3211. Generate Binary Strings Without Adjacent Zeros

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Description

You are given a positive integer n.

A binary string x is valid if all substrings of x of length 2 contain at least one "1".

Return all valid strings with length n, in any order.

 

Example 1:

Input: n = 3

Output: ["010","011","101","110","111"]

Explanation:

The valid strings of length 3 are: "010", "011", "101", "110", and "111".

Example 2:

Input: n = 1

Output: ["0","1"]

Explanation:

The valid strings of length 1 are: "0" and "1".

 

Constraints:

• 1 <= n <= 18

Solutions

Solution 1: DFS

We can enumerate each position $i$ of a binary string of length $n$, and for each position $i$, we can enumerate the possible value $j$ it can take. If $j$ is $0$, then we need to check if its previous position is $1$. If it is $1$, we can continue to recurse further; otherwise, it is invalid. If $j$ is $1$, then we directly recurse further.

The time complexity is $O(n \times 2^n)$, where $n$ is the length of the string. Ignoring the space consumption of the answer array, the space complexity is $O(n)$.

Python3

class Solution:
    def validStrings(self, n: int) -> List[str]:
        def dfs(i: int):
            if i >= n:
                ans.append("".join(t))
                return
            for j in range(2):
                if (j == 0 and (i == 0 or t[i - 1] == "1")) or j == 1:
                    t.append(str(j))
                    dfs(i + 1)
                    t.pop()

        ans = []
        t = []
        dfs(0)
        return ans

Java

class Solution {
    private List<String> ans = new ArrayList<>();
    private StringBuilder t = new StringBuilder();
    private int n;

    public List<String> validStrings(int n) {
        this.n = n;
        dfs(0);
        return ans;
    }

    private void dfs(int i) {
        if (i >= n) {
            ans.add(t.toString());
            return;
        }
        for (int j = 0; j < 2; ++j) {
            if ((j == 0 && (i == 0 || t.charAt(i - 1) == '1')) || j == 1) {
                t.append(j);
                dfs(i + 1);
                t.deleteCharAt(t.length() - 1);
            }
        }
    }
}

C++

class Solution {
public:
    vector<string> validStrings(int n) {
        vector<string> ans;
        string t;
        auto dfs = [&](this auto&& dfs, int i) {
            if (i >= n) {
                ans.emplace_back(t);
                return;
            }
            for (int j = 0; j < 2; ++j) {
                if ((j == 0 && (i == 0 || t[i - 1] == '1')) || j == 1) {
                    t.push_back('0' + j);
                    dfs(i + 1);
                    t.pop_back();
                }
            }
        };
        dfs(0);
        return ans;
    }
};

Go

func validStrings(n int) (ans []string) {
	t := []byte{}
	var dfs func(int)
	dfs = func(i int) {
		if i >= n {
			ans = append(ans, string(t))
			return
		}
		for j := 0; j < 2; j++ {
			if (j == 0 && (i == 0 || t[i-1] == '1')) || j == 1 {
				t = append(t, byte('0'+j))
				dfs(i + 1)
				t = t[:len(t)-1]
			}
		}
	}
	dfs(0)
	return
}

TypeScript

function validStrings(n: number): string[] {
    const ans: string[] = [];
    const t: string[] = [];
    const dfs = (i: number) => {
        if (i >= n) {
            ans.push(t.join(''));
            return;
        }
        for (let j = 0; j < 2; ++j) {
            if ((j == 0 && (i == 0 || t[i - 1] == '1')) || j == 1) {
                t.push(j.toString());
                dfs(i + 1);
                t.pop();
            }
        }
    };
    dfs(0);
    return ans;
}