Easy Solution. Using a left_count, right_count, and underscore_count.
IntuitionMy intuition is to count the amount of L moves and R moves. Also count the amount of underscores. Whatever move (L, R) has the highest count, add the c
# Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
My intuition is to count the amount of L moves and R moves. Also count the amount of underscores. Whatever move (L, R) has the highest count, add the count of underscores to that move, then subtract the other move to find the furthest point.
EX: left = 5 , right = 3, underscore = 3
    return (left + underscore) - right
# Approach
<!-- Describe your approach to solving the problem. -->
```
# Step 1: Create the variables
underscore_count = 0
left_count = 0
right_count = 0
```
```
# Step 2: Loop through the input and count the moves.
for move in moves:
    if move == 'L':
        left_count += 1
    elif move == 'R':
        right_count += 1
    else:
        underscore_count += 1
```
```
# Step 3: Compare the elements to find the furthest moves.
 if left_count > right_count:
        return (left_count + underscore_count) - right_count
    elif left_count < right_count:
        return (right_count + underscore_count) - left_count
    else:
        return underscore_count
```
# Complexity
- Time complexity: O(N) If the length of moves is N, every element will be seen N times. The "for loop" loops through every move in moves exactly once.
<!-- Add your time complexity here, e.g. $$O(n)$$ -->

- Space complexity: O(1) No additional data structures were used.
<!-- Add your space complexity here, e.g. $$O(n)$$ -->

# Code
```python3 []
class Solution:
    def furthestDistanceFromOrigin(self, moves: str) -> int:
        underscore_count = 0
        left_count = 0
        right_count = 0

        for move in moves:
            if move == 'L':
                left_count += 1
            elif move == 'R':
                right_count += 1
            else:
                underscore_count += 1

     
        if left_count > right_count:
            return (left_count + underscore_count) - right_count
        elif left_count < right_count:
            return (right_count + underscore_count) - left_count
        else:
            return underscore_count
```
['Python3']