What is the time complexity of Largest Number After Digit Swaps by Parity?

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

What data structures are used to solve Largest Number After Digit Swaps by Parity?

This problem uses Sorting, Heap (Priority Queue). Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving Largest Number After Digit Swaps by Parity?

Insight 1: Separating even and odd digits is crucial. We cannot swap even and odd digits, so they need to be handled independently. Insight 2: Sorting the even and odd digits in descending order allows us to easily construct the largest possible number by placing the largest even digits in even positions and largest odd digits in odd positions.

Which companies ask LeetCode 2231. Largest Number After Digit Swaps by Parity in interviews?

This problem is frequently asked at ZScaler.

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Largest Number After Digit Swaps by Parity - LeetCode 2231 Solution

Largest Number After Digit Swaps by Parity - Complete Solution Guide

Largest Number After Digit Swaps by Parity is LeetCode problem 2231, 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

You are given a positive integer num . You may swap any two digits of num that have the same parity (i.e. both odd digits or both even digits). Return the largest possible value of num after any number of swaps. Example 1: Input: num = 1234 Output: 3412 Explanation: Swap the digit 3 with the digit 1, this results in the number 3214. Swap the digit 2 with the digit 4, this results in the number 3412. Note that there may be other sequences of swaps but it can be shown that 3412 is the largest poss

Detailed Explanation

The problem asks you to find the largest possible integer that can be formed by swapping digits of the input integer `num` such that only digits of the same parity (both even or both odd) can be swapped. The input is a positive integer, and the output is the largest possible integer achievable through these allowed swaps. For example, if the input is 1234, the largest possible number after swapping digits with the same parity is 3412 (swapping 1 and 3, and 2 and 4).

Solution Approach

The solution uses a greedy approach. It first separates the digits of the input number into two lists: one for odd digits and one for even digits. Then it sorts these lists in descending order. Finally, it iterates through the original digits and replaces them with the largest available odd or even digit from the sorted lists, maintaining the original order of even and odd positions.

Step-by-Step Algorithm

Step 1: Convert the input integer `num` into a string `s` for easier digit access.
Step 2: Create two lists (or arrays): `odds` to store odd digits and `evens` to store even digits from `s`.
Step 3: Sort `odds` and `evens` in descending order.
Step 4: Iterate through the characters of `s`. If a digit is odd, replace it with the next largest odd digit from the sorted `odds` list. If a digit is even, replace it with the next largest even digit from the sorted `evens` list.
Step 5: Concatenate the modified digits to form a new string, and convert it back to an integer, which is the result.

Key Insights

Insight 1: Separating even and odd digits is crucial. We cannot swap even and odd digits, so they need to be handled independently.
Insight 2: Sorting the even and odd digits in descending order allows us to easily construct the largest possible number by placing the largest even digits in even positions and largest odd digits in odd positions.
Insight 3: Converting the integer to a string (or an array of digits) simplifies digit manipulation and comparison.

Complexity Analysis

Time Complexity: O(n log n)

Space Complexity: O(n)

Topics

This problem involves: Sorting, Heap (Priority Queue).

Companies

Asked at: ZScaler.

Largest Number After Digit Swaps by Parity | LeetCode 2231

Easy

Sorting Heap (Priority Queue)

Time: O(n log n)Space: O(n)

Solution Code

Available in:

class Solution {
    public int largestInteger(int num) {
        String s = Integer.toString(num);
        char[] chars = s.toCharArray();
        List<Character> odd = new ArrayList<>();
        List<Character> even = new ArrayList<>();
        for (char c : chars) {
            if ((c - '0') % 2 == 0) {
                even.add(c);
            } else {
                odd.add(c);
            }
        }
        Collections.sort(odd, Collections.reverseOrder());
        Collections.sort(even, Collections.reverseOrder());
        int i = 0, j = 0;
        for (int k = 0; k < chars.length; k++) {
            if ((chars[k] - '0') % 2 == 0) {
                chars[k] = even.get(i++);
            } else {
                chars[k] = odd.get(j++);
            }
        }
        return Integer.parseInt(new String(chars));
    }
}

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Problem Info

Description

Problem 2231: Largest Number After Digit Swaps by Parity - Detailed explanation and solution approaches.

Topics

Sorting Heap (Priority Queue)

Companies

ZScaler

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:2231

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

5 steps

1
Step 1: Convert the input integer `num` into a string `s` for easier digit access.
2
Step 2: Create two lists (or arrays): `odds` to store odd digits and `evens` to store even digits from `s`.
3
Step 3: Sort `odds` and `evens` in descending order.
4
Step 4: Iterate through the characters of `s`. If a digit is odd, replace it with the next largest odd digit from the sorted `odds` list. If a digit is even, replace it with the next largest even digit from the sorted `evens` list.
5
Step 5: Concatenate the modified digits to form a new string, and convert it back to an integer, which is the result.

💡

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 log n)

✨ Log-linear - Great for sorting algorithms!

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The dominant operation is sorting the `odds` and `evens` lists, which takes O(n log n) time, where n is the number of digits in the input number. The space complexity is O(n) because we store the digits in separate lists (or arrays).

📖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

Using a Priority Queue

Instead of sorting, you could use two priority queues (one for odds and one for evens) to efficiently access the largest remaining odd/even digit at each step. This would maintain the same time complexity in the average case.

Time: O(n log n)Space: O(n)

Trade-offs: A priority queue approach might offer slightly better performance in the average case, but the sorting approach is generally simpler to implement and understand.

✨

Key Insights

1Insight 1: Separating even and odd digits is crucial. We cannot swap even and odd digits, so they need to be handled independently.
2Insight 2: Sorting the even and odd digits in descending order allows us to easily construct the largest possible number by placing the largest even digits in even positions and largest odd digits in odd positions.
3Insight 3: Converting the integer to a string (or an array of digits) simplifies digit manipulation and comparison.

📚

Key Terms Explained

For Students

Sorting

Arranging elements in a specific order (ascending/descending).

Edge Cases

Edge case 1: The input number contains only odd or only even digits. The solution still works correctly; one of the lists will be empty, and the algorithm will only use the other list.
Edge case 2: The input number is a single digit. The solution handles this trivially; it returns the input number itself.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to sort the odd and even digits before placing them back into the number. This would lead to an incorrect result.

✗Common mistake 2: Incorrectly handling the indices of the `odds` and `evens` lists during the replacement step. This would lead to incorrect digit assignments.

✅

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?

💼Real-world scenario 1: Optimizing a routing algorithm where you want to select the largest available bandwidth among links of the same type (e.g., fiber or copper).

🏦Real-world scenario 2: In a resource allocation system where you have different types of resources (with some constraints on their usage) and you want to maximize the total resource value allocated.

💡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

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Don't rush. Explain your thought process clearly. Easy problems are often about communication.

⚠️ Watch Out

Always test your solution with edge cases: empty input, single element, and boundary values.

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: August 11, 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

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJun 29

❓

Frequently Asked Questions

What should I learn from easy problems?

Easy problems introduce core patterns that appear in harder problems. Master basic operations (iteration, conditionals, simple data structures), recognize common patterns (counting, searching, basic transformations), and practice explaining your thought process clearly.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Largest Number After Digit Swaps by Parity"?

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

🤖 Need help with this problem?