Backspace String Compare - Math - Easy - LeetCode
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Backspace String Compare - Math - Easy - LeetCode

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  • 1The problem involves comparing two strings with backspace characters represented by '#'.
  • 2Both strings are processed to simulate typing in a text editor, ignoring backspaces.
  • 3The solution checks if the final processed strings are equal, achieving O(N) time complexity.

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"The problem involves comparing two strings with backspace characters represented by '#'."

Backspace String Compare - Math - Easy - LeetCode

Given two strings S and T, return 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 = "#a#c" Output: true Explanation: Both S and T become "c". Example 4:

Input: S = "a#c", T = "b" Output: false Explanation: S becomes "c" while T becomes "b". Note:

1 <= S.length <= 200 1 <= T.length <= 200 S and T only contain lowercase letters and '#' characters. Follow up:

Can you solve it in O(N) time and O(1) space?

public class Solution {
 public bool BackspaceCompare(string S, string T) { 
 var stack1 = new Stack<char>();
 var stack2 = new Stack<char>();
 for(int i=0;i<S.Length;i++){
 if(S[i]=='#'){
 if(stack1.Count!=0){
 stack1.Pop();
 }
 }
 else{
 stack1.Push(S[i]);
 }
 }
 
 for(int i=0;i<T.Length;i++){
 if(T[i]=='#'){
 if(stack2.Count!=0){
 stack2.Pop();
 }
 }
 else{
 stack2.Push(T[i]);
 }
 }
 
 if(stack1.Count != stack2.Count){
 return false;
 }
 
 while(stack1.Count>0){
 if(stack1.Pop()!=stack2.Pop()){
 return false;
 }
 }
 
 return true;
 }
}

Time Complexity: O(n)

Space Complexity: O(n)

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sumitc91

Published on 2 October 2020 · 1 min read · 207 words

Part of AskGif Blog · coding

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Backspace String Compare - Math - Easy - LeetCode | AskGif Blog