What is the time complexity of Monotone Increasing Digits?

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

What data structures are used to solve Monotone Increasing Digits?

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

What are the key insights for solving Monotone Increasing Digits?

Insight 1: The core idea is to scan the number from right to left, and if we find a digit that is smaller than the digit to its left, we need to decrement the left digit. This might create a cascade effect, requiring us to iterate backwards until we find the point where the monotone increasing property is violated. Insight 2: Once a violation is found and the necessary decrements are made, all digits to the right of the decremented digit must be set to '9' to maximize the number while still satisfying the monotone increasing constraint.

Which companies ask LeetCode 738. Monotone Increasing Digits in interviews?

This problem is frequently asked at SAP.

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Monotone Increasing Digits - LeetCode 738 Solution

Q: How to solve Monotone Increasing Digits?

The provided solution converts the integer 'n' into a string (or character array). It then iterates through the string from right to left, looking for violations of the monotone increasing property. If a violation is found (i.e., s[i-1] > s[i]), the digit s[i-1] is decremented, and a 'marker' is set to the current index 'i'. This marker indicates the starting point for converting the remaining digits to the right into '9's. After the initial scan, the algorithm iterates from the 'marker' to the end of the string, setting all digits to '9'. Finally, the string is converted back to an integer and returned.

Monotone Increasing Digits - Complete Solution Guide

Monotone Increasing Digits is LeetCode problem 738, 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

An integer has monotone increasing digits if and only if each pair of adjacent digits x and y satisfy x <= y . Given an integer n , return the largest number that is less than or equal to n with monotone increasing digits . Example 1: Input: n = 10 Output: 9 Example 2: Input: n = 1234 Output: 1234 Example 3: Input: n = 332 Output: 299 Constraints: 0 <= n <= 10 9

Detailed Explanation

The problem asks us to find the largest number less than or equal to a given integer 'n' that has monotone increasing digits. A number has monotone increasing digits if, for any two adjacent digits x and y, x <= y. For example, 1234 is a monotone increasing number, while 332 is not. We need to return the largest monotone increasing number that is no greater than the input 'n'.

Solution Approach

The provided solution converts the integer 'n' into a string (or character array). It then iterates through the string from right to left, looking for violations of the monotone increasing property. If a violation is found (i.e., s[i-1] > s[i]), the digit s[i-1] is decremented, and a 'marker' is set to the current index 'i'. This marker indicates the starting point for converting the remaining digits to the right into '9's. After the initial scan, the algorithm iterates from the 'marker' to the end of the string, setting all digits to '9'. Finally, the string is converted back to an integer and returned.

Step-by-Step Algorithm

Step 1: Convert the input integer 'n' to a string (or character array) 's'.
Step 2: Initialize a 'marker' variable to the length of the string. This variable will store the index from which we need to change digits to '9'.
Step 3: Iterate through the string 's' from right to left (from index length - 1 to 1).
Step 4: In each iteration, compare s[i-1] and s[i]. If s[i-1] > s[i], it means we have a violation of the monotone increasing property.
Step 5: If a violation is found, decrement s[i-1] by 1 (converting it back to a character after decrementing the integer representation). Update the 'marker' to be the current index 'i'.
Step 6: After the first loop finishes, iterate through the string from the 'marker' to the end of the string.
Step 7: In this second loop, set each digit s[i] to '9'.
Step 8: Convert the modified string 's' back to an integer and return the result.

Key Insights

Insight 1: The core idea is to scan the number from right to left, and if we find a digit that is smaller than the digit to its left, we need to decrement the left digit. This might create a cascade effect, requiring us to iterate backwards until we find the point where the monotone increasing property is violated.
Insight 2: Once a violation is found and the necessary decrements are made, all digits to the right of the decremented digit must be set to '9' to maximize the number while still satisfying the monotone increasing constraint.
Insight 3: Converting the number to a string or character array facilitates digit-by-digit manipulation.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Math, Greedy.

Companies

Asked at: SAP.

