Find the direction from given string

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:

1. Use a counter that incremented on seeing R and decremented on seeing L.
2. Finally, use modulo on the counter to get the direction.
3. 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)