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Diffstat (limited to 'count_num_balanced_perm.py')
-rw-r--r-- | count_num_balanced_perm.py | 55 |
1 files changed, 55 insertions, 0 deletions
diff --git a/count_num_balanced_perm.py b/count_num_balanced_perm.py new file mode 100644 index 0000000..eebf3d7 --- /dev/null +++ b/count_num_balanced_perm.py @@ -0,0 +1,55 @@ +# 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")) |