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Print all numbers less than N with at-most 2 unique digits in C++
In this problem, we are given an integer N and we have printed all the number less than N with at-most 2 unique digits i.e. maximum 2 different digits can be used to create the number.
Let’s take an example to understand the problem −
Input: N = 17 Output: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
To solve this problem, we will be generating all numbers that have only two unique digits. Our number generating process starts from 0 and ends when our number is equal to or greater than N. For two unique chosen, recursively generate numbers using num*10+i and num*10+j. There may arise some duplicate numbers in this process. So, we can use set to store numbers to avoid it.
Example
This program shows the implementation of our approach to solve the problem
#include <bits/stdc++.h>
using namespace std;
set<int> numbers;
void generateNumbers(int n, int num, int i, int j){
if (num > 0 && num < n)
numbers.insert(num);
if (num >= n)
return;
if (num*10+i > num)
generateNumbers(n, num*10+i, i, j);
generateNumbers(n, num*10+j, i, j);
}
void printUniqueBitNumber(int n){
for (int i = 0; i <= 9; i++)
for (int j = i + 1; j <= 9; j++)
generateNumbers(n, 0, i, j);
cout<<"The numbers are generated are : ";
while (!numbers.empty()) {
cout<<*numbers.begin()<<" ";
numbers.erase(numbers.begin());
}
}
int main(){
int n = 17;
printUniqueBitNumber(n);
return 0;
}Output
The numbers are generated are : 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
Updated on: 22-Jan-2020
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