How do you solve Total Waviness of Numbers in Range I on LeetCode?

Total Waviness of Numbers in Range I (LeetCode #3751) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Waviness is only defined for numbers with 3 or more digits.

What is the time complexity of Total Waviness of Numbers in Range I?

The optimal solution for Total Waviness of Numbers in Range I runs in O(n) time and uses O(1) space. The outer loop runs for each number in the range, and the inner loop processes digits in constant time, leading to O(n) complexity.

What is the brute force solution for Total Waviness of Numbers in Range I?

The brute force approach for Total Waviness of Numbers in Range I is Brute Force, with O(n²) time complexity and O(1) space complexity. We will iterate through each number in the range and check each digit to count peaks and valleys. This straightforward method ensures we cover all cases, albeit inefficiently. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Total Waviness of Numbers in Range I?

Total Waviness of Numbers in Range I uses the following concepts: Math, Dynamic Programming, Enumeration. The recommended patterns to study are: Dynamic Programming, Array.

What are the interview tips for Total Waviness of Numbers in Range I?

Clarify the definition of peaks and valleys. Discuss edge cases like single and double-digit numbers. Optimize your solution after brute force.

What are common mistakes when solving Total Waviness of Numbers in Range I?

Ignoring numbers with fewer than 3 digits. Incorrectly counting peaks and valleys due to boundary conditions.

#3751

Total Waviness of Numbers in Range I

Medium

MathDynamic ProgrammingEnumerationDynamic ProgrammingArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(1)

We can optimize by directly calculating waviness without converting numbers to strings repeatedly. Instead, we analyze digits mathematically.

⚙️

Algorithm

3 steps

1Step 1: Loop through each number from num1 to num2.
2Step 2: For each number, extract digits and check for peaks and valleys using arithmetic comparisons.
3Step 3: Sum the waviness counts for each number.

solution.py18 lines

1def total_waviness(num1, num2):
2    total = 0
3    for num in range(num1, num2 + 1):
4        if num < 100:
5            continue
6        waviness = 0
7        prev_digit = num % 10
8        num //= 10
9        curr_digit = num % 10
10        while num > 0:
11            next_digit = num % 10
12            if (curr_digit > prev_digit and curr_digit > next_digit) or (curr_digit < prev_digit and curr_digit < next_digit):
13                waviness += 1
14            prev_digit = curr_digit
15            curr_digit = next_digit
16            num //= 10
17        total += waviness
18    return total

ℹ

Complexity note: The outer loop runs for each number in the range, and the inner loop processes digits in constant time, leading to O(n) complexity.

1Waviness is only defined for numbers with 3 or more digits.
2Peaks and valleys are determined by comparing adjacent digits.

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