844. Backspace String Compare
Explanation
To solve this problem, we can use a stack to simulate the typing process for both strings. We iterate over each character in the input strings, and if we encounter a character that is not a backspace ('#'), we push it onto the stack. If we encounter a backspace character, we pop the top character from the stack.
After iterating through both strings, we compare the characters left in the stacks for s and t. If the characters in the stacks are equal, then the strings are equal after applying backspaces.
Time Complexity: O(n) where n is the length of the longer string among s and t.
Space Complexity: O(n)
class Solution {
public boolean backspaceCompare(String s, String t) {
return buildString(s).equals(buildString(t));
}
private String buildString(String str) {
StringBuilder sb = new StringBuilder();
for (char c : str.toCharArray()) {
if (c != '#') {
sb.append(c);
} else if (sb.length() > 0) {
sb.deleteCharAt(sb.length() - 1);
}
}
return sb.toString();
}
}Code Editor (Testing phase)
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