Print shortest path to print a string on screen
Given a screen containing alphabets from A-Z, we can go from one character to another characters using a remote. The remote contains left, right, top and bottom keys.
Find shortest possible path to type all characters of given string using the remote. Initial position is top left and all characters of input string should be printed in order.
Screen:
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: “GEEK” Output: Move Down Move Right Press OK Move Up Move Right Move Right Move Right Press OK Press OK Move Left Move Left Move Left Move Left Move Down Move Down Press OK
The idea is to consider screen as 2D-matrix of characters. Then we consider all characters of given string one by one and print out the shortest path between current character and next character in the matrix. In order to find shortest path, we consider the coordinates of current character and next character in the matrix. Based on the difference between x and y values of current and next character’s coordinates, we move left, right, top or bottom. i.e.
If row difference is negative, we move up If row difference is positive, we move down If column difference is negative, we go left If column difference is positive, we go right
Below is implementation of above idea
C++
// C++ program to print shortest possible path to // type all characters of given string using a remote #include <iostream> using namespace std; // Function to print shortest possible path to // type all characters of given string using a remote void printPath(string str) { int i = 0; // start from character 'A' present at position (0, 0) int curX = 0, curY = 0; while (i < str.length()) { // find coordinates of next character int nextX = (str[i] - 'A' ) / 5; int nextY = (str[i] - 'B' + 1) % 5; // Move Up if destination is above while (curX > nextX) { cout << "Move Up" << endl; curX--; } // Move Left if destination is to the left while (curY > nextY) { cout << "Move Left" << endl; curY--; } // Move down if destination is below while (curX < nextX) { cout << "Move Down" << endl; curX++; } // Move Right if destination is to the right while (curY < nextY) { cout << "Move Right" << endl; curY++; } // At this point, destination is reached cout << "Press OK" << endl; i++; } } // Driver code int main() { string str = "COZY" ; printPath(str); return 0; } |
Java
// Java program to print shortest possible path to // type all characters of given string using a remote class GFG { // Function to print shortest possible path to // type all characters of given string using a remote static void printPath(String str) { int i = 0 ; // start from character 'A' present at position (0, 0) int curX = 0 , curY = 0 ; while (i < str.length()) { // find coordinates of next character int nextX = (str.charAt(i) - 'A' ) / 5 ; int nextY = (str.charAt(i) - 'B' + 1 ) % 5 ; // Move Up if destination is above while (curX > nextX) { System.out.println( "Move Up" ); curX--; } // Move Left if destination is to the left while (curY > nextY) { System.out.println( "Move Left" ); curY--; } // Move down if destination is below while (curX < nextX) { System.out.println( "Move Down" ); curX++; } // Move Right if destination is to the right while (curY < nextY) { System.out.println( "Move Right" ); curY++; } // At this point, destination is reached System.out.println( "Press OK" ); i++; } } // driver program public static void main (String[] args) { String str = "COZY" ; printPath(str); } } // Contributed by Pramod Kumar |
Python3
# Python 3 program to print shortest possible # path to type all characters of given string # using a remote # Function to print shortest possible path # to type all characters of given string # using a remote def printPath( str ): i = 0 # start from character 'A' present # at position (0, 0) curX = 0 curY = 0 while (i < len ( str )): # find coordinates of next character nextX = int (( ord ( str [i]) - ord ( 'A' )) / 5 ) nextY = ( ord ( str [i]) - ord ( 'B' ) + 1 ) % 5 # Move Up if destination is above while (curX > nextX): print ( "Move Up" ) curX - = 1 # Move Left if destination is to the left while (curY > nextY): print ( "Move Left" ) curY - = 1 # Move down if destination is below while (curX < nextX): print ( "Move Down" ) curX + = 1 # Move Right if destination is to the right while (curY < nextY): print ( "Move Right" ) curY + = 1 # At this point, destination is reached print ( "Press OK" ) i + = 1 # Driver code if __name__ = = '__main__' : str = "COZY" printPath( str ) # This code is contributed by # Sanjit_Prasad |
C#
// C# program to print shortest // possible path to type all // characters of given string // using a remote using System; class GFG { // Function to print shortest // possible path to type all // characters of given string // using a remote static void printPath(String str) { int i = 0; // start from character 'A' // present at position (0, 0) int curX = 0, curY = 0; while (i < str.Length) { // find coordinates of // next character int nextX = (str[i] - 'A' ) / 5; int nextY = (str[i] - 'B' + 1) % 5; // Move Up if destination // is above while (curX > nextX) { Console.WriteLine( "Move Up" ); curX--; } // Move Left if destination // is to the left while (curY > nextY) { Console.WriteLine( "Move Left" ); curY--; } // Move down if destination // is below while (curX < nextX) { Console.WriteLine( "Move Down" ); curX++; } // Move Right if destination // is to the right while (curY < nextY) { Console.WriteLine( "Move Right" ); curY++; } // At this point, destination // is reached Console.WriteLine( "Press OK" ); i++; } } // Driver Code public static void Main () { String str = "COZY" ; printPath(str); } } // This Code is contributed by nitin mittal. |
PHP
<?php // PHP program to print shortest possible path to // type all characters of given string using a remote // Function to print shortest possible path to // type all characters of given string using a remote function printPath( $str ) { $i = 0; // start from character 'A' present at position (0, 0) $curX = $curY = 0; while ( $i < strlen ( $str )) { // find coordinates of next character $nextX = (int)((ord( $str [ $i ]) - ord( 'A' )) / 5); $nextY = (ord( $str [ $i ]) - ord( 'B' ) + 1) % 5; // Move Up if destination is above while ( $curX > $nextX ) { echo "Move Up\n" ; $curX --; } // Move Left if destination is to the left while ( $curY > $nextY ) { echo "Move Left\n" ; $curY --; } // Move down if destination is below while ( $curX < $nextX ) { echo "Move Down\n" ; $curX ++; } // Move Right if destination is to the right while ( $curY < $nextY ) { echo "Move Right\n" ; $curY ++; } // At this point, destination is reached echo "Press OK\n" ; $i ++; } } // Driver code $str = "COZY" ; printPath( $str ); // This code is contributed by chandan_jnu ?> |
Javascript
<script> // JavaScript program to print shortest // possible path to type all // characters of given string // using a remote // Function to print shortest // possible path to type all // characters of given string // using a remote function printPath(str) { let i = 0; // start from character 'A' // present at position (0, 0) let curX = 0, curY = 0; while (i < str.length) { // find coordinates of // next character let nextX = parseInt((str[i].charCodeAt() - 'A' .charCodeAt()) / 5, 10); let nextY = (str[i].charCodeAt() - 'B' .charCodeAt() + 1) % 5; // Move Up if destination // is above while (curX > nextX) { document.write( "Move Up" + "</br>" ); curX--; } // Move Left if destination // is to the left while (curY > nextY) { document.write( "Move Left" + "</br>" ); curY--; } // Move down if destination // is below while (curX < nextX) { document.write( "Move Down" + "</br>" ); curX++; } // Move Right if destination // is to the right while (curY < nextY) { document.write( "Move Right" + "</br>" ); curY++; } // At this point, destination // is reached document.write( "Press OK" + "</br>" ); i++; } } let str = "COZY" ; printPath(str); </script> |
Output:
Move Right Move Right Press OK Move Down Move Down Move Right Move Right Press OK Move Left Move Left Move Left Move Left Move Down Move Down Move Down Press OK Move Up Move Right Move Right Move Right Move Right Press OK
Time Complexity: O(n*n), where n represents the size of the given string.
Auxiliary Space: O(1), no extra space is required, so it is a constant.
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