Given two integers N and K, the task is to find the count of N-digit numbers such that the absolute difference of adjacent digits in the number is not greater than K.
Input: N = 2, K = 1
Explanation: The numbers are 10, 11, 12, 21, 22, 23, 32, 33, 34, 43, 44, 45, 54, 55, 56, 65, 66, 67, 76, 77, 78, 87, 88, 89, 98, 99
Input: N = 3, K = 2
Naive Approach and Dynamic Programming Approach: Refer to Count of N-digit numbers with absolute difference of adjacent digits not exceeding K for the simplest solution and the dynamic programming approach that requires O(N) auxiliary space.
Follow the steps below to optimize the above approach:
- In the above approach, a 2D array dp is initialized where dp[i][j] stores the count of numbers having i digits and ending with j.
- It can be observed that the answer for any length i depends only on the count generated for i – 1. So, instead of a 2D array, initialize an array dp of size of all possible digits, and for every i upto N, dp[j] stores count of such numbers of length i ending with digit j.
- Initialize another array next of size of all possible digits.
- Since the count of single digit numbers ending with a value j is always 1, fill dp with 1 at all indices initially.
- Iterate over the range [2, N], and for every value in the range i, check if the last digit is j, then the allowed digits for this place are in the range (max(0, j-k), min(9, j+k)). Perform a range update:
next[l] = next[l] + dp[j]
next[r + 1] = next[r + 1] – dp[j]
where l and r are max(0, j – k) and min(9, j + k) respectively.
- Once the above step is completed for a particular value of i, compute prefix sum of next and update dp with the values of next.
Below is the implementation of the above approach:
Time Complexity: O(N)
Auxiliary Space: O(1)
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