Print all n-digit strictly increasing numbers

Given number of digits n in a number, print all n-digit numbers whose digits are strictly increasing from left to right.

Examples:

Input:  n = 2
Output:  
01 02 03 04 05 06 07 08 09 12 13 14 15 16 17 18 19 23 24 25 26 27 28 
29 34 35 36 37 38 39 45 46 47 48 49 56 57 58 59 67 68 69 78 79 89

Input:  n = 3
Output:  
012 013 014 015 016 017 018 019 023 024 025 026 027 028 029 034 
035 036 037 038 039 045 046 047 048 049 056 057 058 059 067 068 
069 078 079 089 123 124 125 126 127 128 129 134 135 136 137 138 
139 145 146 147 148 149 156 157 158 159 167 168 169 178 179 189 
234 235 236 237 238 239 245 246 247 248 249 256 257 258 259 267 
268 269 278 279 289 345 346 347 348 349 356 357 358 359 367 368 
369 378 379 389 456 457 458 459 467 468 469 478 479 489 567 568 
569 578 579 589 678 679 689 789

Input:  n = 1
Output: 0 1 2 3 4 5 6 7 8 9

The idea is to use recursion. We start from the leftmost position of a possible N-digit number and fill it from set of all digits greater than its previous digit. i.e. fill current position with digits (i to 9] where i is its previous digit. After filling current position, we recurse for next position with strictly increasing numbers.

Below is implementation of above idea –

CPP

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// C++ program to print all n-digit numbers whose digits
// are strictly increasing from left to right
#include <bits/stdc++.h>
using namespace std;
  
// Function to print all n-digit numbers whose digits
// are strictly increasing from left to right.
// out   --> Stores current output number as string
// start --> Current starting digit to be considered
void findStrictlyIncreasingNum(int start, string out, int n)
{
    // If number becomes N-digit, print it
    if (n == 0)
    {
        cout << out << " ";
        return;
    }
  
    // start from (prev digit + 1) till 9
    for (int i = start; i <= 9; i++)
    {
        // append current digit to number
        string str = out + to_string(i);
  
        // recurse for next digit
        findStrictlyIncreasingNum(i + 1, str, n - 1);
    }
}
  
// Driver code
int main()
{
    int n = 3;
    findStrictlyIncreasingNum(0, "", n);
    return 0;
}

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Java

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// Java program to print all n-digit numbers whose digits
// are strictly increasing from left to right
import java.io.*;
  
class Increasing
{
    // Function to print all n-digit numbers whose digits
    // are strictly increasing from left to right.
    // out   --> Stores current output number as string
    // start --> Current starting digit to be considered
    void findStrictlyIncreasingNum(int start, String out, int n)
    {
        // If number becomes N-digit, print it
        if (n == 0)
        {
            System.out.print(out + " ");
            return;
        }
   
        // start from (prev digit + 1) till 9
        for (int i = start; i <= 9; i++)
        {
            // append current digit to number
            String str = out + Integer.toString(i);
   
            // recurse for next digit
            findStrictlyIncreasingNum(i + 1, str, n - 1);
        }
    }
  
    // Driver code for above function
    public static void main(String args[])throws IOException
    {
        Increasing obj = new Increasing();
        int n = 3;
        obj.findStrictlyIncreasingNum(0, " ", n);
    
}

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C#

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// C# program to print all n-digit numbers
// whose digits are strictly increasing 
// from left to right
using System;
  
class GFG {
      
    // Function to print all n-digit numbers
    // whose digits are strictly increasing 
    // from left to right. out --> Stores 
    // current output number as string
    // start --> Current starting digit to
    // be considered
    static void findStrictlyIncreasingNum(int start,
                                  string Out, int n)
    {
          
        // If number becomes N-digit, print it
        if (n == 0)
        {
            Console.Write(Out + " ");
            return;
        }
  
        // start from (prev digit + 1) till 9
        for (int i = start; i <= 9; i++)
        {
              
            // append current digit to number
            string str = Out + Convert.ToInt32(i);
  
            // recurse for next digit
            findStrictlyIncreasingNum(i + 1, str, n - 1);
        }
    }
  
    // Driver code for above function
    public static void Main()
    {
        int n = 3;
        findStrictlyIncreasingNum(0, " ", n);
    
}
  
// This code is contributed by Sam007.

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Output:

012 013 014 015 016 017 018 019 023 024 025 026 027 028 029 034 
035 036 037 038 039 045 046 047 048 049 056 057 058 059 067 068 
069 078 079 089 123 124 125 126 127 128 129 134 135 136 137 138 
139 145 146 147 148 149 156 157 158 159 167 168 169 178 179 189 
234 235 236 237 238 239 245 246 247 248 249 256 257 258 259 267 
268 269 278 279 289 345 346 347 348 349 356 357 358 359 367 368 
369 378 379 389 456 457 458 459 467 468 469 478 479 489 567 568 
569 578 579 589 678 679 689 789

Exercise: Print all n-digit numbers whose digits are strictly decreasing from left to right.

This article is contributed by Aditya Goel. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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Improved By : Sam007



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