LEETCODE 670:DIGIT VECTOR PATTERN:Maximize a Number by Swapping Two Digits:LC 670 Solution Explained

Welcome to our comprehensive walkthrough of LeetCode Problem 670: "Maximum Swap." In this video, we will guide you through the solution to this intriguing problem, ensuring you grasp the concepts and strategies needed to maximize a number by swapping its digits.
Introduction
LeetCode Problem 670 presents a captivating challenge: given a non-negative integer, your task is to maximize it by swapping at most two digits. This problem tests your understanding of number manipulation and your ability to devise an optimal solution within given constraints.
Problem Statement
You are given a non-negative integer, and you need to swap two digits (at most once) to achieve the maximum possible number. The integer is provided in its standard numeric form (e.g., 2736). Your goal is to find the largest possible number that can be obtained by swapping two digits exactly once.
Key Concepts
To tackle this problem effectively, we need to delve into several key concepts:
Digit Manipulation: Understanding how to extract, compare, and swap digits within a number.
Greedy Algorithm: Utilizing a greedy approach to ensure we make the best possible swap to maximize the number.
Optimal Swapping: Determining which digits to swap to achieve the largest possible number.
Step-by-Step Solution
Our solution involves a series of well-defined steps:
Digit Extraction: Convert the number into a list of its digits for easier manipulation.
Identify Swap Candidates: Traverse the list to identify the best candidates for swapping. This involves finding the largest digit that can be swapped with a smaller digit appearing before it in the list.
Perform the Swap: Swap the identified digits to achieve the maximum possible number.
Reconstruct the Number: Convert the modified list of digits back into a number.
Detailed Walkthrough
Let's walk through the solution with an example. Suppose we are given the number 2736:
Digit Extraction: Convert 2736 into the list [2, 7, 3, 6].
Identify Swap Candidates: Traverse the list to find the best swap:
At index 0, the digit is 2.
At index 1, the digit is 7. Since 7 is larger than 2, consider it for swapping.
At index 2, the digit is 3. No swap candidate yet.
At index 3, the digit is 6. Still, no swap candidate.
The largest digit for swapping is 7, but no larger digit appears after it. Thus, no swap involving 7.
Consider digit 2 for swapping: the best candidate is 6 at index 3.
Perform the Swap: Swap digits 2 and 6 to get the list [6, 7, 3, 2].
Reconstruct the Number: Convert [6, 7, 3, 2] back to the number 6732.
Edge Cases
It's crucial to consider edge cases, such as:
Single-digit Numbers: These numbers cannot be swapped, so the output remains the same.
All Digits Identical: No swap can maximize the number, so the output is the original number.
Already Maximum Number: If the number is already in its maximum form, no swap is needed.
Complexity Analysis
The solution has a time complexity of O(n), where n is the number of digits in the number. This is due to the traversal of the digit list to identify swap candidates. The space complexity is O(1) since we only use a fixed amount of extra space for digit manipulation.
Conclusion
By understanding the principles of digit manipulation and utilizing a greedy approach, we can efficiently solve LeetCode Problem 670. This problem not only sharpens your problem-solving skills but also enhances your understanding of number theory and optimization techniques.
Practice and Implementation
We encourage you to try implementing this solution on your own and test it against various test cases. Practice is key to mastering these concepts and becoming proficient in solving similar problems.
Thank you for watching our detailed explanation of LeetCode Problem 670. If you found this video helpful, please like, share, and subscribe to our channel for more insightful tutorials and problem-solving strategies. Leave your comments and questions below, and we'll be happy to help you further. Happy coding!

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