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247. Strobogrammatic Number II (Medium)

Given an integer n, return all the strobogrammatic numbers that are of length n. You may return the answer in any order.

A strobogrammatic number is a number that looks the same when rotated 180 degrees (looked at upside down).

 

Example 1:

Input: n = 2
Output: ["11","69","88","96"]

Example 2:

Input: n = 1
Output: ["0","1","8"]

 

Constraints:

• 1 <= n <= 14

Companies:
Facebook, Uber

Related Topics:
Array, String, Recursion

Similar Questions:

• Strobogrammatic Number (Easy)
• Strobogrammatic Number III (Hard)
• Sum of k-Mirror Numbers (Hard)

Solution 1. DFS

The strobogrammatic numbers are very sparse, so looping through all the n-digit numbers and test if they are strobogrammatic number is inefficient. We should try generating them.

We can use recursion/DFS/backtracking. In each recursion, we fill the ith digit ([0,n/2]), and fill the corresponding j = n-1-ith digit.

// OJ: https://leetcode.com/problems/strobogrammatic-number-ii/
// Author: github.com/lzl124631x
// Time: O(5^(N/2))
// Space: O(N)
const char pairs[5][2] = {{'0','0'},{'1','1'},{'8','8'},{'6','9'},{'9','6'}};
class Solution {
public:
    vector<string> findStrobogrammatic(int n) {
        string s(n, '0');
        vector<string> ans;
        function<void(int)> dfs = [&](int i) {
            if (i == (n + 1) / 2) {
                ans.push_back(s);
                return;
            }
            int j = n - 1 - i;
            for (auto &[a, b] : pairs) {
                if (i == j && a != b) continue;
                if (i == 0 && n > 1 && a == '0') continue;
                s[i] = a;
                s[j] = b;
                dfs(i + 1);
            }
        };
        dfs(0);
        return ans;
    }
};