Monotone Increasing Digits | LeetCode 738

Medium

Math Greedy

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

Solution Code

Available in:

class Solution {
    public int monotoneIncreasingDigits(int n) {
        char[] s = String.valueOf(n).toCharArray();
        int length = s.length;
        int marker = length;

        for (int i = length - 1; i > 0; i--) {
            if (s[i - 1] > s[i]) {
                s[i - 1]--;
                marker = i;
            }
        }

        for (int i = marker; i < length; i++) {
            s[i] = '9';
        }

        return Integer.parseInt(new String(s));
    }
}

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Problem Info

Description

Problem 738: Monotone Increasing Digits - Detailed explanation and solution approaches.

Topics

Math Greedy

Companies

SAP

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:738

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 input integer 'n' to a string (or character array) 's'.
2
Step 2: Initialize a 'marker' variable to the length of the string. This variable will store the index from which we need to change digits to '9'.
3
Step 3: Iterate through the string 's' from right to left (from index length - 1 to 1).
4
Step 4: In each iteration, compare s[i-1] and s[i]. If s[i-1] > s[i], it means we have a violation of the monotone increasing property.
5
Step 5: If a violation is found, decrement s[i-1] by 1 (converting it back to a character after decrementing the integer representation). Update the 'marker' to be the current index 'i'.
6
Step 6: After the first loop finishes, iterate through the string from the 'marker' to the end of the string.
7
Step 7: In this second loop, set each digit s[i] to '9'.
8
Step 8: Convert the modified string 's' back to an integer and return 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)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is linear because the code iterates through the digits a fixed number of times. The space complexity is linear because the size of the string/character array grows linearly with the number of digits in the input number.

📖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 (Less Readable)

It's possible to solve this problem without converting the number to a string by using mathematical operations to extract and manipulate digits. However, this approach would likely be less readable and more complex to implement correctly, especially when considering the cascading decrements.

Time: O(n), where n is the number of digitsSpace: O(1)

Trade-offs: The mathematical approach has a smaller space complexity but is harder to understand and implement.

✨

Key Insights

1Insight 1: The core idea is to scan the number from right to left, and if we find a digit that is smaller than the digit to its left, we need to decrement the left digit. This might create a cascade effect, requiring us to iterate backwards until we find the point where the monotone increasing property is violated.
2Insight 2: Once a violation is found and the necessary decrements are made, all digits to the right of the decremented digit must be set to '9' to maximize the number while still satisfying the monotone increasing constraint.
3Insight 3: Converting the number to a string or character array facilitates digit-by-digit manipulation.

📚

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. In this case, the algorithm will not enter the loop and will return the number itself which satisfies the condition.
Edge case 2: Input is already a monotone increasing number. In this case, no violations will be found, and the algorithm will return the number itself.
Edge case 3: Input starts with a large digit, followed by smaller digits, like 332. The algorithm will correctly decrement the initial digits and fill the remaining digits with 9s (resulting in 299).
Edge case 4: Input is 0. While not explicitly stated, if the constraint was `n >= 0`, handling '0' as input correctly is important. In the given implementations it should function correctly.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the cascading decrement correctly. When a digit is decremented, it might cause further violations to the left, which must be handled recursively or iteratively.

✗Common mistake 2: Forgetting to set the digits to the right of the decremented digit to '9'. This is crucial for maximizing the result.

✗Common mistake 3: Off-by-one errors in the loop conditions when iterating through the digits.

✗Common mistake 4: Not converting the character representation back to an integer correctly after modifications, either by using `atoi()` or `Integer.parseInt()` or equivalent methods depending on the programming language.

✅

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: In financial systems, generating a sequence of increasing transaction IDs or timestamps with specific monotonicity constraints.

🏦Real-world scenario 2: In data logging, ensuring that log entries have monotonically increasing sequence numbers for easier analysis and debugging.

💡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

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ 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: June 19, 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

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

@CodeNinjaJul 13

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 6

❓

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 "Monotone Increasing Digits"?

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?