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

Total Waviness of Numbers in Range II (LeetCode #3753) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n * d^2) time complexity and O(n) space complexity. Key insight: Understanding peaks and valleys is crucial for counting waviness.

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

The optimal solution for Total Waviness of Numbers in Range II runs in O(n * d^2) time and uses O(n) space. The complexity arises from exploring each digit and its states, with memoization reducing redundant calculations.

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

The brute force approach for Total Waviness of Numbers in Range II is Brute Force, with O(n²) time complexity and O(1) space complexity. Check each number in the range and count peaks and valleys by examining each digit. This is straightforward but inefficient for larger ranges. The optimal Optimal Solution approach improves this to O(n * d^2).

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

Total Waviness of Numbers in Range II uses the following concepts: Math, Dynamic Programming. The recommended patterns to study are: Dynamic Programming, Digit DP.

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

Explain your thought process clearly. Discuss time and space complexities. Practice similar problems to build confidence.

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

Not considering edge cases for single or double-digit numbers. Overlooking memoization in dynamic programming.

#3753

Total Waviness of Numbers in Range II

Hard

MathDynamic ProgrammingDynamic ProgrammingDigit DP

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n * d^2)
Space	O(1)	O(n)

💡

Intuition

Time O(n * d^2)Space O(n)

Use digit dynamic programming to efficiently count peaks and valleys without explicitly checking each number. This reduces redundant calculations.

⚙️

Algorithm

3 steps

1Step 1: Define a recursive function with parameters (position, tight, lastDigit, secondLastDigit) to explore digit placements.
2Step 2: Use memoization to store results for previously computed states to avoid re-computation.
3Step 3: Count peaks and valleys based on the current digit and its neighbors.

solution.py16 lines

1def countWaviness(num, pos, tight, lastDigit, secondLastDigit, memo):
2    if pos == len(num): return 0
3    if (pos, tight, lastDigit, secondLastDigit) in memo:
4        return memo[(pos, tight, lastDigit, secondLastDigit)]
5    limit = int(num[pos]) if tight else 9
6    total = 0
7    for digit in range(0, limit + 1):
8        if pos > 1:
9            if (digit > lastDigit and lastDigit > secondLastDigit) or (digit < lastDigit and lastDigit < secondLastDigit):
10                total += 1
11        total += countWaviness(num, pos + 1, tight and (digit == limit), digit, lastDigit, memo)
12    memo[(pos, tight, lastDigit, secondLastDigit)] = total
13    return total
14
15def totalWaviness(num1, num2):
16    return countWaviness(str(num2 + 1), 0, True, -1, -1, {}) - countWaviness(str(num1), 0, True, -1, -1, {})

ℹ

Complexity note: The complexity arises from exploring each digit and its states, with memoization reducing redundant calculations.

1Understanding peaks and valleys is crucial for counting waviness.
2Dynamic programming can significantly optimize counting problems.

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