What is the time complexity of Lexicographically Smallest String After a Swap?

The optimal solution has a time complexity of O(n^2).

What data structures are used to solve Lexicographically Smallest String After a Swap?

This problem uses String, Greedy. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving Lexicographically Smallest String After a Swap?

The problem involves finding the lexicographically smallest string, which implies a greedy approach: we should prioritize swapping smaller digits into earlier positions. We need to iterate through the string and, for each digit, find the smallest digit with the same parity that can be swapped with it.

Which companies ask LeetCode 3216. Lexicographically Smallest String After a Swap in interviews?

This problem is frequently asked at J.P. Morgan.

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Lexicographically Smallest String After a Swap - LeetCode 3216 Solution

Lexicographically Smallest String After a Swap - Complete Solution Guide

Lexicographically Smallest String After a Swap is LeetCode problem 3216, a Easy level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

Given a string s containing only digits, return the lexicographically smallest string that can be obtained after swapping adjacent digits in s with the same parity at most once . Digits have the same parity if both are odd or both are even. For example, 5 and 9, as well as 2 and 4, have the same parity, while 6 and 9 do not. Example 1: Input: s = "45320" Output: "43520" Explanation: s[1] == '5' and s[2] == '3' both have the same parity, and swapping them results in the lexicographically smallest

Detailed Explanation

The problem asks you to find the lexicographically smallest string that can be created by swapping adjacent digits in the input string `s` that have the same parity (both even or both odd). The catch is that you can only perform at most one swap. Lexicographically smallest means the string that would come earliest in a dictionary ordering. For example, "123" is lexicographically smaller than "213". The input string `s` consists only of digits.

Solution Approach

The provided solutions all use a brute-force approach with nested loops. The outer loop iterates through each digit of the string. The inner loop searches for the smallest digit with the same parity to the right of the current digit. If a smaller digit is found, the two digits are swapped. This process guarantees that after the loops, the string is the lexicographically smallest possible string achievable through at most one swap of adjacent digits with the same parity.

Step-by-Step Algorithm

Convert the input string `s` into a list of characters (or use `toCharArray()` in Java).
Iterate through each character of the list using a nested loop. The outer loop index `i` represents the current digit.
Inner loop iterates from `i + 1` to the end of the string, comparing the parity (even or odd) of the current digit `s[i]` with each subsequent digit `s[j]`.
If a digit `s[j]` with the same parity and smaller value than the current smallest found digit is found, update `best_idx` (or keep track of the smallest character).
After the inner loop, if a smaller digit with the same parity was found (`best_idx != i`), swap the digits at indices `i` and `best_idx`.
Finally, convert the modified character list back into a string and return it.

Key Insights

The problem involves finding the lexicographically smallest string, which implies a greedy approach: we should prioritize swapping smaller digits into earlier positions.
We need to iterate through the string and, for each digit, find the smallest digit with the same parity that can be swapped with it.
The constraint of swapping only adjacent digits simplifies the search space and makes a nested loop approach feasible.

Complexity Analysis

Time Complexity: O(n^2)

Space Complexity: O(n)

Topics

This problem involves: String, Greedy.

Companies

Asked at: J.P. Morgan.

Lexicographically Smallest String After a Swap | LeetCode 3216

Easy

String Greedy

Time: O(n^2)Space: O(n)

Solution Code

Available in:

class Solution {
    public String getSmallestString(String s) {
        char[] chars = s.toCharArray();
        int n = chars.length;

        for (int i = 0; i < n; i++) {
            for (int j = i + 1; j < n; j++) {
                if ((chars[i] - '0') % 2 == (chars[j] - '0') % 2 && chars[j] < chars[i]) {
                    char temp = chars[i];
                    chars[i] = chars[j];
                    chars[j] = temp;
                }
            }
        }

        return new String(chars);
    }
}

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Problem Info

Description

Problem 3216: Lexicographically Smallest String After a Swap - Detailed explanation and solution approaches.

Topics

String Greedy

Companies

J.P. Morgan

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:3216

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

6 steps

1
Convert the input string `s` into a list of characters (or use `toCharArray()` in Java).
2
Iterate through each character of the list using a nested loop. The outer loop index `i` represents the current digit.
3
Inner loop iterates from `i + 1` to the end of the string, comparing the parity (even or odd) of the current digit `s[i]` with each subsequent digit `s[j]`.
4
If a digit `s[j]` with the same parity and smaller value than the current smallest found digit is found, update `best_idx` (or keep track of the smallest character).
5
After the inner loop, if a smaller digit with the same parity was found (`best_idx != i`), swap the digits at indices `i` and `best_idx`.
6
Finally, convert the modified character list back into a string and return it.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n^2)

🐢 Quadratic - Gets slow with large inputs. Consider optimization!

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The nested loops dominate the runtime, resulting in a time complexity of O(n^2), where n is the length of the string. The extra space used to store the list of characters is proportional to the string length, giving a space complexity of O(n).

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

✨

Key Insights

1The problem involves finding the lexicographically smallest string, which implies a greedy approach: we should prioritize swapping smaller digits into earlier positions.
2We need to iterate through the string and, for each digit, find the smallest digit with the same parity that can be swapped with it.
3The constraint of swapping only adjacent digits simplifies the search space and makes a nested loop approach feasible.

📚

Key Terms Explained

For Students

Greedy Algorithm

Making the best choice at each step without looking ahead.

String

A sequence of characters. Like a word or sentence in programming!

Edge Cases

Empty string: The solution handles this implicitly because the loops won't execute. Returning an empty string is correct.
String with only one digit: No swap is possible, so the original string is returned, which is correct.
String with no possible swap of adjacent digits with the same parity: The solution correctly returns the original string

⚠️

Common Mistakes to Avoid

✗Forgetting to check for same parity before comparing digits. Swapping digits of different parity is not allowed.

✗Not handling the case where the smallest digit with the same parity is already at the correct position (i.e., the string is already lexicographically smallest).

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼String optimization problems in databases or data processing pipelines where lexicographical ordering is important.

🏦Text processing and information retrieval systems might use similar techniques for efficient searching and sorting.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

String problems often involve sliding windows or two pointers. Consider if the character set is limited (lowercase letters = only 26 possibilities).

🎯 Interview Tip

Even for easy problems, discuss time and space complexity. Interviewers want to see you think analytically.

⚠️ Watch Out

Mutating strings directly in languages where strings are immutable. Use StringBuilder or arrays.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: November 4, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 13

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 6

❓

Frequently Asked Questions

How do I handle string manipulation efficiently?

In languages with immutable strings (Java, Python, C#), use StringBuilder/list to avoid O(n) concatenation cost. Consider using character arrays for in-place modifications. For pattern matching, consider KMP or Rabin-Karp for O(n+m) complexity.

How do I verify if a greedy approach is correct?

Prove the greedy choice property: the locally optimal choice leads to globally optimal solution. Use exchange argument: show that swapping any element with the greedy choice doesn't improve the solution. If you can't prove it, test extensively or consider DP.

How should I approach easy-level algorithm problems?

Easy problems build foundational skills. Focus on: 1) Understanding the problem completely before coding, 2) Writing clean, readable code, 3) Handling all edge cases (empty input, single element, etc.), 4) Analyzing time and space complexity. Don't skip complexity analysis even for simple solutions.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Lexicographically Smallest String After a Swap"?

For this easy-level String problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether greedy techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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