Robot Return to Origin in C++

Suppose there is a robot and its starting position is (0, 0). If we have a sequence of its movements, we have to check whether this robot ends up at (0, 0) after it completes its moves.

The move sequence is given as a string, and the character moves[i] represents its ith move. Symbols are R for right, L for left, U for up, and D for down. If the robot returns to the origin after it finishes all of its moves, return true. Otherwise, return false.

So, if the input is like "RRULLD", then the output will be true, Right two-unit, then go up, then left two unit then again down, so this is the starting position.

To solve this, we will follow these steps −

l := size of moves array
if l is same as 0, then −
- return true
lft := 0, up := 0
for initialize i := 0, when i < l, update (increase i by 1), do −
- if moves[i] is same as 'L', then −
  - (increase lft by 1)
- if moves[i] is same as 'R', then −
  - (decrease lft by 1)
- if moves[i] is same as 'U', then −
  - (increase up by 1)
- if moves[i] is same as 'D', then −
  - (decrease up by 1)
if lft is same as 0 and up is same as 0, then −
- return true
return false

Example

Let us see the following implementation to get a better understanding −

Live Demo

#include <bits/stdc++.h>
using namespace std;
class Solution {
public:
   bool judgeCircle(string moves) {
      int l = moves.length();
      if (l == 0) {
         return true;
      }
      int lft = 0, up = 0;
      for (int i = 0; i < l; i++) {
         if (moves[i] == 'L') {
            lft++;
         }
         if (moves[i] == 'R') {
            lft--;
         }
         if (moves[i] == 'U') {
            up++;
         }
         if (moves[i] == 'D') {
            up--;
         }
      }
      if (lft == 0 && up == 0) {
         return true;
      }
      return false;
   }
};
main(){
   Solution ob;
   cout << (ob.judgeCircle("RRULLD"));
}