844. Backspace String Compare - LeetCode Solution


Stack Two Pointer

Python Code:

class Solution:
    def backspaceCompare(self, S: str, T: str) -> bool:
        s = []
        t = []

        for i in range(len(S)):

            if(S[i] == "#"):
                if(len(s) != 0):

                    s.pop()
            else:
                s.append(S[i])

        for i in range(len(T)):

            if (T[i] == "#"):
                if (len(t) != 0):
                    t.pop()
            else:
                t.append(T[i])


        return s==t
        


Comments

Submit
0 Comments
More Questions

352A - Jeff and Digits
1327A - Sum of Odd Integers
1276A - As Simple as One and Two
812C - Sagheer and Nubian Market
272A - Dima and Friends
1352C - K-th Not Divisible by n
545C - Woodcutters
1528B - Kavi on Pairing Duty
339B - Xenia and Ringroad
189A - Cut Ribbon
1182A - Filling Shapes
82A - Double Cola
45A - Codecraft III
1242A - Tile Painting
1663E - Are You Safe
1663D - Is it rated - 3
1311A - Add Odd or Subtract Even
977F - Consecutive Subsequence
939A - Love Triangle
755A - PolandBall and Hypothesis
760B - Frodo and pillows
1006A - Adjacent Replacements
1195C - Basketball Exercise
1206A - Choose Two Numbers
1438B - Valerii Against Everyone
822A - I'm bored with life
9A - Die Roll
1430B - Barrels
279B - Books
1374B - Multiply by 2 divide by 6