Tactical Conversion Practice Coding Problem

Difficulty:1626

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Tactical Conversion

You are given a binary string $S$ of length $N$ , i.e. each character of $S$ is either $0$ or $1$ .

You would like to convert every character of $S$ into $0$ .
To achieve this, you can perform the following operation:

Choose an index $i$ $(1 \le i \le N)$ such that $S_i = 1$ , and change $S_i$ to $0$ .

However, there is one restriction:
You cannot perform two consecutive operations on adjacent indices.
That is, if you operate on the sequence of indices $i_1, i_2, \ldots, i_k$ , then for each $1 \le j \lt k$ the condition $|i_j - i_{j+1}| \gt 1$ must hold.

Determine whether it is possible to make the entire string consist of only zeros under these conditions.

Input Format

The first line of input contains a single integer $T$ , denoting the number of test cases.
Each test case consists of two lines of input:
- The first line contains a single integer $N$ — the length of the binary string.
- The second line contains the binary string $S$ of length $N$ , consisting only of characters 0 and 1.

Output Format

For each test case, output a single string on a new line — YES if it is possible to convert the entire string to all zeros under the given rule, or NO otherwise.

You may print each character of the string in uppercase or lowercase (for example, the strings YES, yEs, yes, and yeS will all be treated as identical).

Constraints

$1 \leq T \leq 2\cdot 10^5$
$1 \leq N \leq 2\cdot 10^5$
$S$ is a binary string.
The sum of $N$ over all test cases won't exceed $2\cdot 10^5$ .

Sample 1:

Input

Output

Yes
Yes
Yes
No
No

Explanation:

Test case $1$ : No operations are needed.

Test case $2$ : There is a single $1$ at position $3$ .
Simply perform one operation with $i = 3$ , and the string becomes $000$ as desired.

Test case $3$ : There are two $1$ 's, at positions $1$ and $3$ .
Perform one operation with $i = 1$ , and the next operation with $i = 3$ , and we're done.

Test case $4$ : There are three ones, at positions $1, 2, 3$ .
It's not possible to operate on all of them, because:

If our first operation is on index $1$ , the second operation cannot be index $2$ and so must be index $3$ .
But then after index $3$ we cannot operate on index $2$ anyway, so that index will continue to contain a $1$ .
Similarly, the first operation being on index $3$ will leave us unable to operate on index $2$ .
Finally, if the first operation is on index $2$ , then the second operation cannot be on either index $1$ or index $3$ since they're both adjacent to it.

Thus, there's no way to make everything $0$ .

More Info

Time limit1 secs

Memory limit1.5 GB

Source Limit50000 Bytes

Author(s)

archit_adm

Tester(s)

kingmessi

Editorialist(s)

iceknight1093

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                                        #include
                                          <bits/stdc++.h>
                                      
                                        using
                                        namespace
                                        std;
                                      
                                        int
                                        main()
                                        {
                                      
                                        // your code goes here
                                      
                                        }
                                      
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