LeetCode 2434. Using a Robot to Print the Lexicographically Smallest String

原题链接在这里：https://leetcode.com/problems/using-a-robot-to-print-the-lexicographically-smallest-string/

题目：

You are given a string s and a robot that currently holds an empty string t. Apply one of the following operations until s and t are both empty:

• Remove the first character of a string s and give it to the robot. The robot will append this character to the string t.
• Remove the last character of a string t and give it to the robot. The robot will write this character on paper.

Return the lexicographically smallest string that can be written on the paper.

Example 1:

Input: s = "zza"
Output: "azz"
Explanation: Let p denote the written string.
Initially p="", s="zza", t="".
Perform first operation three times p="", s="", t="zza".
Perform second operation three times p="azz", s="", t="".

Example 2:

Input: s = "bac"
Output: "abc"
Explanation: Let p denote the written string.
Perform first operation twice p="", s="c", t="ba". 
Perform second operation twice p="ab", s="c", t="". 
Perform first operation p="ab", s="", t="c". 
Perform second operation p="abc", s="", t="".

Example 3:

Input: s = "bdda"
Output: "addb"
Explanation: Let p denote the written string.
Initially p="", s="bdda", t="".
Perform first operation four times p="", s="", t="bdda".
Perform second operation four times p="addb", s="", t="".

Constraints:

• 1 <= s.length <= 105
• s consists of only English lowercase letters.

题解：

The point is to find the remaining smallest char. e.g."bdda", we need to find "a".

We first need to count the frequency of char in s.

Have a stack, move the current char to stack, if it is equal or smaller than remaining lowest char, then add it to the result.

Time Complexity: O(n). n = s.length(). each char is only in and out of stack once.

Space: O(n).

AC Java:

 1 class Solution {
 2     public String robotWithString(String s) {
 3         Stack<Character> stk = new Stack<>();
 4         StringBuilder sb = new StringBuilder();
 5         int [] map = new int[26];
 6         char [] charArr = s.toCharArray();
 7         for(char c : charArr){
 8             map[c - 'a']++;
 9         }
10         
11         int low = 0;
12         for(char c : charArr){
13             stk.push(c);
14             map[c - 'a']--;
15             while(low < 26 && map[low] == 0){
16                 low++;
17             }
18             
19             while(!stk.isEmpty() && stk.peek() - 'a' <= low){
20                 sb.append(stk.pop());
21             }
22         }
23         
24         return sb.toString();
25     }
26 }