Roomba - Amazon Top Interview Questions
Problem Statement :
A Roomba robot is currently sitting in a Cartesian plane at (0, 0). You are given a list of its moves that it will make, containing NORTH, SOUTH, WEST, and EAST. Return whether after its moves it will end up in the coordinate (x, y). Constraints n ≤ 100,000 where n is length of moves Example 1 Input moves = ["NORTH", "EAST"] x = 1 y = 1 Output True Explanation Moving north moves it to (0, 1) and moving east moves it to (1, 1) Example 2 Input moves = ["WEST", "EAST"] x = 1 y = 0 Output False Explanation This Roomba will end up at (0, 0)
Solution :
Solution in C++ :
bool solve(vector<string>& moves, int x, int y) {
for (auto& move : moves) {
if (move.front() == 'N')
--y;
else if (move.front() == 'S')
++y;
else if (move.front() == 'W')
++x;
else if (move.front() == 'E')
--x;
}
if (!x && !y) return true;
return false;
}
Solution in Java :
import java.util.*;
class Solution {
public boolean solve(String[] moves, int ax, int ay) {
String px = "EAST";
String nx = "WEST";
String py = "NORTH";
String ny = "SOUTH";
int x = 0;
int y = 0;
for (int i = 0; i < moves.length; i++) {
if (moves[i].equals(px)) {
x++;
continue;
}
if (moves[i].equals(nx)) {
x--;
continue;
}
if (moves[i].equals(py)) {
y++;
continue;
}
if (moves[i].equals(ny)) {
y--;
}
}
if (x == ax && ay == y)
return true;
else
return false;
}
}
Solution in Python :
class Solution:
def solve(self, moves, x, y):
currX, currY = 0, 0
d = {
"EAST": (1, 0),
"WEST": (-1, 0),
"NORTH": (0, 1),
"SOUTH": (0, -1),
}
for i in moves:
dx, dy = d[i]
currX += dx
currY += dy
if currX == x and currY == y:
return True
return False
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