# Strobogrammatic Number II in C++

Suppose we have a length n. We have to find all strobogrammatic numbers that are of length n.

As we know that a strobogrammatic number is a number that looks the same when rotated 180 degrees.

So, if the input is like n = 2, then the output will be ["11","69","88","96"]

To solve this, we will follow these steps −

• Define an array ret

• if n is odd, then −

• insert "0" at the end of ret

• insert "1" at the end of ret

• insert "8" at the end of ret

• Otherwise

• insert blank string at the end of ret

• for n > 1, update n := n - 2, do −

• Define an array temp

• for initialize i := 0, when i < size of ret, update (increase i by 1), do −

• s := ret[i]

• if n > 3, then −

• insert "0" concatenate s concatenate "0" at the end of temp

• insert "1" concatenate s concatenate "1" at the end of temp

• insert "8" concatenate s concatenate "8" at the end of temp

• insert "6" concatenate s concatenate "9" at the end of temp

• insert "9" concatenate s concatenate "6" at the end of temp

• ret := temp

• return ret

## Example

Let us see the following implementation to get a better understanding −

Live Demo

#include <bits/stdc++.h>
using namespace std;
void print_vector(vector<auto< v){
cout << "[";
for(int i = 0; i<v.size(); i++){
cout << v[i] << ", ";
}
cout << "]"<<endl;
}
class Solution {
public:
vector<string< findStrobogrammatic(int n) {
vector<string< ret;
if (n & 1) {
ret.push_back("0");
ret.push_back("1");
ret.push_back("8");
}
else {
ret.push_back("");
}
for (; n > 1; n -= 2) {
vector<string< temp;
for (int i = 0; i < ret.size(); i++) {
string s = ret[i];
if (n > 3) {
temp.push_back("0" + s + "0");
}
temp.push_back("1" + s + "1");
temp.push_back("8" + s + "8");
temp.push_back("6" + s + "9");
temp.push_back("9" + s + "6");
}
ret = temp;
}
return ret;
}
};
main(){
Solution ob;
print_vector(ob.findStrobogrammatic(3));
}

## Input

3

## Output

[101, 808, 609, 906, 111, 818, 619, 916, 181, 888, 689, 986, ]

Updated on: 18-Nov-2020

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