A number is called as a Jumping Number if all adjacent digits in it differ by 1. The difference between ‘9’ and ‘0’ is not considered as 1.
All single digit numbers are considered as Jumping Numbers. For example 7, 8987 and 4343456 are Jumping numbers but 796 and 89098 are not.
Given a positive number x, print all Jumping Numbers smaller than or equal to x. The numbers can be printed in any order.
Input: x = 20 Output: 0 1 2 3 4 5 6 7 8 9 10 12 Input: x = 105 Output: 0 1 2 3 4 5 6 7 8 9 10 12 21 23 32 34 43 45 54 56 65 67 76 78 87 89 98 101 Note: Order of output doesn't matter, i,e., numbers can be printed in any order
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One Simple Solution is to traverse all numbers from 0 to x. For every traversed number, check if it is a Jumping number. If yes, then print it. Otherwise ignore it. Time Complexity of this solution is O(x).
Assume that we have a graph where the starting node is 0 and we need to traverse it from the start node to all the reachable nodes.
With the restrictions given in the graph about the jumping numbers, what do you think should be the restrictions defining the next transitions in the graph.
Lets take a example for input x = 90 Start node = 0 From 0, we can move to 1 2 3 4 5 6 7 8 9 [these are not in our range so we don't add it] Now from 1, we can move to 12 and 10 From 2, 23 and 21 From 3, 34 and 32 . . . . . . and so on.
Below is BFS based C++ implementation of above idea.
0 1 10 12 2 21 23 3 32 34 4 5 6 7 8 9
Thanks to Gaurav Ahirwar for above solution.
- Change the above solution to use DFS instead of BFS.
- Extend your solution to print all numbers in sorted order instead of any order.
- Further extend the solution to print all numbers in a given range.
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Improved By : fruit_jam