# Print all numbers in given range having digits in strictly increasing order

Last Updated : 23 May, 2021

Given two positive integers L and R, the task is to print the numbers in the range [L, R] which have their digits in strictly increasing order.

Examples:

Input: L = 10, R = 15
Output: 12 13 14 15
Explanation:
In the range [10, 15], only the numbers {12, 13, 14, 15} have their digits in strictly increasing order.

Input: L = 60, R = 70
Output: 67 68 69
Explanation:
In the range [60, 70], only the numbers {67, 68, 69} have their digits in strictly increasing order.

Approach: The idea is to iterate over the range [L, R] and for each number in this range check if digits of this number are in strictly increasing order or not. If yes then print that number else check for the next number.

Below is the implementation of the above approach:

## C++

 // C++ program for the above approach #include using namespace std;   // Function to print all numbers // in the range [L, R] having digits // in strictly increasing order void printNum(int L, int R) {     // Iterate over the range     for (int i = L; i <= R; i++) {           int temp = i;         int c = 10;         int flag = 0;           // Iterate over the digits         while (temp > 0) {               // Check if the current digit             // is >= the previous digit             if (temp % 10 >= c) {                   flag = 1;                 break;             }               c = temp % 10;             temp /= 10;         }           // If the digits are in         // ascending order         if (flag == 0)             cout << i << " ";     } }   // Driver Code int main() { // Given range L and R     int L = 10, R = 15;   // Function Call     printNum(L, R);     return 0; }

## Java

 // Java program for the above approach import java.util.*;   class GFG{   // Function to print all numbers // in the range [L, R] having digits // in strictly increasing order static void printNum(int L, int R) {           // Iterate over the range     for(int i = L; i <= R; i++)     {         int temp = i;         int c = 10;         int flag = 0;           // Iterate over the digits         while (temp > 0)         {                           // Check if the current digit             // is >= the previous digit             if (temp % 10 >= c)             {                 flag = 1;                 break;             }                           c = temp % 10;             temp /= 10;         }                   // If the digits are in         // ascending order         if (flag == 0)             System.out.print(i + " ");     } }   // Driver code public static void main(String[] args) {           // Given range L and R     int L = 10, R = 15;           // Function call     printNum(L, R); } }   // This code is contributed by offbeat

## Python3

 # Python3 program for the above approach   # Function to print all numbers in # the range [L, R] having digits # in strictly increasing order def printNum(L, R):           # Iterate over the range     for i in range(L, R + 1):         temp = i         c = 10         flag = 0           # Iterate over the digits         while (temp > 0):               # Check if the current digit             # is >= the previous digit             if (temp % 10 >= c):                 flag = 1                 break                           c = temp % 10             temp //= 10                   # If the digits are in         # ascending order         if (flag == 0):             print(i, end = " ")       # Driver Code   # Given range L and R L = 10 R = 15   # Function call printNum(L, R)   # This code is contributed by code_hunt

## C#

 // C# program for the above approach using System;   class GFG{   // Function to print all numbers // in the range [L, R] having digits // in strictly increasing order static void printNum(int L, int R) {           // Iterate over the range     for(int i = L; i <= R; i++)     {         int temp = i;         int c = 10;         int flag = 0;           // Iterate over the digits         while (temp > 0)         {                           // Check if the current digit             // is >= the previous digit             if (temp % 10 >= c)             {                 flag = 1;                 break;             }               c = temp % 10;             temp /= 10;         }           // If the digits are in         // ascending order         if (flag == 0)             Console.Write(i + " ");     } }   // Driver Code public static void Main() {           // Given range L and R     int L = 10, R = 15;       // Function call     printNum(L, R); } }   // This code is contributed by jrishabh99

## Javascript



Output:

12 13 14 15

Time Complexity O(N), N is absolute difference between L and R.
Auxiliary Space: O(1)

Article Tags :
Practice Tags :