Given a string containing only L’s and R’s which represents left rotation and right rotation respectively. The task is to find the final direction of pivot(i.e N/E/S/W). Let a pivot is pointed towards the north(N) in a compass.
Examples:
Input: str = "LLRLRRL" Output: W In this input string we rotate pivot to left when a L char is encountered and right when R is encountered. Input: str = "LL" Output: S
Approach:
- Use a counter that incremented on seeing R and decremented on seeing L.
- Finally, use modulo on the counter to get the direction.
- If the count is negative then directions will be different. Check the code for negative as well.
Below is the implementation of the above approach:
C++
// CPP implementation of above approach #include <bits/stdc++.h> using namespace std;
// Function to find the final direction string findDirection(string s) { int count = 0;
string d = "" ;
for ( int i = 0; i < s.length(); i++) {
if (s[0] == '\n' )
return NULL;
if (s[i] == 'L' )
count--;
else {
if (s[i] == 'R' )
count++;
}
}
// if count is positive that implies
// resultant is clockwise direction
if (count > 0) {
if (count % 4 == 0)
d = "N" ;
else if (count % 4 == 1)
d = "E" ;
else if (count % 4 == 2)
d = "S" ;
else if (count % 4 == 3)
d = "W" ;
}
// if count is negative that implies
// resultant is anti-clockwise direction
if (count < 0) {
if (count % 4 == 0)
d = "N" ;
else if (count % 4 == -1)
d = "W" ;
else if (count % 4 == -2)
d = "S" ;
else if (count % 4 == -3)
d = "E" ;
}
return d;
} // Driver code int main()
{ string s = "LLRLRRL" ;
cout << (findDirection(s)) << endl;
s = "LL" ;
cout << (findDirection(s)) << endl;
} // This code is contributed by // SURENDRA_GANGWAR |
Java
// Java implementation of above approach import java.util.*;
class GFG {
// Function to find the final direction
static String findDirection(String s)
{
int count = 0 ;
String d = "" ;
for ( int i = 0 ; i < s.length(); i++) {
if (s.charAt( 0 ) == '\n' )
return null ;
if (s.charAt(i) == 'L' )
count--;
else {
if (s.charAt(i) == 'R' )
count++;
}
}
// if count is positive that implies
// resultant is clockwise direction
if (count > 0 ) {
if (count % 4 == 0 )
d = "N" ;
else if (count % 4 == 1 )
d = "E" ;
else if (count % 4 == 2 )
d = "S" ;
else if (count % 4 == 3 )
d = "W" ;
}
// if count is negative that implies
// resultant is anti-clockwise direction
if (count < 0 ) {
if (count % 4 == 0 )
d = "N" ;
else if (count % 4 == - 1 )
d = "W" ;
else if (count % 4 == - 2 )
d = "S" ;
else if (count % 4 == - 3 )
d = "E" ;
}
return d;
}
// Driver code
public static void main(String[] args)
{
String s = "LLRLRRL" ;
System.out.println(findDirection(s));
s = "LL" ;
System.out.println(findDirection(s));
}
} |
Python3
# Python3 implementation of # the above approach # Function to find the # final direction def findDirection(s):
count = 0
d = ""
for i in range ( len (s)):
if (s[i] = = 'L' ):
count - = 1
else :
if (s[i] = = 'R' ):
count + = 1
# if count is positive that
# implies resultant is clockwise
# direction
if (count > 0 ):
if (count % 4 = = 0 ):
d = "N"
elif (count % 4 = = 1 ):
d = "E"
elif (count % 4 = = 2 ):
d = "S"
elif (count % 4 = = 3 ):
d = "W"
# if count is negative that
# implies resultant is anti-
# clockwise direction
if (count < 0 ):
count * = - 1
if (count % 4 = = 0 ):
d = "N"
elif (count % 4 = = 1 ):
