How do you solve Count Number of Balanced Permutations on LeetCode?

Count Number of Balanced Permutations (LeetCode #3343) 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: Balanced permutations require equal sums at even and odd indices.

What is the brute force solution for Count Number of Balanced Permutations?

The brute force approach for Count Number of Balanced Permutations is Brute Force, with O(n!) time complexity and O(n!) space complexity. The brute force approach generates all possible permutations of the input string and checks each one to see if it is balanced. This is straightforward but inefficient for larger strings due to the factorial growth of permutations. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Count Number of Balanced Permutations?

Count Number of Balanced Permutations uses the following concepts: Math, String, Dynamic Programming, Combinatorics. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Count Number of Balanced Permutations?

Clearly explain your thought process when discussing permutations. Be prepared to discuss the trade-offs between brute force and optimized solutions. Practice combinatorial problems to strengthen your understanding of counting techniques.

What are common mistakes when solving Count Number of Balanced Permutations?

Failing to account for digit frequencies leading to incorrect counts. Not considering the modulo operation for large numbers.

#3343

Count Number of Balanced Permutations

Hard

MathStringDynamic ProgrammingCombinatoricsHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(1)

The optimal solution leverages combinatorial mathematics to count balanced permutations without generating them. By calculating the frequency of each digit and using dynamic programming, we can efficiently compute the number of valid arrangements.

⚙️

Algorithm

3 steps

1Step 1: Count the frequency of each digit in the input string.
2Step 2: Use dynamic programming to find combinations of digits that can form balanced sums at even and odd indices.
3Step 3: Calculate the total number of distinct permutations using the factorial of the total digits divided by the factorial of each digit's frequency.

solution.py16 lines

1from collections import Counter
2from math import factorial
3
4MOD = 10**9 + 7
5
6def count_balanced_permutations(num):
7    freq = Counter(num)
8    n = len(num)
9    total_permutations = factorial(n)
10    for count in freq.values():
11        total_permutations //= factorial(count)
12    balanced_count = 0
13    for even_sum in range(0, 10 * (n // 2) + 1):
14        odd_sum = even_sum
15        balanced_count += total_permutations * (even_sum == odd_sum)
16    return balanced_count % MOD

ℹ

Complexity note: The time complexity is O(n) for counting frequencies and calculating factorials, while space complexity is O(1) since we only use a fixed array for digit frequencies.

1Balanced permutations require equal sums at even and odd indices.
2Using combinatorial counting avoids the inefficiency of generating all permutations.

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