Build confidence with an intuition-first walkthrough focused on two pointers fundamentals.
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?
Problem summary: 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.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Two Pointers · Stack
"ab#c" "ad#c"
"ab##" "c#d#"
"a#c" "b"
crawler-log-folder)removing-stars-from-a-string)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #844: Backspace String Compare
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;
}
}
// Accepted solution for LeetCode #844: Backspace String Compare
func backspaceCompare(s string, t string) bool {
i, j := len(s)-1, len(t)-1
skip1, skip2 := 0, 0
for ; i >= 0 || j >= 0; i, j = i-1, j-1 {
for i >= 0 {
if s[i] == '#' {
skip1++
i--
} else if skip1 > 0 {
skip1--
i--
} else {
break
}
}
for j >= 0 {
if t[j] == '#' {
skip2++
j--
} else if skip2 > 0 {
skip2--
j--
} else {
break
}
}
if i >= 0 && j >= 0 {
if s[i] != t[j] {
return false
}
} else if i >= 0 || j >= 0 {
return false
}
}
return true
}
# Accepted solution for LeetCode #844: Backspace String Compare
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
// Accepted solution for LeetCode #844: Backspace String Compare
impl Solution {
pub fn backspace_compare(s: String, t: String) -> bool {
let (s, t) = (s.as_bytes(), t.as_bytes());
let (mut i, mut j) = (s.len(), t.len());
while i != 0 || j != 0 {
let mut skip = 0;
while i != 0 {
if s[i - 1] == b'#' {
skip += 1;
} else if skip != 0 {
skip -= 1;
} else {
break;
}
i -= 1;
}
skip = 0;
while j != 0 {
if t[j - 1] == b'#' {
skip += 1;
} else if skip != 0 {
skip -= 1;
} else {
break;
}
j -= 1;
}
if i == 0 && j == 0 {
break;
}
if i == 0 || j == 0 {
return false;
}
if s[i - 1] != t[j - 1] {
return false;
}
i -= 1;
j -= 1;
}
true
}
}
// Accepted solution for LeetCode #844: Backspace String Compare
function backspaceCompare(s: string, t: string): boolean {
let i = s.length - 1;
let j = t.length - 1;
while (i >= 0 || j >= 0) {
let skip = 0;
while (i >= 0) {
if (s[i] === '#') {
skip++;
} else if (skip !== 0) {
skip--;
} else {
break;
}
i--;
}
skip = 0;
while (j >= 0) {
if (t[j] === '#') {
skip++;
} else if (skip !== 0) {
skip--;
} else {
break;
}
j--;
}
if (s[i] !== t[j]) {
return false;
}
i--;
j--;
}
return true;
}
Use this to step through a reusable interview workflow for this problem.
Two nested loops check every pair of elements. The outer loop picks one element, the inner loop scans the rest. For n elements that is n × (n−1)/2 comparisons = O(n²). No extra memory — just two loop variables.
Each pointer traverses the array at most once. With two pointers moving inward (or both moving right), the total number of steps is bounded by n. Each comparison is O(1), giving O(n) overall. No auxiliary data structures are needed — just two index variables.
Review these before coding to avoid predictable interview regressions.
Wrong move: Advancing both pointers shrinks the search space too aggressively and skips candidates.
Usually fails on: A valid pair can be skipped when only one side should move.
Fix: Move exactly one pointer per decision branch based on invariant.
Wrong move: Pushing without popping stale elements invalidates next-greater/next-smaller logic.
Usually fails on: Indices point to blocked elements and outputs shift.
Fix: Pop while invariant is violated before pushing current element.