LeetCode-in-Java

2434. Using a Robot to Print the Lexicographically Smallest String

Medium

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:

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:

Solution

public class Solution {
    public String robotWithString(String s) {
        int n = s.length();
        char[] c = s.toCharArray();
        char[] next = new char[n + 1];
        next[n] = (char) ('z' + 1);
        for (int i = n - 1; i >= 0; i--) {
            next[i] = (char) ('a' + Math.min(c[i] - 'a', next[i + 1] - 'a'));
        }
        char[] stack = new char[n];
        int j = 0;
        int k = 0;
        for (int i = 0; i < n; ++i) {
            if (c[i] == next[i]) {
                c[j++] = c[i];
                while (k > 0 && stack[k - 1] <= next[i + 1]) {
                    c[j++] = stack[--k];
                }
            } else {
                stack[k++] = c[i];
            }
        }
        while (k > 0) {
            c[j++] = stack[--k];
        }
        return new String(c);
    }
}