Is It a Snake
Problem Statement :
One day I was visiting a temple in which snakes were worshiped. I happened to find a golden plate of dimension 2 * n in it. It had 2 rows of n cells each, and so the total number of cells is 2 * n. Each cell of the plate was either white or black, denoted by '.' and '#' respectively. Legend says that a snake was lying on this plate for many years and prayed. So, the cells that were covered by its body have turned black, the rest of the cells were white. Its entire body was supposedly on this plate. Also, you know that a snake likes to make itself comfortable, so none of its parts will be intersecting with its other parts. Usually, I am skeptical of such legends. So, I want to check for myself whether the legend could potentially be true or not. For example, consider the golden plate given below: ## ## Now, the legend could be true. A snake can be on it, and one possible configuration is as follows. The head of the snake is at (1, 1), then the next portion at (1, 2), then at (2, 2) and then the finally the tail at (2, 1). Notice that the parts of the snake can be adjacent only if the corresponding cells have a common side. ##.#.. .###.. There can be a snake on the plate above. ##.## .#.#. The legend is surely false if the plate is as above. These are not marks of a single snake. There could be more than one possible snakes here. But not a single snake. Given the description of the plate, figure out if the legend could be true, or if it is definitely false. Input The first line of the input contains an integer T denoting the number of test cases. The description of T test cases follows. The first line of each test case contains an integer n. The next two lines each contain n characters denoting the rows of the plate. Output For each test case, output a single line containing "yes" or "no" (without quotes) corresponding to the answer of the problem. Constraints 1 ≤ T ≤ 500 1 ≤ n ≤ 500 There will be at least one cell containing the character '#' Example Input 6 2 ## .. 2 ## .# 2 #. .# 7 #.###.. ####### 6 ##.#.. .###.. 5 ##.## .#.#. Output yes yes no no yes no
Solution :
Solution in C :
#include <stdio.h>
#define BLACK 0
#define WHITE 1
int main()
{
int cases,n,i,j,start,end,prev,next;
char c;
scanf("%d",&cases);
label:
while(cases--)
{
scanf("%d\n",&n);
int a[2][n];
for(i=0;i<2;i++)
{
for(j=0;j<n;j++)
{
scanf(" %c",&c);
a[i][j]=(c=='.')?WHITE:BLACK;
}
}
for(i=0;1;i++)
if(a[0][i]==BLACK || a[1][i]==BLACK)
{
start=i;
break;
}
for(i=n-1;1;i--)
if(a[0][i]==BLACK || a[1][i]==BLACK)
{
end=i;
break;
}
for(i=start;i<=end;i++)
if((a[0][i] == WHITE) && a[1][i]==WHITE)
{
printf("no\n");
goto label;
}
for(i=start+1;i<=end;i++)
{
if( (a[0][i]==BLACK && a[1][i-1]==BLACK) && (a[0][i-1]==WHITE && a[1][i]==WHITE) )
{
printf("no\n");
goto label;
}
if( (a[0][i]==WHITE && a[1][i-1]==WHITE) && (a[0][i-1]==BLACK && a[1][i]==BLACK) )
{
printf("no\n");
goto label;
}
}
for(i=start;i<=end;i++)
{
if((a[0][i] == a[1][i]) && a[1][i]==BLACK)
{
if(start==i)
{
while(a[0][i]==BLACK && a[1][i]==BLACK)
i++;
}
else
{
int rectlength=0;
prev=(a[0][i-1]==BLACK)?0:1;
while((a[0][i] == a[1][i]) && a[1][i]==BLACK)
{
if(i==end)
{
printf("yes\n");
goto label;
}
rectlength++;
i++;
}
if(rectlength%2==0 )
{
if(a[prev][i]==WHITE)
{
printf("no\n");
goto label;
}
}
else
{
if(a[prev][i]==BLACK)
{
printf("no\n");
goto label;
}
}
}
}
}
printf("yes\n");
}
}
Solution in C++ :
#include <bits/stdc++.h>
#define I_O ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
//~~~~~~~~~~~~ Sort Form Of Long~~~~~~~~~~~//
#define ll long long int
#define lls int64_t
#define ld long double
#define db double
#define ull unsigned long long int
//~~~~~~~~~~~~~~Pair~~~~~~~~~~~~~~~~~~//
#define pii pair<int, int>
#define pll pair<lls, lls>
#define pdd pair<db,db>
#define psi pair<string,int>
