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Print all numbers less than N with at-most 2 unique digits

  • Last Updated : 14 Jul, 2021
Geek Week

Given a number N(less than 10^9). The task is to print all the numbers less than N which are having a maximum of 2 unique digits. 
Note: Number such as 100, 111, 101 are valid as the number of unique digits are at most 2 but 123 is invalid as it has 3 unique digits.
Examples: 
 

Input: N = 12 
Output: The numbers are: 1 2 3 4 5 6 7 8 9 10 11
Input: N = 154 
Output: The numbers are: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 110 111 112 113 114 115 116 117 118 119 121 122 131 133 141 144 151 
 

 

Approach
 

  • Since we are said a maximum of two unique digits, two loops will give us all combination of two digits.
  • A recursive function is written in terms to generate all the numbers using two digits.
  • Call the function with num initially as 0, and generate numbers using recurrence num*10+a and num*10+b, with the base case being num>=n.
  • A set can be used to store all the numbers as there can be duplicate elements which the recursion function generates.

Below is the implementation of the above approach: 
 



C++




// C++ program to print all the numbers
// less than N which have at most 2 unique digits
#include <bits/stdc++.h>
using namespace std;
 
set<int> st;
 
// Function to generate all possible numbers
void generateNumbers(int n, int num, int a, int b)
{
    // If the number is less than n
    if (num > 0 && num < n)
        st.insert(num);
 
    // If the number exceeds
    if (num >= n)
        return;
 
    // Check if it is not the same number
    if (num * 10 + a > num)
        generateNumbers(n, num * 10 + a, a, b);
 
    generateNumbers(n, num * 10 + b, a, b);
}
 
// Function to print all numbers
void printNumbers(int n)
{
    // All combination of digits
    for (int i = 0; i <= 9; i++)
        for (int j = i + 1; j <= 9; j++)
            generateNumbers(n, 0, i, j);
 
    cout << "The numbers are: ";
 
    // Print all numbers
    while (!st.empty()) {
        cout << *st.begin() << " ";
        st.erase(st.begin());
    }
}
 
// Driver code
int main()
{
    int n = 12;
 
    printNumbers(n);
 
    return 0;
}

Java




// Java program to print all the numbers
// less than N which have at most 2 unique digits
import java.util.*;
class GFG{
   
static Set<Integer> st= new HashSet<Integer>();
   
// Function to generate all possible numbers
static void generateNumbers(int n, int num, int a, int b)
{
    // If the number is less than n
    if (num > 0 && num < n)
        st.add(num);
   
    // If the number exceeds
    if (num >= n)
        return;
   
    // Check if it is not the same number
    if (num * 10 + a > num)
        generateNumbers(n, num * 10 + a, a, b);
   
    generateNumbers(n, num * 10 + b, a, b);
}
   
// Function to print all numbers
static void printNumbers(int n)
{
    // All combination of digits
    for (int i = 0; i <= 9; i++)
        for (int j = i + 1; j <= 9; j++)
            generateNumbers(n, 0, i, j);
   
    System.out.print( "The numbers are: ");
   
    // Print all numbers
    System.out.print(st);
     
    st.clear();
}
   
// Driver code
public static void main(String args[])
{
    int n = 12;
   
    printNumbers(n);
   
}
}
// This code is contributed by Arnab Kundu

Python3




# Python 3 program to print all the
# numbers less than N which have at
# most 2 unique digits
st = set()
 
# Function to generate all possible numbers
def generateNumbers(n, num, a, b):
     
    # If the number is less than n
    if (num > 0 and num < n):
        st.add(num)
 
    # If the number exceeds
    if (num >= n):
        return
 
    # Check if it is not the same number
    if (num * 10 + a > num):
        generateNumbers(n, num * 10 + a, a, b)
 
    generateNumbers(n, num * 10 + b, a, b)
 
# Function to print all numbers
def printNumbers(n):
     
    # All combination of digits
    for i in range(10):
        for j in range(i + 1, 10, 1):
            generateNumbers(n, 0, i, j)
 
    print("The numbers are:", end = " ")
     
    # Print all numbers
    l = list(st)
    for i in l:
        print(i, end = " ")
 
# Driver code
if __name__ == '__main__':
    n = 12
 
    printNumbers(n)
 
# This code is contributed by
# Shashank_Sharma

C#




// C# program to print all the numbers
// less than N which have at most 2 unique digits
using System;
using System.Collections.Generic;
 
class GFG
{
     
static SortedSet<int> st = new SortedSet<int>();
     
// Function to generate all possible numbers
static void generateNumbers(int n, int num, int a, int b)
{
    // If the number is less than n
    if (num > 0 && num < n)
        st.Add(num);
     
    // If the number exceeds
    if (num >= n)
        return;
     
    // Check if it is not the same number
    if (num * 10 + a > num)
        generateNumbers(n, num * 10 + a, a, b);
     
    generateNumbers(n, num * 10 + b, a, b);
}
     
// Function to print all numbers
static void printNumbers(int n)
{
    // All combination of digits
    for (int i = 0; i <= 9; i++)
        for (int j = i + 1; j <= 9; j++)
            generateNumbers(n, 0, i, j);
     
    Console.Write( "The numbers are: ");
     
    // Print all numbers
    foreach(int obj in st)
        Console.Write(obj+" ");
     
    st.Clear();
}
     
// Driver code
public static void Main(String []args)
{
    int n = 12;
     
    printNumbers(n);
}
}
 
// This code has been contributed by 29AjayKumar

Javascript




<script>
// Javascript program to print all the numbers
// less than N which have at most 2 unique digits
 
let st= new Set();
 
// Function to generate all possible numbers
function generateNumbers(n,num,a,b)
{
    // If the number is less than n
    if (num > 0 && num < n)
        st.add(num);
     
    // If the number exceeds
    if (num >= n)
        return;
     
    // Check if it is not the same number
    if (num * 10 + a > num)
        generateNumbers(n, num * 10 + a, a, b);
     
    generateNumbers(n, num * 10 + b, a, b);
}
 
// Function to print all numbers
function printNumbers(n)
{
    // All combination of digits
    for (let i = 0; i <= 9; i++)
        for (let j = i + 1; j <= 9; j++)
            generateNumbers(n, 0, i, j);
     
    document.write( "The numbers are: ");
     
    // Print all numbers
    document.write(Array.from(st).sort(function(a,b){return a-b;}).join(" "));
       
    st.clear();
}
 
// Driver code
let n = 12;
 
printNumbers(n);
 
 
// This code is contributed by rag2127
</script>
Output: 
The numbers are: 1 2 3 4 5 6 7 8 9 10 11

 

Time complexity: O(36* 230)
 




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