# Jumping Numbers

Posted: 8 Jan, 2021
Difficulty: Hard

## PROBLEM STATEMENT

#### Note:

``````The difference between ‘9’ and ‘0’ is not considered as 1.
``````

#### Example:

``````Let’s say 'N' = 25. The jumping numbers less than or equal to 25 are: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 21, 23. In all these numbers the adjacent digits differ by 1.
``````
##### Input Format:
``````The first line of input contains an integer ‘T’ representing the number of test cases.

The first and the only line of every test case contains the positive integer ‘N’.
``````
##### Output Format:
``````For each test case, print a single line containing less than or equal to 'N' integers representing all the jumping numbers in sorted order.
``````
##### Note:
``````You do not need to print anything, it has already been taken care of. Just implement the given function.
``````
##### Constraints:
``````1 <= T <= 100
1 <= N <= 10^8

Time Limit: 1 sec
`````` Approach 1
• A brute force approach could be to iterate over all the integers from 0 to N.
• For each integer check if its adjacent digits differ by 1. If yes, then the number is a jumping number.

The steps are as follows:

• Iterate over the integers, 0 to ‘N’.
• For each integer, iterate over its digits.
• Check if the current digit and the previous digit differ by 1.
• If all the adjacent digits in the current integer differ by 1, add the integer to the list of jumping numbers.