Print shortest path to print a string on screen in C Program.


Given a string, the program must display the shortest path which will print the string over the screen using that shortest path.

Like screen will store alphabets in the format

A B C D E
F G H I J
K L M N O
P Q R S T
U V W X Y
Z

Example

Input: HUP
Output : Move Down
Move Down
Move Down
destination reached
Move Left
Move Left
Move Down
Move Down
Move Down
destination reached
Move Up
destination reached

The approach used here is to store the characters in the n x n matrix and perform the following operation −

If row difference is negative then move up
If row difference is positive then move down
If column difference is negative then go left
If column difference is positive then we go right

Algorithm

START
Step 1 -> Declare Function void printpath(char str[])
   Declare variable int i = 0 and cx=0 and cy=0
   Loop While str[i] != '\0'
      Declare variable as int n1 = (str[i] - 'A') / 5
      Declare variable as int n2 = (str[i] - 'B' + 1) % 5
      Loop while cx > n1
         Print move up
         cx—
      End
      Loop while cy > n2
         Print Move Left
         Cy—
      End
      Loop while cx < n1
         Print move down
         Cx++
      End
      Loop while cy < n2
         Print move down
         Cy++
      End
      Print destination reached
      I++
Step 2 -> in main()
   Declare char str[] = {"HUP"}
   Call printpath(str)
STOP

Example

#include <stdio.h>
void printpath(char str[]){
   int i = 0;
   // start from character 'A' present at position (0, 0)
   int cx = 0, cy = 0;
   while (str[i] != '\0'){
      // find cordinates of next character
      int n1 = (str[i] - 'A') / 5;
      int n2 = (str[i] - 'B' + 1) % 5;
      // Move Up if destination is above
      while (cx > n1){
         printf("Move Up\n");
         cx--;
      }
      // Move Left if destination is to the left
      while (cy > n2){
         printf("Move Left\n");
         cy--;
      }
      // Move down if destination is below
      while (cx < n1){
         printf("Move Down\n");
            cx++;
      }
      // Move Right if destination is to the right
      while (cy < n2){
         printf("Move Down\n");
         cy++;
      }
      // At this point, destination is reached
      printf("destination reached\n");
      i++;
   }
}
int main(int argc, char const *argv[]){
   char str[] = {"HUP"};
   printpath(str);
   return 0;
}

Output

If we run the above program then it will generate the following output −

Move Down
Move Down
Move Down
destination reached
Move Left
Move Left
Move Down
Move Down
Move Down
destination reached
Move Up
destination reached
raja
Published on 22-Aug-2019 12:44:49
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