Given two integers N and K, the task is to count the numbers up to N digits such that no two zeros are adjacents and the range of digits are from 0 to K-1.
Examples:
Input: N = 2, K = 3
Output: 8
Explanation:
There are 8 such numbers such that digits are from 0 to 2 only, without any adjacent 0s: {1, 2, 10, 11, 12, 20, 21, 22}Input: N = 3, K = 3
Output: 22
Approach: The idea is to use Dynamic Programming to solve this problem.
Let DP[i][j] be the number of desirable numbers up to ith digit of the number, and its last digit as j.
Observations:
- The number of ways to fill a place is
- As we know, zero’s can’t be adjacent. So when our last element is 0, means the previous index is filled by 1 way, that is 0. Therefore, current place can only be filled by (K-1) digits.
- If the last place is filled by (K-1) digits, Then current digit place can be filled by either 0 or (K-1) digits.
Base Case:
- If n == 1 and last == K, then we can fill this place by (K-1) digits, return (K-1)
- Else, return 1
Recurrence relation:
When last digit place is not filled by zero then
When Last digit place is filled by zero then
![dp[i][j] = solve(n-1, K)](data:image/svg+xml,%3Csvg%20xmlns='http://www.w3.org/2000/svg'%20viewBox='0%200%20287%2027'%3E%3C/svg%3E "Rendered by QuickLaTeX.com")
![dp[i][j] = (K-1)*solve(n-1, K) + (K-1)*solve(n-1, 1)](https://neveropen.co.uk/wp-content/uploads/2023/11/wardslaus_quicklatex.com-9a553e917ff361d9d7526b0a9e10d191_l3-1.png "Rendered by QuickLaTeX.com")
![dp[i][j] = solve(n-1, K)](https://neveropen.co.uk/wp-content/uploads/2023/11/neveropen_quicklatex.com-f049d77e9e6e546ad9b5bda6aa73bff4_l3.png "Rendered by QuickLaTeX.com")
Below is the implementation of above approach:
C++
// C++ implementation to count the// numbers upto N digits such that// no two zeros are adjacent#include <bits/stdc++.h>using namespace std;int dp[15][10];// Function to count the// numbers upto N digits such that// no two zeros are adjacentint solve(int n, int last, int k){ // Condition to check if only // one element remains if (n == 1) { // If last element is non // zero, return K-1 if (last == k) { return (k - 1); } // If last element is 0 else { return 1; } } // Condition to check if value // calculated already if (dp[n][last]) return dp[n][last]; // If last element is non zero, // then two cases arise, // current element can be either // zero or non zero if (last == k) { // Memoize this case return dp[n][last] = (k - 1) * solve(n - 1, k, k) + (k - 1) * solve(n - 1, 1, k); } // If last is 0, then current // can only be non zero else { // Memoize and return return dp[n][last] = solve(n - 1, k, k); }}// Driver Codeint main(){ // Given N and K int n = 2, k = 3; // Function Call int x = solve(n, k, k) + solve(n, 1, k); cout << x;} |
Java
// Java implementation to count the// numbers upto N digits such that// no two zeros are adjacentimport java.io.*;public class GFG{ static int[][] dp = new int[15][10];// Function to count the numbers // upto N digits such that// no two zeros are adjacentstatic int solve(int n, int last, int k){ // Condition to check if only // one element remains if (n == 1) { // If last element is non // zero, return K-1 if (last == k) { return (k - 1); } // If last element is 0 else { return 1; } } // Condition to check if // value calculated already if (dp[n][last] == 1) return dp[n][last]; // If last element is non zero, // then two cases arise, current // element can be either zero // or non zero if (last == k) { // Memoize this case return dp[n][last] = (k - 1) * solve(n - 1, k, k) + (k - 1) * solve(n - 1, 1, k); } // If last is 0, then current // can only be non zero else { // Memoize and return return dp[n][last] = solve(n - 1, k, k); }}// Driver Codepublic static void main(String[] args){ // Given N and K int n = 2, k = 3; // Function Call int x = solve(n, k, k) + solve(n, 1, k); System.out.print(x);}}// This code is contributed by Ritik Bansal |
