Remove All Adjacent Duplicates In String - Stacks - Easy - LeetCode
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Remove All Adjacent Duplicates In String - Stacks - Easy - LeetCode

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  • 1The algorithm removes adjacent duplicate letters from a string until no duplicates remain.
  • 2Using a stack, the solution efficiently processes each character in linear time complexity O(n).
  • 3The final output string is guaranteed to be unique after all duplicate removals.

AI-generated summary · May not capture all nuances

Key Insight
AskGif

"The algorithm removes adjacent duplicate letters from a string until no duplicates remain."

Remove All Adjacent Duplicates In String - Stacks - Easy - LeetCode

Given a string S of lowercase letters, a duplicate removal consists of choosing two adjacent and equal letters, and removing them.

We repeatedly make duplicate removals on S until we no longer can.

Return the final string after all such duplicate removals have been made. It is guaranteed the answer is unique.

Example 1:

Input: "abbaca" Output: "ca" Explanation: For example, in "abbaca" we could remove "bb" since the letters are adjacent and equal, and this is the only possible move. The result of this move is that the string is "aaca", of which only "aa" is possible, so the final string is "ca".

Note:

1 <= S.length <= 20000 S consists only of English lowercase letters.

public class Solution {
 public string RemoveDuplicates(string S) {
 var stack = new Stack<char>();
 for(int i=0;i<S.Length;i++){
 if(stack.Count()==0){
 stack.Push(S[i]);
 }
 else{
 if(stack.Peek()==S[i]){
 stack.Pop();
 }
 else{
 stack.Push(S[i]);
 }
 }
 }
 
 var sb = new StringBuilder();
 var rev = new Stack<char>();
 while(stack.Count()>0){
 rev.Push(stack.Pop());
 }
 
 while(rev.Count()>0){
 sb.Append(rev.Pop());
 }
 
 return sb.ToString();
 }
}

Time Complexity: O(n)

Space Complexity: O(n)

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sumitc91

Published on 3 October 2020 · 1 min read · 171 words

Part of AskGif Blog · coding

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