#define vi vector<int>
#define vl vector<lls>
#define vd vector<db>
#define vs vector<string>
#define vb vector<bool>
#define vpi vector<pii>
#define vpl vector<pll>
#define vpd vector<pdd>
#define vpsi vector<psi>
#define vvi vector<vector<int>>
//~~~~~~~~~~~~~~Vector~~~~~~~~~~~~~~~~~//
#define pb push_back
#define pf push_front
#define MP make_pair
#define in insert
#define ff first
#define ss second
#define al(v) v.begin(),v.end()
#define alr(v) v.rbegin(), v.rend()
#define srt(v) sort(al(v))
#define srtr(v) sort(al(v), greater<int>());
#define len(x) ((int)(x).size())
#define rev(v) reverse(al(v))
#define btcnt(n) __builtin_popcount(n)
#define acl(v, n) accumulate(al(v), n)
#define eb emplace_back
#define Lrt(s, c) rotate(s.begin(), s.begin() + c, s.end())
#define Rrt(s, c) rotate(s.begin(), s.begin() + s.size() - c, s.end())
#define lb(v, kk) lower_bound(al(v), kk) - v.begin()
#define ub(v, kk) upper_bound(al(v), kk) - v.begin()
#define tpu(str) transform(al(str), str.begin(), ::toupper)
#define tpl(str) transform(al(str), str.begin(), ::tolower)
#define cignr cin.ignore(numeric_limits<streamsize>::max(), '\n');
#define mxv(v) *max_element(al(v))
#define mnv(v) *min_element(al(v))
const int MOD = 1e9 + 7;
const lls INF = 2e18;
const int mxn = 2e9 + 9;
const int mxd = 2e5 + 5;
const int mxa = 2e6 + 5;
//~~~~~~~~~~~~~~~~~~Function~~~~~~~~~~~~~~~~~~~~//
lls gcd(lls a, lls b){ if(b == 0LL) return a; return gcd(b, a % b); }
lls lcm(lls a, lls b){ return (a / gcd(a, b) * b); }
lls maxll(lls x, lls y){ return x > y ? x : y; }
lls minll(lls x, lls y){ return x < y ? x : y; }
//~~~~~~~~~~~~~~~Loops and Short~~~~~~~~~~~~~~~~//
#define PI acos(-1)
#define Cn continue
#define Br break
#define off return 0
#define N '\n'
#define fopen freopen("input.txt", "r", stdin);
#define rep(i, n) for(lls i = 0; (lls)i < n; i++)
#define repn(i, a, b) for(lls i = (lls)(a); i < (lls)(b); i++)
#define repr(i, a, b) for(lls i = (lls)(a) - 1; i >= (lls)(b); i--)
#define test_case() int T; cin >> T; while(T--)
using namespace std;
// ===================================~~~~~~ SOLUTION STARTS HERE ~~~~~~=================================== //
//int a, b, c, d, e, res = 0, ans = 0, prod = 1, n, m, cnt = 0, k, l, r, s, t, x, y, f, i, j, p;
char a[2][501];
bool vis[2][502];
bool dfs(int k, int n, int j) {
bool ans = true;
memset(vis, 0, sizeof(vis));
for (int i = j; i < n; i++) {
if (a[k][i] == '#') {
vis[k][i] = 1;
if (k == 0) {
if (a[k + 1][i] == '#') {
vis[++k][i] = true;
}
} else {
if (a[k - 1][i] == '#') {
vis[--k][i] = true;
}
}
} else break;
}
for (int i = 0; i < n; i++) {
if (a[0][i] == '#' and !vis[0][i]) {
ans = false; break;
}
if (a[1][i] == '#' and !vis[1][i]) {
ans = false; break;
}
}
return ans;
}
void Run_Case()
{
int n, b, f, g;
cin >> n;
rep (i, n) cin >> a[0][i];
rep (i, n) cin >> a[1][i];
int cnt = 0;
for (int i = 0; i < n; i++) {
if (a[0][i] == '#' or a[1][i] == '#') {
b = i; break;
}
}
f = dfs(0, n, b);
g = dfs(1, n, b);
if ((f + g) != 0) cout << "yes\n";
else cout << "no\n";
}
int main()
{
I_O
test_case()
{
Run_Case();
}
off;
}
Solution in Java :
import java.io.*;
import java.util.*;
class ISSNAKE {
public static void main (String[] args) throws IOException {
Scanner sc= new Scanner(System.in);
while (sc.hasNextLine())
{
int t = Integer.parseInt( sc.nextLine() );
testcase:
for ( int z = 0; z < t; z++ ) {
int n = Integer.parseInt( sc.nextLine() );
char[][] snake = new char[2][n];
snake[0] = sc.nextLine().toCharArray();
snake[1] = sc.nextLine().toCharArray();
boolean[] s1 = new boolean[n];
boolean[] s2 = new boolean[n];
int start = 0;
while ( snake[0][start] == '.' && snake[1][start] == '.' )
start++;
s1[start] = snake[0][start] == '#';
s2[start] = snake[1][start] == '#';
boolean ended = false;
for ( int i = start + 1; i < n; i++ ) {
