Imprima todos los números en el rango dado con dígitos en orden estrictamente creciente

Dados dos enteros positivos L y R , la tarea es imprimir los números en el rango [L, R] que tienen sus dígitos en orden estrictamente creciente.

Ejemplos:

Entrada: L = 10, R = 15 
Salida: 12 13 14 15 
Explicación: 
En el rango [10, 15], solo los números {12, 13, 14, 15} tienen sus dígitos en orden estrictamente creciente.

Entrada: L = 60, R = 70 
Salida: 67 68 69 
Explicación: 
En el rango [60, 70], solo los números {67, 68, 69} tienen sus dígitos en orden estrictamente creciente.

Enfoque: la idea es iterar sobre el rango [L, R] y para cada número en este rango verificar si los dígitos de este número están en orden estrictamente creciente o no. En caso afirmativo, imprima ese número; de lo contrario, verifique el siguiente número.

A continuación se muestra la implementación del enfoque anterior:

// C++ program for the above approach
#include <bits/stdc++.h>
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 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 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# 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

<script>
 
// Javascript program for the above approach
 
// Function to print all numbers
// in the range [L, R] having digits
// in strictly increasing order
function printNum(L, R)
{
     
    // Iterate over the range
    for(let i = L; i <= R; i++)
    {
        let temp = i;
        let c = 10;
        let 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)
            document.write(i + " ");
    }
}
 
// Driver code
 
// Given range L and R
let L = 10, R = 15;
 
// Function call
printNum(L, R);
 
// This code is contributed by sravan kumar
 
</script>

12 13 14 15

Complejidad temporal O(N), N es la diferencia absoluta entre L y R. 
Espacio auxiliar: O(1)