2434. Using a Robot to Print the Lexicographically Smallest String

Description

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.

Solutions

Solution 1: Greedy + Stack

The problem can be transformed into: given a string sequence, use an auxiliary stack to convert it into the lexicographically smallest string sequence.

We can use an array \(\textit{cnt}\) to maintain the count of each character in string \(s\), use a stack \(\textit{stk}\) as the auxiliary stack mentioned in the problem, and use a variable \(\textit{mi}\) to keep track of the smallest character not yet traversed in the string.

Traverse the string \(s\). For each character \(c\), first decrement its count in the array \(\textit{cnt}\) and update \(\textit{mi}\). Then push \(c\) onto the stack. At this point, if the top element of the stack is less than or equal to \(\textit{mi}\), repeatedly pop the top element from the stack and add it to the answer.

After the traversal, return the answer.

The time complexity is \(O(n + |\Sigma|)\), and the space complexity is \(O(n)\), where \(n\) is the length of the string \(s\) and \(|\Sigma|\) is the size of the character set, which is \(26\) in this problem.

Python3JavaC++GoTypeScriptRust

class Solution:
    def robotWithString(self, s: str) -> str:
        cnt = Counter(s)
        ans = []
        stk = []
        mi = 'a'
        for c in s:
            cnt[c] -= 1
            while mi < 'z' and cnt[mi] == 0:
                mi = chr(ord(mi) + 1)
            stk.append(c)
            while stk and stk[-1] <= mi:
                ans.append(stk.pop())
        return ''.join(ans)

class Solution {
    public String robotWithString(String s) {
        int[] cnt = new int[26];
        for (char c : s.toCharArray()) {
            ++cnt[c - 'a'];
        }
        StringBuilder ans = new StringBuilder();
        Deque<Character> stk = new ArrayDeque<>();
        char mi = 'a';
        for (char c : s.toCharArray()) {
            --cnt[c - 'a'];
            while (mi < 'z' && cnt[mi - 'a'] == 0) {
                ++mi;
            }
            stk.push(c);
            while (!stk.isEmpty() && stk.peek() <= mi) {
                ans.append(stk.pop());
            }
        }
        return ans.toString();
    }
}

class Solution {
public:
    string robotWithString(string s) {
        int cnt[26] = {0};
        for (char& c : s) ++cnt[c - 'a'];
        char mi = 'a';
        string stk;
        string ans;
        for (char& c : s) {
            --cnt[c - 'a'];
            while (mi < 'z' && cnt[mi - 'a'] == 0) ++mi;
            stk += c;
            while (!stk.empty() && stk.back() <= mi) {
                ans += stk.back();
                stk.pop_back();
            }
        }
        return ans;
    }
};

func robotWithString(s string) string {
    cnt := make([]int, 26)
    for _, c := range s {
        cnt[c-'a']++
    }
    mi := byte('a')
    stk := []byte{}
    ans := []byte{}
    for i := range s {
        cnt[s[i]-'a']--
        for mi < 'z' && cnt[mi-'a'] == 0 {
            mi++
        }
        stk = append(stk, s[i])
        for len(stk) > 0 && stk[len(stk)-1] <= mi {
            ans = append(ans, stk[len(stk)-1])
            stk = stk[:len(stk)-1]
        }
    }
    return string(ans)
}

function robotWithString(s: string): string {
    const cnt = new Map<string, number>();
    for (const c of s) {
        cnt.set(c, (cnt.get(c) || 0) + 1);
    }
    const ans: string[] = [];
    const stk: string[] = [];
    let mi = 'a';
    for (const c of s) {
        cnt.set(c, (cnt.get(c) || 0) - 1);
        while (mi < 'z' && (cnt.get(mi) || 0) === 0) {
            mi = String.fromCharCode(mi.charCodeAt(0) + 1);
        }
        stk.push(c);
        while (stk.length > 0 && stk[stk.length - 1] <= mi) {
            ans.push(stk.pop()!);
        }
    }
    return ans.join('');
}

impl Solution {
    pub fn robot_with_string(s: String) -> String {
        let mut cnt = [0; 26];
        for &c in s.as_bytes() {
            cnt[(c - b'a') as usize] += 1;
        }

        let mut ans = Vec::with_capacity(s.len());
        let mut stk = Vec::new();
        let mut mi = 0;

        for &c in s.as_bytes() {
            cnt[(c - b'a') as usize] -= 1;
            while mi < 26 && cnt[mi] == 0 {
                mi += 1;
            }
            stk.push(c);
            while let Some(&top) = stk.last() {
                if (top - b'a') as usize <= mi {
                    ans.push(stk.pop().unwrap());
                } else {
                    break;
                }
            }
        }

        String::from_utf8(ans).unwrap()
    }
}

2434. Using a Robot to Print the Lexicographically Smallest String

Description

Solutions

Solution 1: Greedy + Stack

Comments