What is the time complexity of Max Difference You Can Get From Changing an Integer?

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

What data structures are used to solve Max Difference You Can Get From Changing an Integer?

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

What are the key insights for solving Max Difference You Can Get From Changing an Integer?

Insight 1: To maximize the difference, `a` should be as large as possible and `b` should be as small as possible. Insight 2: To maximize `a`, replace the first non-9 digit with 9.

Which companies ask LeetCode 1432. Max Difference You Can Get From Changing an Integer in interviews?

This problem is frequently asked at Mercari, Trexquant.

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Max Difference You Can Get From Changing an Integer - LeetCode 1432 Solution

Q: How to solve Max Difference You Can Get From Changing an Integer?

The provided solution first converts the integer `num` to a string `s`. To find the maximum possible number `a`, it iterates through the digits of `s` and replaces the first digit that is not '9' with '9'. To find the minimum possible number `b`, it checks if the first digit of `s` is '1'. If not, it replaces the first digit with '1'. If the first digit is '1', it searches for the first digit other than '0' or '1' and replaces it with '0'. Finally, it calculates the difference between the integer values of the resulting strings `a` and `b` and returns it.

Max Difference You Can Get From Changing an Integer - Complete Solution Guide

Max Difference You Can Get From Changing an Integer is LeetCode problem 1432, 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 will apply the following steps to num two separate times: Pick a digit x (0 <= x <= 9) . Pick another digit y (0 <= y <= 9) . Note y can be equal to x . Replace all the occurrences of x in the decimal representation of num by y . Let a and b be the two results from applying the operation to num independently . Return the max difference between a and b . Note that neither a nor b may have any leading zeros, and must not be 0. Example 1: Input: num = 555 Output:

Detailed Explanation

The problem requires finding the maximum difference achievable by applying a digit replacement operation to a given integer `num` twice independently. In each operation, you choose a digit `x` and another digit `y`, and replace all occurrences of `x` in `num` with `y`. The goal is to find the maximum difference between the two resulting integers, `a` and `b`. Critically, the resulting numbers `a` and `b` cannot have leading zeros and must be greater than 0.

Solution Approach

The provided solution first converts the integer `num` to a string `s`. To find the maximum possible number `a`, it iterates through the digits of `s` and replaces the first digit that is not '9' with '9'. To find the minimum possible number `b`, it checks if the first digit of `s` is '1'. If not, it replaces the first digit with '1'. If the first digit is '1', it searches for the first digit other than '0' or '1' and replaces it with '0'. Finally, it calculates the difference between the integer values of the resulting strings `a` and `b` and returns it.

Step-by-Step Algorithm

Step 1: Convert the input integer `num` to a string `s`.
Step 2: Construct the maximum possible number `a`. Iterate through `s` and find the first digit that is not '9'. If such a digit is found, replace all occurrences of that digit in a copy of `s` with '9'. If no such digit is found, `a` is the same as `s`.
Step 3: Convert the string `a` to an integer.
Step 4: Construct the minimum possible number `b`. If the first digit of `s` is not '1', replace all occurrences of the first digit in a copy of `s` with '1'. Otherwise, iterate through `s` (starting from the second digit) and find the first digit that is neither '0' nor '1'. If such a digit is found, replace all occurrences of that digit in a copy of `s` with '0'. If no such digit is found, `b` is the same as `s`.
Step 5: Convert the string `b` to an integer.
Step 6: Return the difference between `a` and `b`.

Key Insights

Insight 1: To maximize the difference, `a` should be as large as possible and `b` should be as small as possible.
Insight 2: To maximize `a`, replace the first non-9 digit with 9.
Insight 3: To minimize `b`, if the first digit is not 1, replace it with 1. Otherwise, find the first digit other than 0 and 1, and replace it with 0.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Math, Greedy.

Companies

Asked at: Mercari, Trexquant.

