What is the time complexity of Maximum Swap?

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

What data structures are used to solve Maximum Swap?

This problem uses Math, Greedy. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Maximum Swap?

Insight 1: To maximize the number, we need to find the leftmost digit that can be swapped with a larger digit to its right. Insight 2: We need to efficiently track the last occurrence of each digit in the number to find the best candidate for the swap.

Which companies ask LeetCode 670. Maximum Swap in interviews?

This problem is frequently asked at KLA.

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Maximum Swap - LeetCode 670 Solution

Q: How to solve Maximum Swap?

The solution involves first converting the integer into a string (or char array). Then, it creates a data structure (HashMap or array) to store the last occurrence of each digit in the number. After that, it iterates through the string from left to right. For each digit, it checks if there's a larger digit occurring later in the number. If such a digit is found, it performs the swap and returns the resulting integer. If no such digit is found after iterating through the entire number, it means the number is already the largest possible, so it returns the original number.

Maximum Swap - Complete Solution Guide

Maximum Swap is LeetCode problem 670, a Medium 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 an integer num . You can swap two digits at most once to get the maximum valued number. Return the maximum valued number you can get . Example 1: Input: num = 2736 Output: 7236 Explanation: Swap the number 2 and the number 7. Example 2: Input: num = 9973 Output: 9973 Explanation: No swap. Constraints: 0 <= num <= 10 8

Detailed Explanation

The problem asks us to find the maximum possible integer that can be obtained from a given integer `num` by swapping two of its digits at most once. If no swap results in a larger number, we return the original number. The input `num` is constrained to be between 0 and 10^8.

Solution Approach

The solution involves first converting the integer into a string (or char array). Then, it creates a data structure (HashMap or array) to store the last occurrence of each digit in the number. After that, it iterates through the string from left to right. For each digit, it checks if there's a larger digit occurring later in the number. If such a digit is found, it performs the swap and returns the resulting integer. If no such digit is found after iterating through the entire number, it means the number is already the largest possible, so it returns the original number.

Step-by-Step Algorithm

Step 1: Convert the integer `num` to a string or character array.
Step 2: Create a data structure (HashMap/array) to store the last index (rightmost occurrence) of each digit in the string.
Step 3: Iterate through the string from left to right (index `i`).
Step 4: For each digit at index `i`, iterate from '9' to `s_list[i]` to find a larger digit.
Step 5: If a larger digit `d_char` is found, check if it exists in the last occurrence map and if its last occurrence index `j` is greater than `i`. This ensures the larger digit is to the right of the current digit.
Step 6: If a suitable larger digit is found, swap the digits at indices `i` and `j`.
Step 7: Convert the modified string back to an integer and return it.
Step 8: If the loop completes without finding a suitable swap, return the original number.

Key Insights

Insight 1: To maximize the number, we need to find the leftmost digit that can be swapped with a larger digit to its right.
Insight 2: We need to efficiently track the last occurrence of each digit in the number to find the best candidate for the swap.
Insight 3: If the digits are already in descending order from left to right, no swap is necessary.

Complexity Analysis

Time Complexity: O(N)

Space Complexity: O(1)

Topics

This problem involves: Math, Greedy.

Companies

Asked at: KLA.

Maximum Swap | LeetCode 670

Medium

Math Greedy

Time: O(N)Space: O(1)

Solution Code

Available in:

import java.util.HashMap;
import java.util.Map;

class Solution {
    public int maximumSwap(int num) {
        char[] sList = String.valueOf(num).toCharArray();

        Map<Character, Integer> lastOccurrence = new HashMap<>();
        for (int i = 0; i < sList.length; i++) {
            lastOccurrence.put(sList[i], i);
        }

        for (int i = 0; i < sList.length; i++) {
            for (char dChar = '9'; dChar >= '0'; dChar--) {
                if (dChar > sList[i]) {
                    if (lastOccurrence.containsKey(dChar) && lastOccurrence.get(dChar) > i) {
                        int j = lastOccurrence.get(dChar);
                        char temp = sList[i];
                        sList[i] = sList[j];
                        sList[j] = temp;
                        return Integer.parseInt(new String(sList));
                    }
                }
            }
        }

        return num;
    }
}

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Problem Info

Description

Problem 670: Maximum Swap - Detailed explanation and solution approaches.

Topics

Math Greedy

Companies

KLA

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:670

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

8 steps

1
Step 1: Convert the integer `num` to a string or character array.
2
Step 2: Create a data structure (HashMap/array) to store the last index (rightmost occurrence) of each digit in the string.
3
Step 3: Iterate through the string from left to right (index `i`).
4
Step 4: For each digit at index `i`, iterate from '9' to `s_list[i]` to find a larger digit.
5
Step 5: If a larger digit `d_char` is found, check if it exists in the last occurrence map and if its last occurrence index `j` is greater than `i`. This ensures the larger digit is to the right of the current digit.
6
Step 6: If a suitable larger digit is found, swap the digits at indices `i` and `j`.
7
Step 7: Convert the modified string back to an integer and return it.
8
Step 8: If the loop completes without finding a suitable swap, return the original number.

💡

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)

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is dominated by the nested loops. However, the inner loop's number of iterations is constant, making the time complexity linear to the number of digits in the input number. The space complexity is constant as the data structure to keep track of the digits is bounded to the 10 digits (0-9).

📖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

Brute Force

Generate all possible numbers by swapping every pair of digits and find the maximum. This involves generating all combinations of swaps.

Time: O(N^2 * N) or O(N^3) - O(N^2) to generate swaps and O(N) to find maximum.Space: O(N), to store numbers after swap

Trade-offs: This is inefficient compared to the original solution. Avoid this approach due to higher time complexity.

✨

Key Insights

1Insight 1: To maximize the number, we need to find the leftmost digit that can be swapped with a larger digit to its right.
2Insight 2: We need to efficiently track the last occurrence of each digit in the number to find the best candidate for the swap.
3Insight 3: If the digits are already in descending order from left to right, no swap is necessary.

📚

Key Terms Explained

For Students

Greedy Algorithm

Making the best choice at each step without looking ahead.

Edge Cases

Edge case 1: Input is a single-digit number (e.g., 5). The solution correctly returns the original number because there are no other digits to swap with.
Edge case 2: Input is already the maximum possible number (e.g., 9876). The solution correctly returns the original number because no swap will produce a larger number.
Edge case 3: Input has repeated digits (e.g., 1992). The solution correctly finds the rightmost occurrence of the largest digit (9) and swaps it with the leftmost digit that is smaller.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not considering the rightmost occurrence of the larger digit. Swapping with an earlier occurrence might not result in the largest possible number.

✗Common mistake 2: Forgetting to handle the case where no swap is needed, and incorrectly returning a modified number.

✗Common mistake 3: Using an inefficient data structure to find the last occurrence of digits, which can increase the time complexity.

✅

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 numerical data by rearranging digits to maximize its value within certain constraints.

🏦Real-world scenario 2: In financial systems, manipulating digits to increase a value (hypothetically, and illegally, of course).

💡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

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ Watch Out

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

👨‍💻 From Experience

This problem pattern appears frequently in real codebases. I've used similar logic when building production systems.

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: October 9, 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

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 14

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

@AlgoMasterJul 7

❓

Frequently Asked Questions

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.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

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 "Maximum Swap"?

For this medium-level Math 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.

🤖 Need help with this problem?