Count of N-digit numbers with absolute difference of adjacent digits not exceeding K
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.
Examples:
Input: N = 2, K = 1
Output: 26
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, 99Input: N = 3, K = 2
Output: 188
Naive Approach
The simplest approach is to iterate over all N digit numbers and check for every number if the adjacent digits have an absolute difference less than or equal to K.
Time Complexity: O(10N * N)
Efficient Approach:
To optimize the above approach, we need to use a Dynamic Programming approach along with Range Update
- Initialize a DP[][] array where dp[i][j] stores the count of numbers having i digits and ending with j.
- Iterate the array from 2 to N and check if the last digit was j, then the allowed digits for this place are in the range (max(0, j-k), min(9, j+k)). Perform a range update on this range.
- Now use Prefix Sum to get the actual answer.
Below is the implementation of the above approach:
C++
// C++ implementation of // the above approach #include <bits/stdc++.h> using namespace std; // Function to return count // of N-digit numbers with // absolute difference of // adjacent digits not // exceeding K long long getCount( int n, int k) { // For 1-digit numbers, // the count is 10 if (n == 1) return 10; long long dp[n + 1][11]; // dp[i][j] stores the number // of such i-digit numbers // ending in j for ( int i = 0; i <= n; i++) { for ( int j = 0; j < 11; j++) dp[i][j] = 0; } // Initialize count for // 1-digit numbers for ( int i = 1; i <= 9; i++) dp[1][i] = 1; // Compute values for count of // digits greater than 1 for ( int i = 2; i <= n; i++) { for ( int j = 0; j <= 9; j++) { // Find the range of allowed // numbers if last digit is j int l = max(0, j - k); int r = min(9, j + k); // Perform Range update dp[i][l] += dp[i - 1][j]; dp[i][r + 1] -= dp[i - 1][j]; } // Prefix sum to find actual // values of i-digit numbers // ending in j for ( int j = 1; j <= 9; j++) dp[i][j] += dp[i][j - 1]; } // Stores the final answer long long count = 0; for ( int i = 0; i <= 9; i++) count += dp[n][i]; return count; } // Driver Code int main() { int N = 2, K = 1; cout << getCount(N, K); } |
Java
// Java Program to implement // the above approach import java.util.*; class GFG { // Function to return count of such numbers public static long getCount( int n, int k) { // For 1-digit numbers, the count // is 10 irrespective of K if (n == 1 ) return 10 ; // dp[i][j] stores the number // of such i-digit numbers // ending in j long dp[][] = new long [n + 1 ][ 11 ]; // Initialize count for // 1-digit numbers for ( int i = 1 ; i <= 9 ; i++) dp[ 1 ][i] = 1 ; // Compute values for count of // digits greater than 1 for ( int i = 2 ; i <= n; i++) { for ( int j = 0 ; j <= 9 ; j++) { // Find the range of allowed // numbers if last digit is j int l = Math.max( 0 , j - k); int r = Math.min( 9 , j + k); // Perform Range update dp[i][l] += dp[i - 1 ][j]; dp[i][r + 1 ] -= dp[i - 1 ][j]; } // Prefix sum to find actual values // of i-digit numbers ending in j for ( int j = 1 ; j <= 9 ; j++) dp[i][j] += dp[i][j - 1 ]; } // Stores the final answer long count = 0 ; for ( int i = 0 ; i <= 9 ; i++) count += dp[n][i]; return count; } // Driver Code public static void main(String[] args) { int n = 2 , k = 1 ; System.out.println(getCount(n, k)); } } |
Python3
# Python 3 Program to implement # the above approach # Function to return count # of N-digit numbers with # absolute difference of # adjacent digits not # exceeding K def getCount(n, k): # For 1-digit numbers, the # count is 10 if n = = 1 : return 10 # dp[i][j] stores the count of # i-digit numbers ending with j dp = [[ 0 for x in range ( 11 )] for y in range (n + 1 )]; # Initialize count for # 1-digit numbers for i in range ( 1 , 10 ): dp[ 1 ][i] = 1 # Compute values for count # of digits greater than 1 for i in range ( 2 , n + 1 ): for j in range ( 0 , 10 ): # Find the range of allowed # numbers if last digit is j l = max ( 0 , j - k) r = min ( 9 , j + k) # Perform Range update dp[i][l] = dp[i][l] + dp[i - 1 ][j] dp[i][r + 1 ] = dp[i][r + 1 ] - dp[i - 1 ][j] # Prefix sum to find count of # of i-digit numbers ending with j for j in range ( 1 , 10 ): dp[i][j] = dp[i][j] + dp[i][j - 1 ] # Stores the final answer count = 0 for i in range ( 0 , 10 ): count = count + dp[n][i] return count # Driver Code n, k = 2 , 1 print (getCount(n, k)) |
C#
// C# Program to implement // the above approach using System; class GFG { // Function to return the // count of N-digit numbers // with absolute difference of // adjacent digits not exceeding K static long getCount( int n, int k) { // For 1-digit numbers, the // count is 10 if (n == 1) return 10; // dp[i][j] stores the count of // i-digit numbers ending with j long [, ] dp = new long [n + 1, 11]; // Initialize count for // 1-digit numbers for ( int i = 1; i <= 9; i++) dp[1, i] = 1; // Compute values for count of // digits greater than 1 for ( int i = 2; i <= n; i++) { for ( int j = 0; j <= 9; j++) { // Find the range of allowed // numbers with last digit j int l = Math.Max(0, j - k); int r = Math.Min(9, j + k); // Perform Range update dp[i, l] += dp[i - 1, j]; dp[i, r + 1] -= dp[i - 1, j]; } // Prefix sum to count i-digit // numbers ending in j for ( int j = 1; j <= 9; j++) dp[i, j] += dp[i, j - 1]; } // Stores the final answer long count = 0; for ( int i = 0; i <= 9; i++) count += dp[n, i]; return count; } // Driver Code public static void Main() { int n = 2, k = 1; Console.WriteLine(getCount(n, k)); } } |
Javascript
<script> // Javascript implementation of // the above approach // Function to return count // of N-digit numbers with // absolute difference of // adjacent digits not // exceeding K function getCount(n, k) { // For 1-digit numbers, the count // is 10 irrespective of K if (n == 1) return 10; // dp[i][j] stores the number // of such i-digit numbers // ending in j var dp = new Array(n + 1); for ( var i = 0; i < dp.length; i++) dp[i] = Array(11).fill(0); // Initialize count for // 1-digit numbers for (i = 1; i <= 9; i++) dp[1][i] = 1; // Compute values for count of // digits greater than 1 for (i = 2; i <= n; i++) { for (j = 0; j <= 9; j++) { // Find the range of allowed // numbers if last digit is j var l = Math.max(0, j - k); var r = Math.min(9, j + k); // Perform Range update dp[i][l] += dp[i - 1][j]; dp[i][r + 1] -= dp[i - 1][j]; } // Prefix sum to find actual values // of i-digit numbers ending in j for (j = 1; j <= 9; j++) dp[i][j] += dp[i][j - 1]; } // Stores the final answer var count = 0; for (i = 0; i <= 9; i++) count += dp[n][i]; return count; } // Driver Code var n = 2, k = 1; document.write(getCount(n, k)); // This code is contributed by umadevi9616 </script> |
26
Time Complexity: O(N)
Auxiliary Space: O(N)