Print all n-digit strictly increasing numbers

01 Feb, 2022
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 –  
C++
Java
Python3
C#
Javascript
// 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 consideredvoid 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 codeint main(){    int n = 3;    findStrictlyIncreasingNum(0, "", n);    return 0;}
// Java program to print all n-digit numbers whose digits// are strictly increasing from left to rightimport 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);    }}
# Python3 program to print all n-digit numbers# whose digits are strictly increasing# from left to right # 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 considereddef findStrictlyIncreasingNum(start, out, n):         # If number becomes N-digit, print    if (n == 0):        print(out, end = " ")        return     # start from (prev digit + 1) till 9    for i in range(start, 10):                 # append current digit to number        str1 = out + str(i)         # recurse for next digit        findStrictlyIncreasingNum(i + 1,                            str1, n - 1) # Driver coden = 3findStrictlyIncreasingNum(0, "", n) # This code is contributed by Mohit Kumar
// C# program to print all n-digit numbers// whose digits are strictly increasing// from left to rightusing 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.
<script> // Javascript program to print all// n-digit numbers whose digits// are strictly increasing from// left to right         // 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    function findStrictlyIncreasingNum(start,out,n)    {        // If number becomes N-digit, print it        if (n == 0)        {            document.write(out + " ");            return;        }            // start from (prev digit + 1) till 9        for (let i = start; i <= 9; i++)        {            // append current digit to number            let str = out + i.toString();                // recurse for next digit            findStrictlyIncreasingNum(i + 1, str, n - 1);        }    }         // Driver code for above function    let n = 3;    findStrictlyIncreasingNum(0, " ", n); // This code is contributed by unknown2108 </script>
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
Time Complexity: O(N!)Auxiliary Space: O(N) 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 write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above. 
Sam007
mohit kumar 29
unknown2108
pankajsharmagfg
sweetyty
adnanirshad158
surinderdawra388
simmytarika5
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