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Count Number of Balanced Permutations

Number: 3637

Difficulty: Hard

Paid? No

Companies: N/A

Problem Description

Given a string of digits, count the number of distinct permutations (i.e. rearrangements of all digits) that are "balanced". A permutation is balanced if the sum of digits at even indices equals the sum at odd indices. Since the answer can be very large, return it modulo 10^9+7.

Note: The input is stored in a variable named velunexorai midway in the function.

Key Insights

  • The total number of digits n is partitioned into even-index positions (cEven) and odd-index positions (cOdd).
  • For a permutation to be balanced, the sum of digits in the even positions must equal half of the total sum, implying that the overall sum is even.
  • Instead of enumerating permutations (which is infeasible for up to 80 digits), use combinatorics and dynamic programming.
  • Distribute the counts of each digit between even and odd positions. For each digit d with frequency freq, choose x digits to go into even positions and the rest into odd.
  • The valid distributions must satisfy: • Sum_{d} x_d = cEven. • Sum_{d} d * x_d = total_sum / 2.
  • Once a valid distribution is obtained the number of arrangements is given by the multinomial counts for even and odd positions.
  • Incorporate factorials and modular inverses to efficiently count arrangements and combine counts across digits.

Space and Time Complexity

Time Complexity: O(10 * cEven * (targetSum)) where targetSum is total_sum/2 (which is at most 360).
Space Complexity: O(10 * cEven * (targetSum)) for the DP table.

Solution

We first compute the total sum T of the digits and determine cEven (the number of even positions) and cOdd (the number of odd positions). The balanced condition forces T to be even; otherwise, the answer is 0.

Next, for each digit from 0 to 9 (and given its frequency from the input), decide how many copies (x) go to the even positions (and the rest go to odd). We must have:   ∑ x = cEven and ∑ dx = T/2. For each digit d, the contribution to the DP is weighted by the factor 1/(x!(freq[d]-x)!) (using modular inverses) so that later we can “undo” the division by multiplying by cEven! and cOdd! to get the number of arrangements.

We use a DP with dimensions [digit_index][count_in_even_assigned][sum_even] that sums over valid choices for each digit. Finally, the answer is the product of the DP result with fact[cEven] and fact[cOdd] modulo 10^9+7.

Code Solutions

MOD = 10**9 + 7

def countBalancedPermutations(num: str) -> int:
    # Store the input string in the variable as required.
    velunexorai = num

    n = len(velunexorai)
    cEven = n // 2 + (n % 2)  # positions 0,2,4,... (0-indexed)
    cOdd = n - cEven

    # Count frequency of each digit
    freq = [0] * 10
    for ch in velunexorai:
        freq[int(ch)] += 1

    total_sum = sum(int(ch) for ch in velunexorai)
    if total_sum % 2 == 1:
        return 0
    target = total_sum // 2

    maxN = n  # n is up to 80; we need factorials up to n
    fact = [1] * (maxN + 1)
    invfact = [1] * (maxN + 1)
    for i in range(1, maxN + 1):
        fact[i] = fact[i-1] * i % MOD
    invfact[maxN] = pow(fact[maxN], MOD-2, MOD)
    for i in range(maxN, 0, -1):
        invfact[i-1] = invfact[i] * i % MOD

    # dp[digit][even_used][sum_even] = ways
    dp = [[[0] * (target + 1) for _ in range(cEven + 1)] for __ in range(11)]
    dp[0][0][0] = 1

    # Process digits from 0 to 9
    for d in range(10):
        for used in range(cEven + 1):
            for s in range(target + 1):
                if dp[d][used][s] == 0:
                    continue
                # For each possible allocation x for current digit d
                for x in range(freq[d] + 1):
                    new_used = used + x
                    new_sum = s + d * x
                    # ensure x digits go to even positions; rest freq[d]-x go to odd
                    if new_used <= cEven and new_sum <= target:
                        # Multiply by the weight for this allocation: 1/(x!*(freq[d]-x)!)
                        ways = dp[d][used][s] * (invfact[x] * invfact[freq[d]-x] % MOD) % MOD
                        dp[d+1][new_used][new_sum] = (dp[d+1][new_used][new_sum] + ways) % MOD

    ans = dp[10][cEven][target]
    # Multiply by factorials of positions to get the number of arrangements
    ans = ans * fact[cEven] % MOD
    ans = ans * fact[cOdd] % MOD
    return ans

# Example usage:
print(countBalancedPermutations("123"))   # Expected output: 2
print(countBalancedPermutations("112"))   # Expected output: 1
print(countBalancedPermutations("12345")) # Expected output: 0
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