# 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
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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++

 `// C++ program to print all n-digit numbers whose digits ` `// are strictly increasing from left to right ` `#include ` `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; ` `} `

## Java

 `// 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); ` `    ``}  ` `} `

## Python3

 `# Python3 program to prall n-digit numbers  ` `# whose digits are str1ictly increasing  ` `# from left to right ` ` `  `# Function to prall n-digit numbers  ` `# whose digits are str1ictly increasing ` `# from left to right. ` `# out --> Stores current output  ` `#         number as str1ing ` `# start --> Current starting digit  ` `#           to be considered ` `def` `findStrictlyIncreasingNum(start, out, n): ` `     `  `    ``# If number becomes N-digit, prit ` `    ``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 code ` `n ``=` `3` `findStrictlyIncreasingNum(``0``, "", n) ` ` `  `# This code is contributed by Mohit Kumar `

## C#

 `// 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. `

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.

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Improved By : Sam007, mohit kumar 29

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