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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 −
#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, ]
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