LeetCode #899 — HARD

Orderly Queue

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

You are given a string s and an integer k. You can choose one of the first k letters of s and append it at the end of the string.

Return the lexicographically smallest string you could have after applying the mentioned step any number of moves.

Example 1:

Input: s = "cba", k = 1
Output: "acb"
Explanation: 
In the first move, we move the 1^st character 'c' to the end, obtaining the string "bac".
In the second move, we move the 1^st character 'b' to the end, obtaining the final result "acb".

Example 2:

Input: s = "baaca", k = 3
Output: "aaabc"
Explanation: 
In the first move, we move the 1^st character 'b' to the end, obtaining the string "aacab".
In the second move, we move the 3^rd character 'c' to the end, obtaining the final result "aaabc".

Constraints:

1 <= k <= s.length <= 1000
s consist of lowercase English letters.

Step 01

Brute Force Baseline

Problem summary: You are given a string s and an integer k. You can choose one of the first k letters of s and append it at the end of the string. Return the lexicographically smallest string you could have after applying the mentioned step any number of moves.

Baseline thinking

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

Pattern signal: Math

Example 1

"cba"
1

Example 2

"baaca"
3

Step 02

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

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

Largest constraint values

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 #899: Orderly Queue
class Solution {
    public String orderlyQueue(String s, int k) {
        if (k == 1) {
            String ans = s;
            StringBuilder sb = new StringBuilder(s);
            for (int i = 0; i < s.length() - 1; ++i) {
                sb.append(sb.charAt(0)).deleteCharAt(0);
                if (sb.toString().compareTo(ans) < 0) {
                    ans = sb.toString();
                }
            }
            return ans;
        }
        char[] cs = s.toCharArray();
        Arrays.sort(cs);
        return String.valueOf(cs);
    }
}

// Accepted solution for LeetCode #899: Orderly Queue
func orderlyQueue(s string, k int) string {
	if k == 1 {
		ans := s
		for i := 0; i < len(s)-1; i++ {
			s = s[1:] + s[:1]
			if s < ans {
				ans = s
			}
		}
		return ans
	}
	t := []byte(s)
	sort.Slice(t, func(i, j int) bool { return t[i] < t[j] })
	return string(t)
}

# Accepted solution for LeetCode #899: Orderly Queue
class Solution:
    def orderlyQueue(self, s: str, k: int) -> str:
        if k == 1:
            ans = s
            for _ in range(len(s) - 1):
                s = s[1:] + s[0]
                ans = min(ans, s)
            return ans
        return "".join(sorted(s))

// Accepted solution for LeetCode #899: Orderly Queue
struct Solution;

impl Solution {
    fn orderly_queue(s: String, k: i32) -> String {
        let mut s: Vec<char> = s.chars().collect();
        let n = s.len();
        if k > 1 {
            s.sort_unstable();
            s.into_iter().collect()
        } else {
            let mut res = s.to_vec();
            for i in 0..n {
                let mut t = vec![];
                for j in i..n {
                    t.push(s[j]);
                }
                for j in 0..i {
                    t.push(s[j]);
                }
                if t < res {
                    res = t;
                }
            }
            res.into_iter().collect()
        }
    }
}

#[test]
fn test() {
    let s = "cba".to_string();
    let k = 1;
    let res = "acb".to_string();
    assert_eq!(Solution::orderly_queue(s, k), res);
    let s = "baaca".to_string();
    let k = 3;
    let res = "aaabc".to_string();
    assert_eq!(Solution::orderly_queue(s, k), res);
}

// Accepted solution for LeetCode #899: Orderly Queue
function orderlyQueue(s: string, k: number): string {
    if (k > 1) {
        return [...s].sort().join('');
    }
    const n = s.length;
    let min = s;
    for (let i = 1; i < n; i++) {
        const t = s.slice(i) + s.slice(0, i);
        if (t < min) {
            min = t;
        }
    }
    return min;
}

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^2)

Space

O(n)

Approach Breakdown

ITERATIVE

O(n) time

O(1) space

Simulate the process step by step — multiply n times, check each number up to n, or iterate through all possibilities. Each step is O(1), but doing it n times gives O(n). No extra space needed since we just track running state.

MATH INSIGHT

O(log n) time

O(1) space

Math problems often have a closed-form or O(log n) solution hidden behind an O(n) simulation. Modular arithmetic, fast exponentiation (repeated squaring), GCD (Euclidean algorithm), and number theory properties can dramatically reduce complexity.

Shortcut: Look for mathematical properties that eliminate iteration. Repeated squaring → O(log n). Modular arithmetic avoids overflow.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.