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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.
1using System;
2using System.Collections.Generic;
3
4class Program {
5 static string BuildFinalString(string str) {
6 var stack = new Stack<char>();
7 foreach (char ch in str) {
8 if (ch != '#') {
9 stack.Push(ch);
10 } else if (stack.Count > 0) {
11 stack.Pop();
12 }
13 }
14 return new string(stack.ToArray());
15 }
16
17 static bool BackspaceCompare(string s, string t) {
18 return BuildFinalString(s) == BuildFinalString(t);
19 }
20
21 static void Main() {
22 Console.WriteLine(BackspaceCompare("ab#c", "ad#c")); // true
23 Console.WriteLine(BackspaceCompare("ab##", "c#d#")); // true
24 Console.WriteLine(BackspaceCompare("a#c", "b")); // false
25 }
26}The C# implementation utilizes a Stack collection to mimic the effect of applying backspaces. The stack stores characters unless a backspace character is encountered, in which case the stack's top is removed if not empty. The complete post-backspace string is then assembled to compare the equality of the 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#include <string>
using namespace std;
bool backspaceCompare(string s, string t) {
int si = s.length() - 1, ti = t.length() - 1;
int skipS = 0, skipT = 0;
while (si >= 0 || ti >= 0) {
while (si >= 0) {
if (s[si] == '#') { skipS++; si--; }
else if (skipS > 0) { skipS--; si--; }
else break;
}
while (ti >= 0) {
if (t[ti] == '#') { skipT++; ti--; }
else if (skipT > 0) { skipT--; ti--; }
else break;
}
if ((si >= 0 && ti >= 0 && s[si] != t[ti]) || ((si >= 0) != (ti >= 0))) {
return false;
}
si--; ti--;
}
return true;
}
int main() {
cout << backspaceCompare("ab#c", "ad#c") << endl; // true
cout << backspaceCompare("ab##", "c#d#") << endl; // true
cout << backspaceCompare("a#c", "b") << endl; // false
return 0;
}The C++ implementation employs two pointers to iteratively move backwards through the strings, adjusting their position and incrementing or decrementing the backspace counter as needed to simulate the typing effect. We directly compare the characters the pointers land on after properly ignoring the backspaced characters.