d = "W"
elif (count % 4 = = 2 ):
d = "S"
elif (count % 4 = = 3 ):
d = "E"
return d
# Driver code if __name__ = = '__main__' :
s = "LLRLRRL"
print (findDirection(s))
s = "LL"
print (findDirection(s))
# This code is contributed by 29AjayKumar |
C#
// C# implementation of above approach using System;
class GFG {
// Function to find the final direction
static String findDirection(String s)
{
int count = 0;
String d = "" ;
for ( int i = 0; i < s.Length; i++) {
if (s[0] == '\n' )
return null ;
if (s[i] == 'L' )
count--;
else {
if (s[i] == 'R' )
count++;
}
}
// if count is positive that implies
// resultant is clockwise direction
if (count > 0) {
if (count % 4 == 0)
d = "N" ;
else if (count % 4 == 1)
d = "E" ;
else if (count % 4 == 2)
d = "S" ;
else if (count % 4 == 3)
d = "W" ;
}
// if count is negative that implies
// resultant is anti-clockwise direction
if (count < 0) {
if (count % 4 == 0)
d = "N" ;
else if (count % 4 == -1)
d = "W" ;
else if (count % 4 == -2)
d = "S" ;
else if (count % 4 == -3)
d = "E" ;
}
return d;
}
// Driver code
public static void Main()
{
String s = "LLRLRRL" ;
Console.WriteLine(findDirection(s));
s = "LL" ;
Console.WriteLine(findDirection(s));
}
} // This code is contributed by Shashank |
PHP
<?php // PHP implementation of above approach // Function to find the final direction function findDirection( $s )
{ $count = 0;
$d = "" ;
for ( $i = 0;
$i < strlen ( $s ); $i ++)
{
if ( $s [0] == '\n' )
return null;
if ( $s [ $i ] == 'L' )
$count -= 1;
else
{
if ( $s [ $i ] == 'R' )
$count += 1;
}
}
// if count is positive that implies
// resultant is clockwise direction
if ( $count > 0)
{
if ( $count % 4 == 0)
$d = "N" ;
else if ( $count % 4 == 1)
$d = "E" ;
else if ( $count % 4 == 2)
$d = "S" ;
else if ( $count % 4 == 3)
$d = "W" ;
}
// if count is negative that
// implies resultant is
// anti-clockwise direction
if ( $count < 0)
{
if ( $count % 4 == 0)
$d = "N" ;
else if ( $count % 4 == -1)
$d = "W" ;
else if ( $count % 4 == -2)
$d = "S" ;
else if ( $count % 4 == -3)
$d = "E" ;
}
return $d ;
} // Driver code $s = "LLRLRRL" ;
echo findDirection( $s ). "\n" ;
$s = "LL" ;
echo findDirection( $s ). "\n" ;
// This code is contributed // by ChitraNayal ?> |
Javascript
<script> // Javascript implementation of above approach
// Function to find the final direction
function findDirection(s)
{
let count = 0;
let d = "" ;
for (let i = 0; i < s.length; i++) {
if (s[0] == '\n' )
return null ;
if (s[i] == 'L' )
count--;
else {
if (s[i] == 'R' )
count++;
}
}
// if count is positive that implies
// resultant is clockwise direction
if (count > 0) {
if (count % 4 == 0)
d = "N" ;
else if (count % 4 == 1)
d = "E" ;
else if (count % 4 == 2)
d = "S" ;
else if (count % 4 == 3)
d = "W" ;
}
// if count is negative that implies
// resultant is anti-clockwise direction
if (count < 0) {
if (count % 4 == 0)
d = "N" ;
else if (count % 4 == -1)
d = "W" ;
else if (count % 4 == -2)
d = "S" ;
else if (count % 4 == -3)
d = "E" ;
}
return d;
}
let s = "LLRLRRL" ;
document.write(findDirection(s) + "</br>" );
s = "LL" ;
document.write(findDirection(s));
// This code is contributed by divyeshrabadiya07.
</script> |
Output
W S
Complexity Analysis:
- Time Complexity: O(N) where N is the length of the string
- Auxiliary Space: O(1)
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