Computer Game

Problem Statement - link #

Monocarp is playing a computer game. Now he wants to complete the first level of this game.

A level is a rectangular grid of 2 rows and n columns. Monocarp controls a character, which starts in cell (1,1) — at the intersection of the 1-st row and the 1-st column.

Monocarp’s character can move from one cell to another in one step if the cells are adjacent by side and/or corner. Formally, it is possible to move from cell (x1,y1) to cell (x2,y2) in one step if |x1−x2|≤1 and |y1−y2|≤1. Obviously, it is prohibited to go outside the grid.

There are traps in some cells. If Monocarp’s character finds himself in such a cell, he dies, and the game ends.

To complete a level, Monocarp’s character should reach cell (2,n) — at the intersection of row 2 and column n.

Help Monocarp determine if it is possible to complete the level.

Input #

The first line contains a single integer t (1≤t≤100) — the number of test cases. Then the test cases follow. Each test case consists of three lines.

The first line contains a single integer n (3≤n≤100) — the number of columns.

The next two lines describe the level. The i-th of these lines describes the i-th line of the level — the line consists of the characters '0' and '1'. The character '0' corresponds to a safe cell, the character '1' corresponds to a trap cell.

Additional constraint on the input: cells (1,1) and (2,n) are safe.

Output #

For each test case, output YES if it is possible to complete the level, and NO otherwise.

Examples #

Example 1:

Input: 
4
3
000
000
4
0011
1100
4
0111
1110
6
010101
101010
Output: 
YES
YES
NO
YES

Note #

Consider the example from the statement.

In the first test case, one of the possible paths is (1,1)→(2,2)→(2,3)

In the second test case, one of the possible paths is (1,1)→(1,2)→(2,3)→(2,4)

In the fourth test case, one of the possible paths is (1,1)→(2,2)→(1,3)→(2,4)→(1,5)→(2,6)

Solution #

#include<bits/stdc++.h>
using namespace std;

int t,n,i;
bool check;
vector<string> row(2);

void solution(){
    cin >> n ;
    cin >> row[0] ;
    cin >> row[1] ;
    check = false;

    for(i=0; i<n; i++)
        if(row[0][i] == '1' && row[1][i] == '1')
            check = true;
    
    cout << (check?"NO":"YES") << endl;  
}

int main(){
    ios_base::sync_with_stdio(0), cin.tie(0);
    cin>>t;
    while(t--) 
        solution();
    return 0;
}