Count of numbers upto N digits formed using digits 0 to K-1 without any adjacent 0s

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 

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 adjacent
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])
        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 Code
int 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 adjacent
class GFG{

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 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 Code
public 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 K
n = 2
k = 3

# Function call
x = 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 adjacent
using 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 adjacent
public 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 Code
public 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 adjacent

var dp = Array.from(Array(15),
() => Array(10).fill(0));

// Function to count the
// numbers upto N digits such that
// no two zeros are adjacent
function 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 K
var n = 2, k = 3;
// Function Call
var x = solve(n, k, k)
        + solve(n, 1, k);
document.write(x);


</script>

Output:

8

Time Complexity: O(N)
Auxiliary Space: O(N*10)