0)return -1;
		}
		return inbuf[ptrbuf++];
	}
	
	private static boolean isSpaceChar(int c) { return !(c >= 33 && c <= 126); }
	private static int skip() { int b; while((b = readByte()) != -1 && isSpaceChar(b)); return b; }
	
	private static double nd() { return Double.parseDouble(ns()); }
	private static char nc() { return (char)skip(); }
	
	private static String ns()
	{
		int b = skip();
		StringBuilder sb = new StringBuilder();
		while(!(isSpaceChar(b))){ // when nextLine, (isSpaceChar(b) && b != ' ')
			sb.appendCodePoint(b);
			b = readByte();
		}
		return sb.toString();
	}
	
	private static char[] ns(int n)
	{
		char[] buf = new char[n];
		int b = skip(), p = 0;
		while(p < n && !(isSpaceChar(b))){
			buf[p++] = (char)b;
			b = readByte();
		}
		return n == p ? buf : Arrays.copyOf(buf, p);
	}
	
	private static char[][] nm(int n, int m)
	{
		char[][] map = new char[n][];
		for(int i = 0;i < n;i++)map[i] = ns(m);
		return map;
	}
	
	private static int[] na(int n)
	{
		int[] a = new int[n];
		for(int i = 0;i < n;i++)a[i] = ni();
		return a;
	}
	
	private static int ni()
	{
		int num = 0, b;
		boolean minus = false;
		while((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'));
		if(b == '-'){
			minus = true;
			b = readByte();
		}
		
		while(true){
			if(b >= '0' && b <= '9'){
				num = num * 10 + (b - '0');
			}else{
				return minus ? -num : num;
			}
			b = readByte();
		}
	}
	
	private static long nl()
	{
		long num = 0;
		int b;
		boolean minus = false;
		while((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'));
		if(b == '-'){
			minus = true;
			b = readByte();
		}
		
		while(true){
			if(b >= '0' && b <= '9'){
				num = num * 10 + (b - '0');
			}else{
				return minus ? -num : num;
			}
			b = readByte();
		}
	}
	
	private static void tr(Object... o) { if(INPUT.length() != 0)System.out.println(Arrays.deepToString(o)); }
}



Python 3 Brick Tiling HackerRank Solution


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def memoize(func):
    pool = {}
    def wrapper(*arg):
        if arg not in pool:
            pool[arg] = func(*arg)
        return pool[arg]
    return wrapper

mod = 1000000007
shapes = (\
    ((1,0),(2,0),(2,1)),\
    ((0,1),(0,2),(-1,2)),\
    ((0,1),(1,1),(2,1)),\
    ((1,0),(0,1),(0,2)),\
    ((0,1),(-1,1),(-2,1)),\
    ((0,1),(0,2),(1,2)),\
    ((1,0),(2,0),(0,1)),\
    ((1,0),(1,1),(1,2)))

for case in range(int(input())):
    Y,X = map(int,input().split())
    mx = [int(''.join('0' if c =='.' else '1' for c in input().rstrip()), 2) for i in range(Y)]
    mx = mx + 3*[0]
    full = (1<<X)-1

    @memoize
    def rec(y,first,second,third):
        if y==Y:
            return 1 if first == second and second == third and third == 0 else 0
        if first == full:
            return rec(y+1,second,third,mx[y+3])

        def can_fit(rows,shape,x_offset):
            res = rows[:]
            for x,y in shape:
                x += x_offset
                if x < 0 or x >= X or y < 0 or y >= Y:
                    return None
                if res[y] & (1<<x) != 0:
                    return None
                res[y] |= (1<<x)
            return res

        free = 0
        while (first & (1<<free)) != 0:
            free += 1
        rows = [first | (1<<free),second,third]
        ans = 0
        for shape in shapes:
            nrows = can_fit(rows,shape,free)
            if nrows != None:
                ans = (ans + rec(y, *nrows)) % mod
        return ans

    print(rec(0,mx[0],mx[1],mx[2]))



Python 2 Brick Tiling HackerRank Solution


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# Enter your code here. Read input from STDIN. Print output to STDOUT
#!/usr/bin/python

def memo(func):
    save = {}
    def f(*args):
        if args not in save:
            save[args] = func(*args)
        return save[args]
    return f

shapes = (
        # x
        # o
        # o o
        ((0,0), (1,0), (2,0), (2,1)),
        # x o o
        # o
        ((0,0), (0,1), (0,2), (1,0)),
        # x o
        #   o
        #   o
        ((0,0), (0,1), (1,1), (2,1)),
        #     x
        # o o o
        ((0,0), (1,-2), (1,-1), (1,0)),
        # x o
        # o
        # o
        ((0,0), (0,1), (1,0), (2,0)),
        # x
        # o o o
        ((0,0), (1,0), (1,1), (1,2)),
        #   x
        #   o
        # o o
        ((0,0), (1,0), (2,-1), (2,0)),
        # x o o
        #     o
        ((0,0), (0,1), (0,2), (1,2)),
        )

def get(layout, n, m):
    full = (1 << m) - 1
    for x in xrange(3):
        layout.append(full)

