Mobile Numeric Keypad Problem | Set 2

Given the mobile numeric keypad. You can only press buttons that are up, left, right or down to the current button or can choose to press the same button again. Round corner buttons (i.e. * and # ) are invalid moves.

Mobile-keypad

Given a number N, You have to find the distinct numbers of length N you can dial by starting from any number from 0-9, under the constraint that you can only move up, left, right or down from the last number you pressed, or you can choose to press the same button again.



Examples:

Input: N = 1
Output: 10
0, 1, 2, 3, 4, 5, 6, 7, 8 and 9 are the possible numbers.

Input: N = 2
Output: 36
All the possible numbers are 00, 08, 11, 12, 14, 21, 22, 23, 25, …

Input: N = 6
Output: 7990

Approach: We have seen many solutions to solve this problem here.

Let Xni be the count of n digit numbers ending with i.
So, by this notation,

X10 = 1 which is {0}
X11 = 1 which is {1}
X12 = 1 which is {2}
X13 = 1 which is {3}
and
X20 = 2 which are {00, 80}
X21 = 3 which are {11, 21, 41}
X22 = 4 which are {22, 12, 32, 52}

The central idea is, if you know Xni, what information you can get about Xn + 1j

Let’s see with the help of an example:

Suppose we know, X21 = 3 which are {11, 21, 41}
Because the last digit of all these numbers is 1, let’s look at the possible moves that we can have from 1:


  1. Pressing 1 again.
  2. Pressing 2 (moving right).
  3. Pressing 4 (moving down)

Now we can pick any element from our set {11, 21, 41} and make any valid move:

  1. {111, 211, 411} can be achieved with the first move.
  2. {112, 212, 412} with the second move.
  3. And {114, 214, 414} with the third.

We can see that making any possible move, we get the set of the same size with every move. i.e. There were 3 elements in the set of two digits numbers ending with 1, we got set of the same size(3) with every possible move from 1.

So, it can be seen that X21 contributes in 3 digits numbers as follow:

X31 += X21
X32 += X21
X34 += X21

So, in general if we know about Xni, we know the count it contributes to Xn+1j, where j is every possible move from i.

      \[{X_{n+1}}^{i} = \sum {X_{n}}^{j}\]

where 0<=j<=9 and from j we can have a valid to i

The idea is to first enumerate all possible direction from every given key, and maintain an array of 10 elements where the element at each index store the count of numbers ending with that index.

E.g. Initial values of array are:

Value 1 1 1 1 1 1 1 1 1 1
Index 0 1 2 3 4 5 6 7 8 9


And initial result for n = 1 is Sum of all elements in the array i.e 1+1+1+1+1+1+1+1+1+1 = 10, there are 10 number of 1 digit that can be dialed.

How to update the array for n > 1 ?

Let’s enumerate all directions for all given numbers first:

From To (Possible moves)
0 0, 8
1 1, 2, 4
2 2, 1, 3, 5
3 3, 6, 2
4 4, 1, 7, 5
5 5, 4, 6, 2, 8
6 6, 3, 5, 9
7 7, 4, 8
8 8, 5, 0, 7, 9
9 9, 6, 8

The first row of the table listed above indicates that, if the last digit in the number was zero, we can move to 0 or 8.

Let’s have a detailed look at the approach for N = 2

For N = 1, Arr[10] is {1, 1, 1, 1, 1, 1, 1, 1, 1, 1} indicating there are Arr[i] numbers ending with index i

 {X_{1}}^{i} = 1 \, \, \: \: 0\leq i\leq 9

Lets Create a new array, say Arr2[10] = {0, 0, 0, 0, 0, 0, 0, 0, 0, 0}

Now, for 0, possible moves are 0 and 8.
we already know
     {X_{1}}^{0} = 1

That would contribute 1 to {0, 8} i.e.
     {X_{2}}^{0} += 1
     {X_{2}}^{8} += 1


Arr2[10] = {1, 0, 0, 0, 0, 0, 0, 0, 1, 0}

For 1, Possible moves are 1, 2, 4

we already know
     {X_{1}}^{1} = 1

That would contribute 1 to {1, 2, 4} i.e.

     {X_{2}}^{1} += 1
     {X_{2}}^{2} += 1
     {X_{2}}^{4} += 1

Arr2[10] = {1, 1, 1, 0, 1, 0, 0, 0, 1, 0}

For 2, Possible moves are 2, 1, 3, 4

we already know
     {X_{1}}^{2} = 1

That would contribute 1 to {2, 1, 3, 4}

     {X_{2}}^{2} += 1
     {X_{2}}^{1} += 1
     {X_{2}}^{3} += 1
     {X_{2}}^{4} += 1


Arr2[10] = {1, 2, 2, 1, 1, 0, 0, 0, 1, 0}

and so on ….

Arr2[10] = {2, 3, 4, 3, 4, 5, 4, 3, 5, 3}

Sum = 2+3+4+3+4+5+4+3+5+3 = 36 (For N=2)

Arr2 now holds the values for {X_{2}}^{i} \, \, \: \: 0\leq i\leq 9 and can be considered as starting point for n=3 and the process continues.

