How do you solve Monotone Increasing Digits on LeetCode?

Monotone Increasing Digits (LeetCode #738) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Understanding how to manipulate digits directly can lead to more efficient solutions.

What is the brute force solution for Monotone Increasing Digits?

The brute force approach for Monotone Increasing Digits is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves generating all numbers from 0 to n and checking each one to see if it has monotone increasing digits. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Monotone Increasing Digits?

Monotone Increasing Digits uses the following concepts: Math, Greedy. The recommended patterns to study are: Greedy, Digit Manipulation.

What are the interview tips for Monotone Increasing Digits?

Think about how to break down the problem into manageable parts. Practice identifying patterns in digit manipulation problems. Always consider edge cases and test your solution against them.

What are common mistakes when solving Monotone Increasing Digits?

Not considering edge cases where digits need to be adjusted multiple times. Overlooking the need to set subsequent digits to 9 after a decrement.

#738

Monotone Increasing Digits

Medium

MathGreedyGreedyDigit Manipulation

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal solution involves modifying the digits of n directly to ensure they are monotone increasing. This approach is efficient because it avoids unnecessary checks and directly constructs the result.

⚙️

Algorithm

4 steps

1Step 1: Convert the number n to a string to access each digit.
2Step 2: Identify the first position where a digit is greater than the next digit.
3Step 3: Decrease the digit at this position by 1 and set all subsequent digits to 9.
4Step 4: Ensure that the modified number is still less than or equal to n.

solution.py16 lines

1# Full working Python code
2
3def monotoneIncreasingDigits(n):
4    s = list(str(n))
5    length = len(s)
6    mark = length
7
8    for i in range(length - 1, 0, -1):
9        if s[i] < s[i - 1]:
10            mark = i
11            s[i - 1] = str(int(s[i - 1]) - 1)
12
13    for i in range(mark, length):
14        s[i] = '9'
15
16    return int(''.join(s))

ℹ

Complexity note: The time complexity is O(n) because we traverse the digits of n a couple of times, and the space complexity is O(n) due to the storage of the digits in an array.

1Understanding how to manipulate digits directly can lead to more efficient solutions.
2Recognizing patterns in digit comparisons is crucial for optimization.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.