LeetCode #3722 — MEDIUM

Lexicographically Smallest String After Reverse

Move from brute-force thinking to an efficient approach using two pointers strategy.

Solve on LeetCode

The Problem

Problem Statement

You are given a string s of length n consisting of lowercase English letters.

You must perform exactly one operation by choosing any integer k such that 1 <= k <= n and either:

reverse the first k characters of s, or
reverse the last k characters of s.

Return the lexicographically smallest string that can be obtained after exactly one such operation.

Example 1:

Input: s = "dcab"

Output: "acdb"

Explanation:

Choose k = 3, reverse the first 3 characters.
Reverse "dca" to "acd", resulting string s = "acdb", which is the lexicographically smallest string achievable.

Example 2:

Input: s = "abba"

Output: "aabb"

Explanation:

Choose k = 3, reverse the last 3 characters.
Reverse "bba" to "abb", so the resulting string is "aabb", which is the lexicographically smallest string achievable.

Example 3:

Input: s = "zxy"

Output: "xzy"

Explanation:

Choose k = 2, reverse the first 2 characters.
Reverse "zx" to "xz", so the resulting string is "xzy", which is the lexicographically smallest string achievable.

Constraints:

1 <= n == s.length <= 1000
s consists of lowercase English letters.

Step 01

Brute Force Baseline

Problem summary: You are given a string s of length n consisting of lowercase English letters. You must perform exactly one operation by choosing any integer k such that 1 <= k <= n and either: reverse the first k characters of s, or reverse the last k characters of s. Return the lexicographically smallest string that can be obtained after exactly one such operation.

Baseline thinking

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

Pattern signal: Two Pointers · Binary Search

Example 1

"dcab"

Example 2

"abba"

Example 3

"zxy"

Step 02

Core Insight

What unlocks the optimal approach

Use bruteforce

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 #3722: Lexicographically Smallest String After Reverse
class Solution {
    public String lexSmallest(String s) {
        String ans = s;
        int n = s.length();
        for (int k = 1; k <= n; ++k) {
            String t1 = new StringBuilder(s.substring(0, k)).reverse().toString() + s.substring(k);
            String t2 = s.substring(0, n - k)
                + new StringBuilder(s.substring(n - k)).reverse().toString();
            if (t1.compareTo(ans) < 0) {
                ans = t1;
            }
            if (t2.compareTo(ans) < 0) {
                ans = t2;
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #3722: Lexicographically Smallest String After Reverse
func lexSmallest(s string) string {
	ans := s
	n := len(s)
	for k := 1; k <= n; k++ {
		t1r := []rune(s[:k])
		slices.Reverse(t1r)
		t1 := string(t1r) + s[k:]

		t2r := []rune(s[n-k:])
		slices.Reverse(t2r)
		t2 := s[:n-k] + string(t2r)

		ans = min(ans, t1, t2)
	}
	return ans
}

# Accepted solution for LeetCode #3722: Lexicographically Smallest String After Reverse
class Solution:
    def lexSmallest(self, s: str) -> str:
        ans = s
        for k in range(1, len(s) + 1):
            t1 = s[:k][::-1] + s[k:]
            t2 = s[:-k] + s[-k:][::-1]
            ans = min(ans, t1, t2)
        return ans

// Accepted solution for LeetCode #3722: Lexicographically Smallest String After Reverse
fn lex_smallest(s: String) -> String {
    let cs: Vec<_> = s.chars().collect();
    let len = s.len();

    let mut ret = cs.clone();
    for i in 2..=len {
        let mut tmp = cs.clone();
        let limit2 = i / 2;
        for j in 0..limit2 {
            tmp.swap(j, i - 1 - j);
        }

        ret = std::cmp::min(ret, tmp);
    }
    for i in 2..=len {
        let mut tmp = cs.clone();
        let limit2 = i / 2;
        for j in 0..limit2 {
            tmp.swap(len - 1 - j, len - i + j);
        }

        ret = std::cmp::min(ret, tmp);
    }

    ret.into_iter().collect()
}

fn main() {
    let ret = lex_smallest("dcab".to_string());
    println!("ret={ret}");
}

#[test]
fn test() {
    assert_eq!(lex_smallest("a".to_string()), "a");
    assert_eq!(lex_smallest("abc".to_string()), "abc");
    assert_eq!(lex_smallest("dcba".to_string()), "abcd");
    assert_eq!(lex_smallest("dcab".to_string()), "acdb");
    assert_eq!(lex_smallest("abba".to_string()), "aabb");
    assert_eq!(lex_smallest("zxy".to_string()), "xzy");
}

// Accepted solution for LeetCode #3722: Lexicographically Smallest String After Reverse
function lexSmallest(s: string): string {
    let ans = s;
    const n = s.length;
    for (let k = 1; k <= n; ++k) {
        const t1 = reverse(s.slice(0, k)) + s.slice(k);
        const t2 = s.slice(0, n - k) + reverse(s.slice(n - k));
        ans = [ans, t1, t2].sort()[0];
    }
    return ans;
}

function reverse(s: string): string {
    return s.split('').reverse().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^2)

Space

O(n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair of elements. The outer loop picks one element, the inner loop scans the rest. For n elements that is n × (n−1)/2 comparisons = O(n²). No extra memory — just two loop variables.

TWO POINTERS

O(n) time

O(1) space

Each pointer traverses the array at most once. With two pointers moving inward (or both moving right), the total number of steps is bounded by n. Each comparison is O(1), giving O(n) overall. No auxiliary data structures are needed — just two index variables.

Shortcut: Two converging pointers on sorted data → O(n) time, O(1) space.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Moving both pointers on every comparison

Wrong move: Advancing both pointers shrinks the search space too aggressively and skips candidates.

Usually fails on: A valid pair can be skipped when only one side should move.

Fix: Move exactly one pointer per decision branch based on invariant.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.