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+# 3343. Count Number of Balanced Permutations
+from collections import Counter
+from math import comb
+
+"""
+you are given a string 'num'. a string of digits is called balanced if the
+sum of the digits at even indices is equal to the sum of the digits at odd
+indices. return the number of distinct permutations of 'num' that are
+balanced. since the answer may be very large, return it modulo 10e9+7. a
+permutation is a rearrangement of all the characters of a string.
+"""
+
+
+class Solution(object):
+ def countBalancedPermutations(self, num):
+ """
+ :type num: str
+ :rtype: int
+ """
+ mod, n = 10**9 + 7, len(num)
+ total = sum(int(i) for i in num)
+ if total % 2:
+ return 0
+ fac = [1] * (n + 1)
+ ifac = [1] * (n + 1)
+ inv = [1] * (n + 1)
+ for i in range(1, n + 1):
+ fac[i] = fac[i - 1] * i % mod
+ for i in range(2, n + 1):
+ inv[i] = mod - (mod // i) * inv[mod % i] % mod
+ for i in range(1, n + 1):
+ ifac[i] = ifac[i - 1] * inv[i] % mod
+ half_sum = total // 2
+ half_len = n // 2
+ dp = [[0] * (half_len + 1) for _ in range(half_sum + 1)]
+ dp[0][0] = 1
+ digits = [0] * 10
+ for c in num:
+ d = int(c)
+ digits[d] += 1
+ for i in range(half_sum, d - 1, -1):
+ for j in range(half_len, 0, -1):
+ dp[i][j] = (dp[i][j] + dp[i - d][j - 1]) % mod
+ ans = dp[half_sum][half_len]
+ ans = ans * fac[half_len] % mod * fac[n - half_len] % mod
+ for cnt in digits:
+ ans = ans * ifac[cnt] % mod
+ return ans
+
+
+if __name__ == "__main__":
+ obj = Solution()
+ print(obj.countBalancedPermutations(num="123"))
+ print(obj.countBalancedPermutations(num="112"))
+ print(obj.countBalancedPermutations(num="12345"))