Python3
# Python3 implementation to count the # numbers upto N digits such that # no two zeros are adjacent dp = [[0] * 10 for j in range(15)]# Function to count the numbers# upto N digits such that no two# zeros are adjacent def solve(n, last, k): # Condition to check if only # one element remains if (n == 1): # If last element is non # zero, return K-1 if (last == k): return (k - 1) # If last element is 0 else: return 1 # Condition to check if value # calculated already if (dp[n][last]): return dp[n][last] # If last element is non zero, # then two cases arise, current # element can be either zero or # non zero if (last == k): # Memoize this case dp[n][last] = ((k - 1) * solve(n - 1, k, k) + (k - 1) * solve(n - 1, 1, k)) return dp[n][last] # If last is 0, then current # can only be non zero else: # Memoize and return dp[n][last] = solve(n - 1, k, k) return dp[n][last]# Driver code# Given N and Kn = 2k = 3# Function callx = solve(n, k, k) + solve(n, 1, k)print(x)# This code is contributed by himanshu77 |
C#
// C# implementation to count the// numbers upto N digits such that// no two zeros are adjacentusing System;class GFG{ public static int [,]dp = new int[15, 10];// Function to count the numbers // upto N digits such that// no two zeros are adjacentpublic static int solve(int n, int last, int k){ // Condition to check if only // one element remains if (n == 1) { // If last element is non // zero, return K-1 if (last == k) { return (k - 1); } // If last element is 0 else { return 1; } } // Condition to check if // value calculated already if (dp[n, last] == 1) return dp[n, last]; // If last element is non zero, // then two cases arise, current // element can be either zero // or non zero if (last == k) { // Memoize this case return dp[n, last] = (k - 1) * solve(n - 1, k, k) + (k - 1) * solve(n - 1, 1, k); } // If last is 0, then current // can only be non zero else { // Memoize and return return dp[n, last] = solve(n - 1, k, k); }}// Driver Codepublic static void Main(string[] args){ // Given N and K int n = 2, k = 3; // Function Call int x = solve(n, k, k) + solve(n, 1, k); Console.WriteLine(x);}}// This code is contributed by SoumikMondal |
Javascript
<script>// Javascript implementation to count the// numbers upto N digits such that// no two zeros are adjacentvar dp = Array.from(Array(15), () => Array(10).fill(0));// Function to count the// numbers upto N digits such that// no two zeros are adjacentfunction solve(n, last, k){ // Condition to check if only // one element remains if (n == 1) { // If last element is non // zero, return K-1 if (last == k) { return (k - 1); } // If last element is 0 else { return 1; } } // Condition to check if value // calculated already if ((dp[n][last])!=0) return dp[n][last]; // If last element is non zero, // then two cases arise, // current element can be either // zero or non zero if (last == k) { // Memoize this case return dp[n][last] = (k - 1) * solve(n - 1, k, k) + (k - 1) * solve(n - 1, 1, k); } // If last is 0, then current // can only be non zero else { // Memoize and return dp[n][last] = solve(n - 1, k, k); return dp[n][last]; }}// Driver Code// Given N and Kvar n = 2, k = 3;// Function Callvar x = solve(n, k, k) + solve(n, 1, k);document.write(x);</script> |
Output:
8
Time Complexity: O(N)
Auxiliary Space: O(N*10)
Another approach: Using DP Tabulation method ( Iterative approach )
The approach to solve this problem is same but DP tabulation(bottom-up) method is better then Dp + memoization(top-down) because memoization method needs extra stack space of recursion calls.
Steps to solve this problem :
- Create a table DP to store the solution of the subproblems.
- Initialize the table with base cases
- Now Iterate over subproblems to get the value of current problem form previous computation of subproblems stored in DP.
- At last return and print the final solution stored in x , where x = dp[n][k] + dp[n][1] .
