Today I have done leetcode's October LeetCoding Challenge with cpp.

October LeetCoding Challenge 19

Description

Backspace String Compare

Given two strings s and t, return true if they are equal when both are typed into empty text editors. '#' means a backspace character.

Note that after backspacing an empty text, the text will continue empty.

Example 1:

Input: s = "ab#c", t = "ad#c"
Output: true
Explanation: Both s and t become "ac".

Example 2:

Input: s = "ab##", t = "c#d#"
Output: true
Explanation: Both s and t become "".

Example 3:

Input: s = "a#c", t = "b"
Output: false
Explanation: s becomes "c" while t becomes "b".

Constraints:

• 1 <= s.length, t.length <= 200
• s and t only contain lowercase letters and '#' characters.

Follow up: Can you solve it in O(n) time and O(1) space?

Solution

class Solution {
  char nextChar(const string &s, int &pos) {
    int backspace = 0;
    while(pos >= 0 && (backspace || s[pos] == '#')) {
      if(s[pos] == '#') {
        backspace += 1;
      } else {
        backspace -= 1;
      }
      pos -= 1;
    }
    if(pos < 0) return -1;
    return s[pos];
  }
public:
  bool backspaceCompare(string s, string t) {
    int sPos = s.length() - 1;
    int tPos = t.length() - 1;
    while(sPos >= 0 || tPos >= 0) {
      char cS = nextChar(s, sPos);
      char cT = nextChar(t, tPos);
      if(cS != cT) return false;
      sPos -= 1;
      tPos -= 1;
    }
    return true;
  }
};

// Accepted
// 114/114 cases passed (0 ms)
// Your runtime beats 100 % of cpp submissions
// Your memory usage beats 54.78 % of cpp submissions (6.5 MB)