Given an integer n, using a phone keypad find out all the possible strings that can be made using digits of input n.
Return empty string for numbers 0 and 1.
Note: The order of strings is not important.
Input format:
Integer n
Output format:
All possible strings in different lines
Constraints:
1 <= n <= 10^6
Sample Input:
23
Sample Output:
ad ae af bd be bf cd ce cf
Code:
#include <string>
#include<iostream>
using namespace std;
int ans(string* arr, string* output, int n)
{
if(n==0)
{
return 0;
}
if(n==1)
{
int k = 0;
for(int i=0; i<arr[0].size(); i++)
{
output[k++] = arr[0][i];
}
return k;
}
int small_output_size = ans(arr + 1, output, n-1);
int k = small_output_size;
while(k<small_output_size*arr[0].size())
{
output[k++] = output[k - small_output_size];
}
string s="";
int i=0;
while(i<arr[0].size())
{
int count=0;
while(count!=small_output_size)
{
s+=arr[0][i];
count++;
}
i++;
}
for(int i=0; i<k; i++)
{
output[i]+=s[i];
}
return k;
}
int keypad(long long n, string* output)
{
string* temp = new string[10];
temp[0] = "";
temp[1] = "";
temp[2] = "abc";
temp[3] = "def";
temp[4] = "ghi";
temp[5] = "jkl";
temp[6] = "mno";
temp[7] = "pqrs";
temp[8] = "tuv";
temp[9] = "wxyz";
int size = 0;
long long copied_number = n;
while (copied_number != 0)
{
size += 1;
copied_number /= 10;
}
string* arr = new string[size];
int k = 0;
while (n != 0)
{
arr[k] = temp[n % 10];
n /= 10;
k++;
}
return ans(arr, output, size);
}
int main(){
int num;
cin >> num;
string output[10000];
int count = keypad(num, output);
for(int i = 0; i < count && i < 10000; i++){
cout << output[i] << endl;
}
return 0;
}
Code language: C++ (cpp)