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <iostream>
#include <list>
using namespace std;
#define MAX 10
  
// Function to return the count of numbers possible
int getCount(int n)
{
    // Array of list storing possible direction
    // for each number from 0 to 9
    // mylist[i] stores possible moves from index i
    list<int> mylist[MAX];
  
    // Initializing list
    mylist[0].assign({ 0, 8 });
    mylist[1].assign({ 1, 2, 4 });
    mylist[2].assign({ 2, 1, 3, 5 });
    mylist[3].assign({ 3, 6, 2 });
    mylist[4].assign({ 4, 1, 7, 5 });
    mylist[5].assign({ 5, 4, 6, 2, 8 });
    mylist[6].assign({ 6, 3, 5, 9 });
    mylist[7].assign({ 7, 4, 8 });
    mylist[8].assign({ 8, 5, 0, 7, 9 });
    mylist[9].assign({ 9, 6, 8 });
  
    // Storing values for n = 1
    int Arr[MAX] = { 1, 1, 1, 1, 1, 1, 1, 1, 1, 1 };
  
    for (int i = 2; i <= n; i++) {
  
        // To store the values for n = i
        int Arr2[MAX] = { 0 };
  
        // Loop to iterate throuh each index
        for (int j = 0; j < MAX; j++) {
  
            // For each move possible from j
            // Increment the value of possible
            // move positions by Arr[j]
            for (int x : mylist[j]) {
                Arr2[x] += Arr[j];
            }
        }
  
        // Update Arr[] for next iteration
        for (int j = 0; j < MAX; j++)
            Arr[j] = Arr2[j];
    }
  
    // Find the count of numbers possible
    int sum = 0;
    for (int i = 0; i < MAX; i++)
        sum += Arr[i];
  
    return sum;
}
  
// Driver code
int main()
{
    int n = 2;
  
    cout << getCount(n);
  
    return 0;
}

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Java

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// Java implementation of the approach
class GFG
{
    static int MAX = 10;
  
    // Function to return the count of numbers possible
    static int getCount(int n)
    {
        // Array of list storing possible direction
        // for each number from 0 to 9
        // list[i] stores possible moves from index i
          
        int [][] list = new int[MAX][];
          
        // Initializing list
        list[0] = new int [] { 0, 8 };
        list[1] = new int [] { 1, 2, 4 };
        list[2] = new int [] { 2, 1, 3, 5 };
        list[3] = new int [] { 3, 6, 2 };
        list[4] = new int [] { 4, 1, 7, 5 };
        list[5] = new int [] { 5, 4, 6, 2, 8 };
        list[6] = new int [] { 6, 3, 5, 9 };
        list[7] = new int [] { 7, 4, 8 };
        list[8] = new int [] { 8, 5, 0, 7, 9 };
        list[9] = new int [] { 9, 6, 8 };
      
        // Storing values for n = 1
        int Arr[] = new int [] { 1, 1, 1, 1, 1, 1, 1, 1, 1, 1 };
      
        for (int i = 2; i <= n; i++) 
        {
      
            // To store the values for n = i
            int Arr2[] = new int [MAX];
      
            // Loop to iterate throuh each index
            for (int j = 0; j < MAX; j++)
            {
      
                // For each move possible from j
                // Increment the value of possible
                // move positions by Arr[j]
                for (int x = 0; x < list[j].length; x++)
                {
                    Arr2[list[j][x]] += Arr[j];
                }
            }
      
            // Update Arr[] for next iteration
            for (int j = 0; j < MAX; j++)
                Arr[j] = Arr2[j];
        }
      
        // Find the count of numbers possible
        int sum = 0;
        for (int i = 0; i < MAX; i++)
            sum += Arr[i];
      
        return sum;
    }
      
    // Driver code
    public static void main (String[] args) 
    {
      
        int n = 2;
      
        System.out.println(getCount(n));
    }
}
  
// This code is contributed by ihritik

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C#

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// C# implementation of the approach
using System;
  
class GFG
{
    static int MAX = 10;
  
    // Function to return the count of numbers possible
    static int getCount(int n)
    {
        // Array of list storing possible direction
        // for each number from 0 to 9
        // list[i] stores possible moves from index i
        int [][] list = new int[MAX][];
          
        // Initializing list
        list[0] = new int [] { 0, 8 };
        list[1] = new int [] { 1, 2, 4 };
        list[2] = new int [] { 2, 1, 3, 5 };
        list[3] = new int [] { 3, 6, 2 };
        list[4] = new int [] { 4, 1, 7, 5 };
        list[5] = new int [] { 5, 4, 6, 2, 8 };
        list[6] = new int [] { 6, 3, 5, 9 };
        list[7] = new int [] { 7, 4, 8 };
        list[8] = new int [] { 8, 5, 0, 7, 9 };
        list[9] = new int [] { 9, 6, 8 };
      
        // Storing values for n = 1
        int [] Arr = new int [] { 1, 1, 1, 1, 1, 1, 1, 1, 1, 1 };
      
        for (int i = 2; i <= n; i++) 
        {
      
            // To store the values for n = i
            int [] Arr2 = new int [MAX];
      
            // Loop to iterate throuh each index
            for (int j = 0; j < MAX; j++)
            {
      
                // For each move possible from j
                // Increment the value of possible
                // move positions by Arr[j]
                for (int x = 0; x < list[j].Length; x++)
                {
                    Arr2[list[j][x]] += Arr[j];
                }
            }
      
            // Update Arr[] for next iteration
            for (int j = 0; j < MAX; j++)
                Arr[j] = Arr2[j];
        }
      
        // Find the count of numbers possible
        int sum = 0;
        for (int i = 0; i < MAX; i++)
            sum += Arr[i];
      
        return sum;
    }
      
    // Driver code
    public static void Main () 
    {
      
        int n = 2;
      
        Console.WriteLine(getCount(n));
    }
}
  
// This code is contributed by ihritik

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Output:

36

Time Complexity: O(N)
Space Complexity: O(1)



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