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Print all numbers in given range having digits in strictly increasing order
• Difficulty Level : Hard
• 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)

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