#3343

Hard

auto_awesomeState transition dynamic programming

LeetCode Problem Workspace

Count Number of Balanced Permutations

Determine how many distinct permutations of a digit string are balanced using state transition dynamic programming efficiently.

math string dynamic programming combinatorics

Problem Statement

You are given a string num containing digits from 0 to 9. A permutation of num is considered balanced if the sum of the digits at even indices equals the sum at odd indices. Your task is to compute the number of distinct permutations that satisfy this condition.

Return the total number of balanced permutations modulo 10^9+7. The input string can be up to 80 digits long, so an efficient algorithm using state transition dynamic programming and combinatorics is required to handle all possibilities.

Examples

Example 1

Input: num = "123"

Output: 2

Example 2

Input: num = "112"

Output: 1

Example 3

Input: num = "12345"

Output: 0

Constraints

2 <= num.length <= 80
num consists of digits '0' to '9' only.

Solution Approach

Count Digit Frequencies

Calculate how many times each digit appears in num. This helps define the DP states and reduces redundant permutations by considering identical digits together.

Define DP States and Transitions

Use a DP table where states represent how many digits have been placed at even and odd indices and their sum differences. Transition by placing available digits at the next index and updating the sums accordingly.

Apply Combinatorics with Modulo

Multiply DP state counts by combinatorial choices for identical digits to avoid overcounting. Apply modulo 10^9+7 after each transition to handle large numbers efficiently.

Complexity Analysis

Metric	Value
Time	O(n^2 \cdot S)
Space	O(n^2 + nS)

The time complexity is O(n^2 \cdot S) where n is the string length and S is the maximum sum difference of digits. Space complexity is O(n^2 + nS) to store DP states and combinatorial counts.

What Interviewers Usually Probe

Notice that simple permutation generation will time out for n up to 80.
Expect you to leverage digit frequencies to reduce redundant states.
They may ask about handling modulo operations during state accumulation.

Common Pitfalls or Variants

Common pitfalls

Ignoring identical digits leads to overcounting permutations.
Not updating DP states correctly when switching between even and odd indices.
Forgetting to apply modulo 10^9+7 consistently causes overflow errors.

Follow-up variants

Count balanced permutations where odd and even index sums differ by exactly k.
Compute balanced permutations for strings with alphabetic characters representing numeric values.
Find the maximum balanced sum achievable from any permutation instead of counting.

FAQ

What is a balanced permutation in Count Number of Balanced Permutations?

A balanced permutation is one where the sum of digits at even indices equals the sum at odd indices.

Why use state transition dynamic programming for this problem?

It efficiently tracks partial sums and placement counts, avoiding factorial-time full permutation enumeration.

How do I handle identical digits in DP calculations?

Count the frequency of each digit and multiply DP states by combinatorial coefficients to prevent overcounting.

What is the modulo constraint and why is it necessary?

All results are returned modulo 10^9+7 to prevent integer overflow and maintain consistency for large counts.

Can this approach scale to the maximum string length of 80?

Yes, using digit frequency-based DP and sum tracking allows handling strings up to length 80 efficiently.

terminal

Solution

Solution 1: Memoization Search + Combinatorial Mathematics

First, we count the occurrences of each digit in the string $\textit{num}$ and record them in the array $\textit{cnt}$, then calculate the total sum $\textit{s}$ of the string $\textit{num}$.

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class Solution:
    def countBalancedPermutations(self, num: str) -> int:
        @cache
        def dfs(i: int, j: int, a: int, b: int) -> int:
            if i > 9:
                return (j | a | b) == 0
            if a == 0 and j:
                return 0
            ans = 0
            for l in range(min(cnt[i], a) + 1):
                r = cnt[i] - l
                if 0 <= r <= b and l * i <= j:
                    t = comb(a, l) * comb(b, r) * dfs(i + 1, j - l * i, a - l, b - r)
                    ans = (ans + t) % mod
            return ans

        nums = list(map(int, num))
        s = sum(nums)
        if s % 2:
            return 0
        n = len(nums)
        mod = 10**9 + 7
        cnt = Counter(nums)
        return dfs(0, s // 2, n // 2, (n + 1) // 2)

Continue Practicing

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