Print all n digit patterns formed by mobile Keypad in C++

C++Server Side ProgrammingProgramming

In this problem, we are given a number n and we have to print all N digit patterns formed by pressing the mobile keypad button. While pressing the buttons, we can press only nearby buttons of the current button i.e. only keys left, right, up, down can be pressed.

Let’s see how the old keypad looks like −

12
ABC
3
DEF
4
GHI
5
JKL
6
MNO
7
PQRS
8
TUV
9
WXYZ
*0#

Let’s take an example to understand the problem

Input: n=2
Output: 12 14 21 23 25 32 36 41 45 47 52 54 56 58 63 65 69 74 78 85 87 89 80 96 98

To solve this problem, we will use a depth-first search (DFS). in DFS, we will select one by one all keys of our keypad as the first digit of the number. Now, we will generate the rest digits of the number by using DFS (which allows left, right, up or down key stokes).

Example

Program to implement the above solution −

 Live Demo

#include <bits/stdc++.h>
using namespace std;
bool isSafe(int x, int y, bool Visited[][3]) {
   return (x >= 0 && x < 4 && y >= 0 && y < 3 && !Visited[x][y]);
}
void searchNumber(bool visited[][3], int Keypad[][3], int n, string pattern, int x, int y) {
   pattern.push_back((Keypad[x][y] + '0'));
   if (pattern.size() == n) {
      cout<<pattern<<"\t";
      return;
   }
   static int row[] = { 0, 1, 0, -1 };
   static int col[] = { 1, 0, -1, 0 };
   visited[x][y] = true;
   for (int k = 0; k < 4; k++)
      if (isSafe(x + row[k], y + col[k], visited) && Keypad[x + row[k]][y + col[k]] != -1)
   searchNumber(visited, Keypad, n, pattern, x + row[k], y + col[k]);
   visited[x][y] = false;
   pattern.pop_back();
}
void GenerateNDigitNumber(int Keypad[][3], int n) {
   bool visited[4][3];
   memset(visited, false, sizeof(visited));
   for (int i = 0; i < 4; i++)
      for (int j = 0; j < 3; j++)
         if (Keypad[i][j] != -1)
            searchNumber(visited, Keypad, n, "", i, j);
}
int main() {
   int Keypad[4][3] ={
      { 1, 2, 3 },
      { 4, 5, 6 },
      { 7, 8, 9 },
      { -1, 0, -1 }
   };
   int n = 2;
   cout<<"All "<<n<<" digit number generated from keypad are :\n";
   GenerateNDigitNumber(Keypad, n);
   return 0;
}

Output

All 2 digit number generated from keypad are −
12 14 23 25 21 36 32 45 47 41 56 58 54 52 69 65 63 78 74 89 80 87 85 98 96 08
raja
Published on 22-Jan-2020 15:16:34
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