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 –
C++
#include <bits/stdc++.h>
using namespace std;
void findStrictlyIncreasingNum( int start, string out, int n)
{
if (n == 0)
{
cout << out << " " ;
return ;
}
for ( int i = start; i <= 9; i++)
{
string str = out + to_string(i);
findStrictlyIncreasingNum(i + 1, str, n - 1);
}
}
int main()
{
int n = 3;
findStrictlyIncreasingNum(0, "" , n);
return 0;
}
|
Java
import java.io.*;
class Increasing
{
void findStrictlyIncreasingNum( int start, String out, int n)
{
if (n == 0 )
{
System.out.print(out + " " );
return ;
}
for ( int i = start; i <= 9 ; i++)
{
String str = out + Integer.toString(i);
findStrictlyIncreasingNum(i + 1 , str, n - 1 );
}
}
public static void main(String args[]) throws IOException
{
Increasing obj = new Increasing();
int n = 3 ;
obj.findStrictlyIncreasingNum( 0 , " " , n);
}
}
|
Python3
def findStrictlyIncreasingNum(start, out, n):
if (n = = 0 ):
print (out, end = " " )
return
for i in range (start, 10 ):
str1 = out + str (i)
findStrictlyIncreasingNum(i + 1 ,
str1, n - 1 )
n = 3
findStrictlyIncreasingNum( 0 , "", n)
|
C#
using System;
class GFG {
static void findStrictlyIncreasingNum( int start,
string Out, int n)
{
if (n == 0)
{
Console.Write(Out + " " );
return ;
}
for ( int i = start; i <= 9; i++)
{
string str = Out + Convert.ToInt32(i);
findStrictlyIncreasingNum(i + 1, str, n - 1);
}
}
public static void Main()
{
int n = 3;
findStrictlyIncreasingNum(0, " " , n);
}
}
|
Javascript
<script>
function findStrictlyIncreasingNum(start,out,n)
{
if (n == 0)
{
document.write(out + " " );
return ;
}
for (let i = start; i <= 9; i++)
{
let str = out + i.toString();
findStrictlyIncreasingNum(i + 1, str, n - 1);
}
}
let n = 3;
findStrictlyIncreasingNum(0, " " , n);
</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.
Last Updated :
01 Feb, 2022
Like Article
Save Article
Share your thoughts in the comments
Please Login to comment...