Given a string S having N directions in which a person travels. The task is to check if he/she will be able to return to the same place where he/she started. On Day i(1 <= i <= N), he will travel a positive distance in the following direction:
North if the i-th letter of str is N
West if the i-th letter of str is W
South if the i-th letter of str is S
East if the i-th letter of str is E
If he can return back to the place where he starts after nth day, print “YES” else print “NO”.
Examples:
Input: str = “NNNWEWESSS”
Output: YES
On the 1st, 2nd, and 3rd day he goes to north and on the 4th day he goes west, then eventually
returns where he was standing on the 3rd day on the 5th day, then on the 6th day he again goes to
west.On the 7th day he again return exactly to where he was standing on the 5th day.And on the
10th day he returns home safely.
Input: str = “NW”
Output: NO
Approach: There has to be a same number of N as there are a number of S and also the same number of E as there is W. So, count each type of directions given and just check if they are equal or not.
Below is the implementation of the above approach:
C++
// C++ implementation of above approach#include<bits/stdc++.h>using namespace std;int main() { string st = "NNNWEWESSS" ; int len = st.length(); int n = 0 ; // Count of North int s = 0 ; // Count of South int e = 0 ; // Count of East int w = 0 ; // Count of West for (int i = 0; i < len ; i++ ) { if(st[i]=='N') n += 1; if(st[i] == 'S') s += 1; if(st[i] == 'W') w+= 1 ; if(st[i] == 'E') e+= 1 ; } if(n == s && w == e) cout<<("YES")<<endl; else cout<<("NO")<<endl; } // This code is contributed by // Sahil_Shelangia |
Java
// Java implementation of above approachpublic class GFG { public static void main(String args[]) { String st = "NNNWEWESSS" ; int len = st.length(); int n = 0 ; // Count of North int s = 0 ; // Count of South int e = 0 ; // Count of East int w = 0 ; // Count of West for (int i = 0; i < len ; i++ ) { if(st.charAt(i)=='N') n+= 1 ; if(st.charAt(i) == 'S') s+= 1 ; if(st.charAt(i) == 'W') w+= 1 ; if(st.charAt(i) == 'E') e+= 1 ; } if(n == s && w == e) System.out.println("YES"); else System.out.println("NO") ; } // This Code is contributed by ANKITRAI1} |
Python
# Python implementation of above approachst = "NNNWEWESSS"length = len(st)n = 0 # Count of Norths = 0 # Count of Southe = 0 # Count of Eastw = 0 # Count of Westfor i in range(length): if(st[i]=="N"): n+= 1 if(st[i]=="S"): s+= 1 if(st[i]=="W"): w+= 1 if(st[i]=="E"): e+= 1if(n == s and w == e): print("YES")else: print("NO") |
C#
// C# implementation of above approach using System;class GFG { // Main Method public static void Main() { string st = "NNNWEWESSS" ; int len = st.Length; int n = 0 ; // Count of North int s = 0 ; // Count of South int e = 0 ; // Count of East int w = 0 ; // Count of West for (int i = 0; i < len ; i++ ) { if(st[i]=='N') n += 1 ; if(st[i] == 'S') s += 1 ; if(st[i] == 'W') w += 1 ; if(st[i] == 'E') e += 1 ; } if(n == s && w == e) Console.WriteLine("YES"); else Console.WriteLine("NO") ; } } // This code is contributed by Subhadeep |
PHP
<?php// PHP implementation of above approach$st = "NNNWEWESSS";$len = strlen($st); $n = 0; // Count of North $s = 0; // Count of South $e = 0; // Count of East $w = 0; // Count of West for ($i = 0; $i < $len; $i++ ){ if($st[$i] == 'N') $n += 1; if($st[$i] == 'S') $s += 1; if($st[$i] == 'W') $w += 1 ; if($st[$i] == 'E') $e += 1;}if($n == $s && $w == $e) echo "YES\n"; else echo "NO\n";// This code is contributed by// Rajput-Ji?> |
Javascript
<script>// JavaScript implementation of the approach// driver code let st = "NNNWEWESSS" ; let len = st.length; let n = 0 ; // Count of North let s = 0 ; // Count of South let e = 0 ; // Count of East let w = 0 ; // Count of West for (let i = 0; i < len ; i++ ) { if(st[i]=='N') n += 1 ; if(st[i] == 'S') s += 1 ; if(st[i] == 'W') w += 1 ; if(st[i] == 'E') e += 1 ; } if(n == s && w == e) document.write("YES"); else document.write("NO") ; </script> |
YES
Time Complexity: O(N), where N is the length of the given string.
Auxiliary Space: O(1) as constant space is used.
Approach 2:
- The program initializes a string variable moves with a sequence of characters representing moves (north, south, east, west).
- It initializes two integer variables x and y to 0, which represent the current position of the person.
- The program then iterates through each character in the moves string using a range-based for loop.
- Inside the loop, the program checks the value of each character and increments or decrements x or y accordingly.
- If the final value of x and y are both 0, the program prints “YES” to indicate that the person has returned to the starting point, otherwise it prints “NO”.
- Finally, the program returns 0 to indicate successful completion.
Here is the code of above approach:
C++
#include <bits/stdc++.h>using namespace std;int main() { string moves = "NNNWEWESSS"; int x = 0, y = 0; for (char move : moves) { if (move == 'N') y++; else if (move == 'S') y--; else if (move == 'E') x++; else if (move == 'W') x--; } if (x == 0 && y == 0) cout << "YES" << endl; else cout << "NO" << endl; return 0;} |
Python3
# Declare and initialize variablesmoves = "NNNWEWESSS"x = 0y = 0# Loop through each move and update the player's positionfor move in moves: if move == 'N': y += 1 elif move == 'S': y -= 1 elif move == 'E': x += 1 elif move == 'W': x -= 1# Check if the player has returned to the starting positionif x == 0 and y == 0: print("YES")else: print("NO") |
Javascript
// Declare and initialize variableslet moves = "NNNWEWESSS"; let x = 0, y = 0; // Loop through each move and update the player's positionfor (let move of moves) { if (move == 'N') y++; else if (move == 'S') y--; else if (move == 'E') x++; else if (move == 'W') x--; }// Check if the player has returned to the starting positionif (x == 0 && y == 0) { console.log("YES"); }else { console.log("NO");} |
Output:
YES
Time Complexity: O(N), where N is the length of the given string.
Auxiliary Space: O(1) as constant space is used.
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