Given an integer N, the task is to count the numbers up to N having an absolute difference of at most K between any two adjacent digits.
Note: Count of integer 0 for any digits is considerable.
Input: N = 20, K = 2
The required numbers are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13 and 20. Notice that 14, 15, 16, 17, 18 and 19 all have adjacent digit’s absolute difference greater than K = 2 and thus they are not counted.
Input: N = 30, K = 3
All numbers upto 30 except 15, 16, 17, 18, 19, 26, 27, 28, 29 are accepted.
Naive Approach: The idea is to iterate till N and check for all the numbers that the difference of K exists or not. If yes then count it otherwise skip the number and keep iterating.
Time Complexity: O(N)
Auxiliary Space: O(1)
Efficient Approach: This problem can be optimized using Digit Dynamic Programming. Following are the detailed dp states for the given problem.
- In Digit Dp, we consider our number as a sequence of digits so a state position is needed so mark at which state we are currently at. In each recursive call, try to build the sequence from left to right by placing a digit from 0 to 9 and increment the position.
- Previous digit stores only those digits are put which have an absolute difference of atmost K from the previous digit. So, another state needed for the previous digit.
- Tight, state tells whether the number we are trying to build has already become smaller than N, so that in the upcoming recursive calls we can place any digit from 0 to 9. Otherwise, we can place till the digit of N at the current position.
- Initiliaze a boolean variable Start which tells whether the number has started or not. If the number has not yet started, we can start the number by placing digits from 1 to the upper limit with respect to tight in the current position. Otherwise, recur forward without starting the number.
- In each recursive call, set the current digit with respect to the previous digit such that the absolute difference between them never exceeds K. In the base case, return 1 if reached the last position.
Below is the implementation of the above approach:
Time Complexity: O( D * 10 * 2 * 2 * 10), considering N has D digits.
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