How do you solve Number of Balanced Integers in a Range on LeetCode?

Number of Balanced Integers in a Range (LeetCode #3791) 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 * d^2) space complexity. Key insight: Balanced integers require at least two digits.

What is the time complexity of Number of Balanced Integers in a Range?

The optimal solution for Number of Balanced Integers in a Range runs in O(n * d^2) time and uses O(n * d^2) space. The time complexity is O(n * d^2) where n is the number of digits and d is the range of possible sums, due to the recursive nature and memoization.

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

The brute force approach for Number of Balanced Integers in a Range is Brute Force, with O(n²) time complexity and O(1) space complexity. We can check each number in the range to see if it is balanced by calculating the sums of digits at odd and even positions. This is straightforward but inefficient for large ranges. The optimal Optimal Solution approach improves this to O(n * d^2).

What algorithm or data structure is used to solve Number of Balanced Integers in a Range?

Number of Balanced Integers in a Range uses the following concepts: Dynamic Programming. The recommended patterns to study are: Digit Dynamic Programming, Recursion with Memoization.

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

Explain your thought process clearly. Consider edge cases and constraints. Practice dynamic programming problems.

What are common mistakes when solving Number of Balanced Integers in a Range?

Forgetting to check the length of digits. Not handling leading zeros correctly.

#3791

Number of Balanced Integers in a Range

Hard

Dynamic ProgrammingDigit Dynamic ProgrammingRecursion with Memoization

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

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

Using digit dynamic programming, we can efficiently count balanced integers up to a number by considering the sums of digits at odd and even positions recursively.

⚙️

Algorithm

3 steps

1Step 1: Define a recursive function that counts balanced integers up to a given number, maintaining the current position, sums of odd/even digits, and whether the number is still 'tight' to the original number.
2Step 2: Use memoization to store results for previously computed states to avoid redundant calculations.
3Step 3: Calculate f(high) - f(low - 1) to get the final count of balanced integers in the range.

solution.py14 lines

1def count_balanced(low, high):
2    def dp(pos, odd_sum, even_sum, tight, s):
3        if pos == len(s):
4            return 1 if odd_sum == even_sum else 0
5        if (pos, odd_sum, even_sum, tight) in memo:
6            return memo[(pos, odd_sum, even_sum, tight)]
7        limit = int(s[pos]) if tight else 9
8        count = 0
9        for digit in range(0, limit + 1):
10            count += dp(pos + 1, odd_sum + (digit if pos % 2 == 0 else 0), even_sum + (digit if pos % 2 == 1 else 0), tight and (digit == limit), s)
11        memo[(pos, odd_sum, even_sum, tight)] = count
12        return count
13    memo = {}
14    return dp(0, 0, 0, True, str(high)) - dp(0, 0, 0, True, str(low - 1))

ℹ

Complexity note: The time complexity is O(n * d^2) where n is the number of digits and d is the range of possible sums, due to the recursive nature and memoization.

1Balanced integers require at least two digits.
2The position of digits matters for summation.

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