How do you solve Number of Beautiful Integers in the Range on LeetCode?

Number of Beautiful Integers in the Range (LeetCode #2827) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n * d) time complexity and O(n * d) space complexity. Key insight: Beautiful integers require both even and odd digit counts to be equal.

What is the brute force solution for Number of Beautiful Integers in the Range?

The brute force approach for Number of Beautiful Integers in the Range is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking each number in the range from low to high to see if it meets the criteria of being beautiful. This is straightforward but inefficient for large ranges. The optimal Optimal Solution approach improves this to O(n * d).

What algorithm or data structure is used to solve Number of Beautiful Integers in the Range?

Number of Beautiful Integers in the Range uses the following concepts: Math, Dynamic Programming. The recommended patterns to study are: Dynamic Programming, Digit Counting.

What are the interview tips for Number of Beautiful Integers in the Range?

Clearly define what makes a number beautiful before diving into code. Consider edge cases and how they might affect your solution. Use memoization effectively to optimize recursive solutions.

What are common mistakes when solving Number of Beautiful Integers in the Range?

Not checking for edge cases such as single-digit numbers. Overlooking the condition of equal counts of even and odd digits.

#2827

Number of Beautiful Integers in the Range

Hard

MathDynamic ProgrammingDynamic ProgrammingDigit Counting

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

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

The optimal solution leverages dynamic programming and digit counting to efficiently calculate the number of beautiful integers without checking each number individually. This reduces the time complexity significantly.

⚙️

Algorithm

5 steps

1Step 1: Define a recursive function that counts beautiful numbers up to a given number n.
2Step 2: Use memoization to store results for previously computed states to avoid redundant calculations.
3Step 3: Count the number of even and odd digits dynamically as we build the number.
4Step 4: Check divisibility by k at each valid number formation.
5Step 5: Use the helper function to compute f(high) - f(low - 1) to get the final result.

solution.py23 lines

1# Full working Python code
2
3def count_beautiful_integers(low, high, k):
4    def dp(pos, even_count, odd_count, tight, num_str):
5        if pos == len(num_str):
6            return 1 if even_count == odd_count and int(''.join(num_str)) % k == 0 else 0
7        if (pos, even_count, odd_count, tight) in memo:
8            return memo[(pos, even_count, odd_count, tight)]
9        limit = int(num_str[pos]) if tight else 9
10        count = 0
11        for digit in range(0, limit + 1):
12            num_str[pos] = str(digit)
13            count += dp(pos + 1, even_count + (digit % 2 == 0), odd_count + (digit % 2 != 0), tight and (digit == limit), num_str)
14        memo[(pos, even_count, odd_count, tight)] = count
15        return count
16
17    memo = {}
18    num_str_high = list(str(high))
19    num_str_low = list(str(low - 1))
20    return dp(0, 0, 0, True, num_str_high) - dp(0, 0, 0, True, num_str_low)
21
22# Example usage
23print(count_beautiful_integers(10, 20, 3))

ℹ

Complexity note: The time complexity is O(n * d) where d is the number of digits in the number, as we are exploring all digit combinations. The space complexity is O(n * d) due to memoization.

1Beautiful integers require both even and odd digit counts to be equal.
2Divisibility by k is a crucial constraint that must be checked for each candidate.

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