    '''
    def p(x):
        s = bin(x)
        a = len(s) - 2
        return "0000000000"[:m-a] + s[2:]
    '''

    @memo
    def f(row, fst, snd, thr):
        '''
        print 'row', row
        print p(fst)
        print p(snd)
        print p(thr)
        '''
        if row == n: return 1
        if fst == full: return f(row+1, snd, thr, layout[row+3])
        pos = 0
        while pos < m:
            if (fst & (1 << (m-1-pos))) == 0:
                break
            pos += 1

        def go(shape):
            ret = [fst, snd, thr]
            for x,y in shape:
                r, c = row + x, pos + y
                if r < 0 or c < 0 or r >= n or c >= m: return None
                if ret[x] & (1<<(m-1-c)): return None
                ret[x] |= 1 << (m-1-c)
            return ret

        ret = 0
        for x in shapes:
            g = go(x)
            if g is not None:
                ret += f(row, *g)
        return ret % 1000000007

    return f(0, layout[0], layout[1], layout[2])

t = input()
for x in xrange(t):
    n, m = map(int, raw_input().strip().split())
    layout = []
    empty = 0
    for y in xrange(n):
        here = 0
        row = raw_input()
        for x in row:
            if x == '.':
                empty += 1
                here = here * 2
            else:
                here = here * 2 + 1
        layout.append(here)
    if empty == 0:
        print 1
    elif empty % 4 != 0:
        print 0
    else:
        print get(layout, n, m)



C Brick Tiling HackerRank Solution


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#include "assert.h"
#include "stdio.h"

#define rep(i,n) for(i=0;i<(n);i++)

#define MOD 1000000007
#define N 25
#define M 8
#define B 8

int n, m, tm, tm2, grid[N], ways[N][1<<(2*M)], width[B] = { 1, 1, 1, 1, 2, 2, 3, 3 }, b[M][B];
char buf[M+1];

int make_brick( int a, int b, int c, int d )
{
	return ( tm2 * a + tm * b + c ) << d;
}

void rec( int x, int y, int s0, int s1 )
{
	if( y == m )
	{
		ways[x+1][s1] = ( ways[x+1][s1] + ways[x][s0] ) % MOD;
		return;
	}

	if( ( ( s0 / tm ) >> y ) & 1 )
	{
		rec( x, y + 1, s0, s1 );
		return;
	}

	int i;
	rep(i,B) if( b[y][i] && ( ( b[y][i] / tm ) & s0 ) == 0 && ( b[y][i] & s1 ) == 0 )
		rec( x, y + width[i], s0, s1 ^ ( b[y][i] % tm2 ) );
}

void run()
{
	int i, j;
	scanf( "%d%d", &n, &m );
	assert( 1 <= n && n <= 20 && 1 <= m && m <= 8 );
	tm = 1<<m, tm2 = 1<<(2*m);

	rep(i,m)
	{
		b[i][0] = i + 2 <= m ? make_brick( 1, 1, 3, i ) : 0;
		b[i][1] = i - 1 >= 0 ? make_brick( 2, 2, 3, i - 1 ) : 0;
		b[i][2] = i + 3 <= m ? make_brick( 1, 7, 0, i ) : 0;
		b[i][3] = i - 2 >= 0 ? make_brick( 4, 7, 0, i - 2 ) : 0;
		b[i][4] = i + 2 <= m ? make_brick( 3, 1, 1, i ) : 0;
		b[i][5] = i + 2 <= m ? make_brick( 3, 2, 2, i ) : 0;
		b[i][6] = i + 3 <= m ? make_brick( 7, 1, 0, i ) : 0;
		b[i][7] = i + 3 <= m ? make_brick( 7, 4, 0, i ) : 0;
	}

	rep(i,n)
	{
		grid[i] = 0;
		scanf( "%s", buf );
		rep(j,m) assert( buf[j] == '#' || buf[j] == '.' );
		rep(j,m) if( buf[j] == '#' ) grid[i] ^= 1<<j;
	}
	grid[n] = grid[n+1] = tm-1;

	rep(i,n+1) rep(j,tm2) ways[i][j] = 0;
	ways[0][ tm * grid[0] + grid[1] ] = 1;
	rep(i,n) rep(j,tm2) if( ways[i][j] ) rec( i, 0, j, ( j % tm ) * tm + grid[i+2] );

	printf( "%d\n", ways[n][tm2-1] );
}

int main()
{
	int t;
	scanf( "%d", &t );
	assert( 1 <= t && t <= 50 );
	while( t-- ) run();
	return 0;
}

 

Warmup
Implementation
Strings
Sorting
Search
Graph Theory
Greedy
Dynamic Programming
Constructive Algorithms
Bit Manipulation
Recursion
Game Theory
NP Complete
Debugging

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