Sponsored
Sponsored
This approach simulates the typing process using two stacks, one for each string. We iterate over each string, using the stack to build the string as it would appear after considering the backspace characters ('#'). By comparing the final stack representations, we can determine if the two strings are equivalent.
Time Complexity: O(n + m), where n and m are the lengths of s and t. Each character is processed once.
Space Complexity: O(n + m) due to the auxiliary arrays used to store the processed results of s and t.
1function backspaceCompare(s, t) {
2 function buildFinalString(str) {
3 const stack = [];
4 for (let ch of str) {
5 if (ch !== '#') {
6 stack.push(ch);
7 } else if (stack.length > 0) {
8 stack.pop();
9 }
10 }
11 return stack.join('');
12 }
13 return buildFinalString(s) === buildFinalString(t);
14}
15
16// Example usage:
17console.log(backspaceCompare("ab#c", "ad#c")); // true
18console.log(backspaceCompare("ab##", "c#d#")); // true
19console.log(backspaceCompare("a#c", "b")); // falseThe JavaScript solution uses an array to simulate stack behavior. For each character, we push it to the array if it's not a '#', and pop from the array if a '#' is encountered and the array isn't empty. We finally join the array into a string and compare the two processed strings.
This approach avoids extra space by using two pointers to traverse the strings backwards. By skipping over characters that are effectively backspaced due to a '#' character, we can compare corresponding positions in each string without actually building the resultant strings.
Time Complexity: O(n + m), where n and m are the lengths of s and t.
Space Complexity: O(1), since we only use constant space.
1
In this Python implementation, two pointers are maintained for the two strings, each moving from the end towards the start. Using a counter to track backspaces, the pointers skip over characters as needed to handle backspaces, thereby allowing a direct comparison of the effectively remaining characters.