if ( snake[0][i] == '#' && snake[1][i] == '#' ) {
s1[i] = s2[i - 1];
s2[i] = s1[i - 1];
}
else if ( snake[0][i] == '#' ) {
s1[i] = s1[i - 1];
}
else if ( snake[1][i] == '#' ) {
s2[i] = s2[i - 1];
}
if ( ! s1[i] && ! s2[i] )
ended = true;
if ( ended && ( snake[0][i] == '#' || snake[1][i] == '#' ) ) {
System.out.println( "no" );
continue testcase;
}
}
System.out.println( "yes" );
}
}
}
}
Solution in Python :
def strtolist(a:str):
b = []
for i in range(0, n):
b.append(a[i])
return b
def mainthing():
p = -1
l = 0
for i in range(0, n * 2):
if p == -1:
if l1[i] == '#':
p = i
l = 1
l1[p] = 0
elif l2[i] == '#':
p = i
l = 2
l2[p] = 0
else:
if l == 1:
if l2[p] == '#':
l = 2
l2[p] = 0
elif p != n - 1 and l1[p + 1] == '#':
p += 1
l1[p] = 0
else:
break
elif l == 2:
if l1[p] == '#':
l = 1
l1[p] = 0
elif p != n - 1 and l2[p + 1] == '#':
p += 1
l2[p] = 0
else:
break
t = int(input())
for j in range(0, t):
n = int(input())
a = str(input())
l1 = strtolist(a)
l1b = strtolist(a)
a = str(input())
l2 = strtolist(a)
l2b = strtolist(a)
mainthing()
# This part checks if there are any black tiles left.
r = False
for i in range(0, n):
if l1[i] == '#' or l2[i] == '#':
r = True
break
l1 = l2b
l2 = l1b
mainthing()
# This part checks if there are any black tiles left.
r2 = False
for i in range(0, n):
if l1[i] == '#' or l2[i] == '#':
r2 = True
break
if not r or not r2:
print('yes')
else:
print('no')
View More Similar Problems
Tree: Preorder Traversal
Complete the preorder function in the editor below, which has 1 parameter: a pointer to the root of a binary tree. It must print the values in the tree's preorder traversal as a single line of space-separated values. Input Format Our test code passes the root node of a binary tree to the preOrder function. Constraints 1 <= Nodes in the tree <= 500 Output Format Print the tree's
View Solution →Tree: Postorder Traversal
Complete the postorder function in the editor below. It received 1 parameter: a pointer to the root of a binary tree. It must print the values in the tree's postorder traversal as a single line of space-separated values. Input Format Our test code passes the root node of a binary tree to the postorder function. Constraints 1 <= Nodes in the tree <= 500 Output Format Print the
View Solution →Tree: Inorder Traversal
In this challenge, you are required to implement inorder traversal of a tree. Complete the inorder function in your editor below, which has 1 parameter: a pointer to the root of a binary tree. It must print the values in the tree's inorder traversal as a single line of space-separated values. Input Format Our hidden tester code passes the root node of a binary tree to your $inOrder* func
View Solution →Tree: Height of a Binary Tree
The height of a binary tree is the number of edges between the tree's root and its furthest leaf. For example, the following binary tree is of height : image Function Description Complete the getHeight or height function in the editor. It must return the height of a binary tree as an integer. getHeight or height has the following parameter(s): root: a reference to the root of a binary
View Solution →Tree : Top View
Given a pointer to the root of a binary tree, print the top view of the binary tree. The tree as seen from the top the nodes, is called the top view of the tree. For example : 1 \ 2 \ 5 / \ 3 6 \ 4 Top View : 1 -> 2 -> 5 -> 6 Complete the function topView and print the resulting values on a single line separated by space.
View Solution →Tree: Level Order Traversal
Given a pointer to the root of a binary tree, you need to print the level order traversal of this tree. In level-order traversal, nodes are visited level by level from left to right. Complete the function levelOrder and print the values in a single line separated by a space. For example: 1 \ 2 \ 5 / \ 3 6 \ 4 F
View Solution →