Implementation :
C++
// C++ program for above approach#include <bits/stdc++.h>using namespace std;int dp[15][10];// Function to count the// numbers upto N digits such that// no two zeros are adjacentint solve(int n, int k){ // Initialize the base cases for (int i = 1; i <= k; i++) { dp[1][i] = (i == k) ? (k - 1) : 1; } // Fill the table using previously computed values for (int i = 2; i <= n; i++) { for (int j = 1; j <= k; j++) { if (j == k) { dp[i][j] = (k - 1) * dp[i - 1][k] + (k - 1) * dp[i - 1][1]; } else { dp[i][j] = dp[i - 1][k]; } } } // Compute the final result int x = dp[n][k] + dp[n][1]; return x;}// Driver Codeint main(){ // Given N and K int n = 2, k = 3; // Function Call int x = solve(n, k); cout << x;}// this code contributed by bhardwajji |
Java
import java.util.*;public class Main { static int dp[][] = new int[15][10]; // Function to count the // numbers upto N digits such that // no two zeros are adjacent static int solve(int n, int k) { // Initialize the base cases for (int i = 1; i <= k; i++) { dp[1][i] = (i == k) ? (k - 1) : 1; } // Fill the table using previously computed values for (int i = 2; i <= n; i++) { for (int j = 1; j <= k; j++) { if (j == k) { dp[i][j] = (k - 1) * dp[i - 1][k] + (k - 1) * dp[i - 1][1]; } else { dp[i][j] = dp[i - 1][k]; } } } // Compute the final result int x = dp[n][k] + dp[n][1]; return x; } // Driver Code public static void main(String[] args) { // Given N and K int n = 2, k = 3; // Function Call int x = solve(n, k); System.out.println(x); }} |
Python3
# Python program for above approach# Function to count the# numbers upto N digits such that# no two zeros are adjacentdef solve(n, k): dp = [[0 for i in range(10)] for j in range(15)] # Initialize the base cases for i in range(1, k + 1): dp[1][i] = k - 1 if i == k else 1 # Fill the table using previously computed values for i in range(2, n + 1): for j in range(1, k + 1): if j == k: dp[i][j] = (k - 1) * dp[i - 1][k] + (k - 1) * dp[i - 1][1] else: dp[i][j] = dp[i - 1][k] # Compute the final result x = dp[n][k] + dp[n][1] return x# Driver Codeif __name__ == "__main__": # Given N and K n = 2 k = 3 # Function Call x = solve(n, k) print(x) |
Javascript
// JavaScript program for above approachlet dp = Array.from({length: 15}, () => Array.from({length: 10}).fill(0));// Function to count the// numbers upto N digits such that// no two zeros are adjacentfunction solve(n, k){// Initialize the base casesfor (let i = 1; i <= k; i++) {dp[1][i] = (i == k) ? (k - 1) : 1;}// Fill the table using previously computed valuesfor (let i = 2; i <= n; i++) { for (let j = 1; j <= k; j++) { if (j == k) { dp[i][j] = (k - 1) * dp[i - 1][k] + (k - 1) * dp[i - 1][1]; } else { dp[i][j] = dp[i - 1][k]; } }}// Compute the final resultlet x = dp[n][k] + dp[n][1];return x;}// Driver Codelet n = 3, k = 3;// Function Calllet x = solve(n, k);console.log(x); |
C#
// C# program for above approachusing System;class Program{static int[,] dp = new int[15, 10]; // Function to count the// numbers upto N digits such that// no two zeros are adjacentstatic int Solve(int n, int k){ // Initialize the base cases for (int i = 1; i <= k; i++) { dp[1, i] = (i == k) ? (k - 1) : 1; } // Fill the table using previously computed values for (int i = 2; i <= n; i++) { for (int j = 1; j <= k; j++) { if (j == k) { dp[i, j] = (k - 1) * dp[i - 1, k] + (k - 1) * dp[i - 1, 1]; } else { dp[i, j] = dp[i - 1, k]; } } } // Compute the final result int x = dp[n, k] + dp[n, 1]; return x;}// Driver Codestatic void Main(string[] args){ // Given N and K int n = 2, k = 3; // Function Call int x = Solve(n, k); Console.WriteLine(x);}} |
Output
8
Time Complexity: O(N*K)
Auxiliary Space: O(N*10)
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