https://leetcode.com/problems/push-dominoes/description/?envType=daily-question&envId=2025-05-02
initial
So theres so many conditions to account if current idx is ./L/R. We wanna make the main logic as simple as possible and dont want to handle edge cases.
The mini trick is appending L to left and R to right of the initial string. and use 2 pointer approach of appending length of (right-left-1) of whatever logic. It is right-left-1 cuz we are appending the previous result at the current iteration of for loop
sol
class Solution:
def pushDominoes(self, dominoes: str) -> str:
ans=""
dominoes = 'L'+dominoes+'R'
left=0
for right in range(1,len(dominoes)):
if dominoes[right]=='.':
continue
mid = right-left-1
if left:
ans+=dominoes[left]
if dominoes[left]==dominoes[right]:
ans+=dominoes[left]*mid
elif dominoes[left]=='L' and dominoes[right]=='R':
ans+='.'*mid
else:
ans+='R'*(mid//2)+'.'*(mid%2)+'L'*(mid//2)
left=right
return ans
complexity
n time
n space
