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

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++

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// 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;
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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

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Output:

The numbers are: 1 2 3 4 5 6 7 8 9 10 11

Time complexity: O(36* 230)



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