Make The String Great - String - Easy - LeetCode
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Make The String Great - String - Easy - LeetCode

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  • 1A good string does not have adjacent characters that are the same letter in different cases.
  • 2To make a string good, remove adjacent bad character pairs until the string is valid.
  • 3The solution guarantees a unique output for the given constraints, with time complexity O(n).

AI-generated summary · May not capture all nuances

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"A good string does not have adjacent characters that are the same letter in different cases."

Make The String Great - String - Easy - LeetCode

Given a string s of lower and upper case English letters.

A good string is a string which doesn't have two adjacent characters s[i] and s[i + 1] where:

0 <= i <= s.length - 2 s[i] is a lower-case letter and s[i + 1] is the same letter but in upper-case or vice-versa. To make the string good, you can choose two adjacent characters that make the string bad and remove them. You can keep doing this until the string becomes good.

Return the string after making it good. The answer is guaranteed to be unique under the given constraints.

Notice that an empty string is also good.

Example 1:

Input: s = "leEeetcode" Output: "leetcode" Explanation: In the first step, either you choose i = 1 or i = 2, both will result "leEeetcode" to be reduced to "leetcode". Example 2:

Input: s = "abBAcC" Output: "" Explanation: We have many possible scenarios, and all lead to the same answer. For example: "abBAcC" --> "aAcC" --> "cC" --> "" "abBAcC" --> "abBA" --> "aA" --> "" Example 3:

Input: s = "s" Output: "s"

Constraints:

1 <= s.length <= 100 s contains only lower and upper case English letters.

public class Solution {
 public string MakeGood(string s) {
 if(s.Length==0){
 return s;
 }
 
 var stack = new Stack<char>(); 
 stack.Push(s[0]);
 for(int i=1;i<s.Length;i++){
 if(stack.Count>0){
 if(IsUpperLower(stack.Peek(),s[i])){
 stack.Pop(); 
 }
 else{
 stack.Push(s[i]);
 }
 }
 else{
 stack.Push(s[i]);
 }
 }
 
 var temp = new Stack<char>();
 while(stack.Count>0){
 temp.Push(stack.Pop());
 }
 
 var sb = new StringBuilder();
 while(temp.Count>0){
 sb.Append(temp.Pop());
 }
 
 return sb.ToString();
 }
 
 private bool IsUpperLower(char x, char y){
 return (Math.Abs(x-y)==32);
 }
}

Time Complexity: O(n)

Space Complexity: O(n)

Where n is the size of the string.

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sumitc91

Published on 2 October 2020 · 1 min read · 277 words

Part of AskGif Blog · coding

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Make The String Great - String - Easy - LeetCode | AskGif Blog