LeetCode #2434 — MEDIUM

Using a Robot to Print the Lexicographically Smallest String

Move from brute-force thinking to an efficient approach using hash map strategy.

The Problem

Problem Statement

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.

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Hash Map · Stack · Greedy

Example 1

"zza"

Example 2

"bac"

Example 3

"bdda"

Core Insight

What unlocks the optimal approach

If there are some character “a” ’ s in the string, they can be written on paper before anything else.
Every character in the string before the last “a” should be written in reversed order.
After the robot writes every “a” on paper, the same holds for other characters “b”, ”c”, …etc.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2434: Using a Robot to Print the Lexicographically Smallest String
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();
    }
}

// Accepted solution for LeetCode #2434: Using a Robot to Print the Lexicographically Smallest String
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)
}

# Accepted solution for LeetCode #2434: Using a Robot to Print the Lexicographically Smallest String
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)

// Accepted solution for LeetCode #2434: Using a Robot to Print the Lexicographically Smallest String
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()
    }
}

// Accepted solution for LeetCode #2434: Using a Robot to Print the Lexicographically Smallest String
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('');
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n + |\Sigma|)

Space

O(n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

For each element, scan left (or right) to find the next greater/smaller element. The inner scan can visit up to n elements per outer iteration, giving O(n²) total comparisons. No extra space needed beyond loop variables.

MONOTONIC STACK

O(n) time

O(n) space

Each element is pushed onto the stack at most once and popped at most once, giving 2n total operations = O(n). The stack itself holds at most n elements in the worst case. The key insight: amortized O(1) per element despite the inner while-loop.

Shortcut: Each element pushed once + popped once → O(n) amortized. The inner while-loop does not make it O(n²).

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

Breaking monotonic invariant

Wrong move: Pushing without popping stale elements invalidates next-greater/next-smaller logic.

Usually fails on: Indices point to blocked elements and outputs shift.

Fix: Pop while invariant is violated before pushing current element.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.