Max Difference You Can Get From Changing an Integer | LeetCode 1432

Medium

Math Greedy

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

Solution Code

Available in:

class Solution {
    public int maxDiff(int num) {
        String s = String.valueOf(num);
        
        String s_a = s;
        char digit_to_change_a = '\0';
        for (int i = 0; i < s.length(); i++) {
            char char_i = s.charAt(i);
            if (char_i != '9') {
                digit_to_change_a = char_i;
                break;
            }
        }
        
        if (digit_to_change_a != '\0') {
            s_a = s.replace(digit_to_change_a, '9');
        }
        
        int a = Integer.parseInt(s_a);
            
        String s_b = s;
        if (s.charAt(0) != '1') {
            char digit_to_change_b = s.charAt(0);
            s_b = s.replace(digit_to_change_b, '1');
        } else {
            char digit_to_change_b = '\0';
            for (int i = 1; i < s.length(); i++) {
                char char_i = s.charAt(i);
                if (char_i != '0' && char_i != '1') {
                    digit_to_change_b = char_i;
                    break;
                }
            }
            
            if (digit_to_change_b != '\0') {
                s_b = s.replace(digit_to_change_b, '0');
            }
        }
        
        int b = Integer.parseInt(s_b);
                
        return a - b;
    }
}

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Problem Info

Description

Problem 1432: Max Difference You Can Get From Changing an Integer - Detailed explanation and solution approaches.

Topics

Math Greedy

Companies

MercariTrexquant

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:1432

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
Step 1: Convert the input integer `num` to a string `s`.
2
Step 2: Construct the maximum possible number `a`. Iterate through `s` and find the first digit that is not '9'. If such a digit is found, replace all occurrences of that digit in a copy of `s` with '9'. If no such digit is found, `a` is the same as `s`.
3
Step 3: Convert the string `a` to an integer.
4
Step 4: Construct the minimum possible number `b`. If the first digit of `s` is not '1', replace all occurrences of the first digit in a copy of `s` with '1'. Otherwise, iterate through `s` (starting from the second digit) and find the first digit that is neither '0' nor '1'. If such a digit is found, replace all occurrences of that digit in a copy of `s` with '0'. If no such digit is found, `b` is the same as `s`.
5
Step 5: Convert the string `b` to an integer.
6
Step 6: Return the difference between `a` and `b`.

💡

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)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is determined by the number of iterations required to find the digits to be replaced and to perform the replacements. Since the replacements are done using string replace functions or similar loops, and we perform these a constant number of times, the time complexity is proportional to the length of the input number. The space complexity comes from storing the number as a string and creating potentially two new strings `s_a` and `s_b`.

📖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

Mathematical Approach

Instead of string manipulation, a purely mathematical approach might be possible. However, it would require more complex logic to handle leading zeros and digit replacements without creating intermediate strings. The mathematical approach would require determining the place value to add or subtract to maximize or minimize the number.

Time: Potentially O(n) but might be more complex to derive.Space: O(1) if implemented without strings.

Trade-offs: Could be more efficient in terms of space if implemented carefully without strings, but might be harder to understand and implement correctly. The string-based solution is more readable.

✨

Key Insights

1Insight 1: To maximize the difference, `a` should be as large as possible and `b` should be as small as possible.
2Insight 2: To maximize `a`, replace the first non-9 digit with 9.
3Insight 3: To minimize `b`, if the first digit is not 1, replace it with 1. Otherwise, find the first digit other than 0 and 1, and replace it with 0.

📚

Key Terms Explained

For Students

Greedy Algorithm

Making the best choice at each step without looking ahead.

Edge Cases

Edge case 1: `num` is a single-digit number, e.g., 9. The algorithm correctly handles this by creating a = 9 and b = 1.
Edge case 2: `num` consists of all 9s, e.g., 9999. The algorithm sets `a` to 9999 and `b` to 1000 (if the problem description allowed leading zeros for b) or tries to find another digit. But, in this case, the first digit is '1' if minimizing, then looking for other digits which are not 0 or 1 becomes important
Edge case 3: `num` starts with '1' and subsequent digits are '0' or '1', like '10111'. The algorithm needs to correctly identify a digit other than '0' or '1' to replace with '0' for the minimum value.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Failing to handle the case where the first digit for minimization is '1' and all other digits are '0' or '1'. The code must search for the first digit that is neither '0' nor '1' to replace with '0'.

✗Common mistake 2: Incorrectly handling leading zeros in the minimized number `b`. The logic must ensure that `b` does not start with a zero.

✗Common mistake 3: Not handling cases where all digits are 9's for the maximizing part of the problem.

✅

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 numerical configuration setting. For example, increasing a setting to its maximum possible value or decreasing it to its minimum possible value by changing specific digits.

🏦Real-world scenario 2: Financial modeling where you might want to simulate the best-case and worst-case scenarios based on changing some parameters (represented as digits in a number).

🎮Real-world scenario 3: Data anonymization, replacing certain digits with predefined values for security purposes while maximizing the range after transformation.

💡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.

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 20, 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

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

@InterviewProJul 14

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

@CodeNinjaJul 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 "Max Difference You Can Get From Changing an Integer"?

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?