2833. Furthest Point From Origin

String
Greedy

TIME

O(n)

SPACE

O(1)

Intuition

To maximize distance from origin, use all '_' moves in the same direction that extends the furthest.

Solution

Count L, R, and _ characters
Fixed displacement = |L_count - R_count|
Use all _ to extend in the dominant direction

return abs(L_count - R_count) + underscore_count

Why This Works

L's and R's have fixed directions, giving net displacement |L - R|
All underscores should move in whichever direction is already dominant (or either if tied)
This maximizes the absolute distance from origin

Complexity

Time: O(n)
Space: O(1)

Edge Cases

All underscores → distance = n
No underscores → distance = |L - R|
Equal L and R → distance = underscore_count

Code

class Solution:
    def furthestDistanceFromOrigin(self, moves: str) -> int:
        """
        2833. Furthest Point From Origin
        Time: O(n)
        Space: O(1)
        """
        left = moves.count('L')
        right = moves.count('R')
        underscore = moves.count('_')

        return abs(left - right) + underscore

/**
 * 2833. Furthest Point From Origin
 * Time: O(n)
 * Space: O(1)
 */
class Solution {
    public int furthestDistanceFromOrigin(String moves) {
        int left = 0, right = 0, underscore = 0;

        for (char c : moves.toCharArray()) {
            if (c == 'L') left++;
            else if (c == 'R') right++;
            else underscore++;
        }

        return Math.abs(left - right) + underscore;
    }
}