Description
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 <= 200s and t only contain lowercase letters and '#' characters.
Follow up: Can you solve it in O(n) time and O(1) space?
Solutions
Solution 1
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29 | class Solution:
def backspaceCompare(self, s: str, t: str) -> bool:
i, j, skip1, skip2 = len(s) - 1, len(t) - 1, 0, 0
while i >= 0 or j >= 0:
while i >= 0:
if s[i] == '#':
skip1 += 1
i -= 1
elif skip1:
skip1 -= 1
i -= 1
else:
break
while j >= 0:
if t[j] == '#':
skip2 += 1
j -= 1
elif skip2:
skip2 -= 1
j -= 1
else:
break
if i >= 0 and j >= 0:
if s[i] != t[j]:
return False
elif i >= 0 or j >= 0:
return False
i, j = i - 1, j - 1
return True
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38 | class Solution {
public boolean backspaceCompare(String s, String t) {
int i = s.length() - 1, j = t.length() - 1;
int skip1 = 0, skip2 = 0;
for (; i >= 0 || j >= 0; --i, --j) {
while (i >= 0) {
if (s.charAt(i) == '#') {
++skip1;
--i;
} else if (skip1 > 0) {
--skip1;
--i;
} else {
break;
}
}
while (j >= 0) {
if (t.charAt(j) == '#') {
++skip2;
--j;
} else if (skip2 > 0) {
--skip2;
--j;
} else {
break;
}
}
if (i >= 0 && j >= 0) {
if (s.charAt(i) != t.charAt(j)) {
return false;
}
} else if (i >= 0 || j >= 0) {
return false;
}
}
return true;
}
}
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34 | class Solution {
public:
bool backspaceCompare(string s, string t) {
int i = s.size() - 1, j = t.size() - 1;
int skip1 = 0, skip2 = 0;
for (; i >= 0 || j >= 0; --i, --j) {
while (i >= 0) {
if (s[i] == '#') {
++skip1;
--i;
} else if (skip1) {
--skip1;
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