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authorJack Sangdahl <[email protected]>2025-05-08 23:54:32 -0600
committerJack Sangdahl <[email protected]>2025-05-08 23:54:32 -0600
commit67d6c0ec318b5e9726ce680cb17df94e55510ba2 (patch)
tree444484bf0b7171a8cec399587eb58466eed72efd
parent9b8f993a101cb3f7ab3fdb753c3956a6e9a256c2 (diff)
downloadleetcode-67d6c0ec318b5e9726ce680cb17df94e55510ba2.tar.gz
leetcode-67d6c0ec318b5e9726ce680cb17df94e55510ba2.zip
3343. count number of balanced permutations
-rw-r--r--count_num_balanced_perm.c78
-rw-r--r--count_num_balanced_perm.py55
2 files changed, 133 insertions, 0 deletions
diff --git a/count_num_balanced_perm.c b/count_num_balanced_perm.c
new file mode 100644
index 0000000..203f8c2
--- /dev/null
+++ b/count_num_balanced_perm.c
@@ -0,0 +1,78 @@
+// 3343. Count Number of Balanced Permutations
+#include "leetcode.h"
+
+/*
+ * 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.
+ */
+
+static int mod = 1e9 + 7;
+
+long long mod_exp(long long base, long long exp) {
+ long long res = 1;
+ base %= mod;
+ while (exp) {
+ if (exp & 1)
+ res = (res * base) % mod;
+ base = (base * base) % mod;
+ exp >>= 1;
+ }
+ return res;
+}
+
+int countBalancedPermutations(char *num) {
+ int n = strlen(num), freq[10] = {0};
+ for (int i = 0; i < n; i++)
+ freq[num[i] - '0']++;
+ int even_cnt = (n + 1) / 2, odd_cnt = n - even_cnt, sum = 0;
+ for (int i = 0; i < 10; i++)
+ sum += i * freq[i];
+ if (sum & 1)
+ return 0;
+ int target = sum / 2, max = n;
+ long long *fac = (long long *)malloc((max + 1) * sizeof(long long));
+ long long *ifac = (long long *)malloc((max + 1) * sizeof(long long));
+ fac[0] = 1;
+ for (int i = 1; i <= max; i++)
+ fac[i] = (fac[i - 1] * i) % mod;
+ ifac[max] = mod_exp(fac[max], mod - 2);
+ for (int i = max - 1; i >= 0; i--)
+ ifac[i] = (ifac[i + 1] * (i + 1)) % mod;
+ long long dp[11][42][730] = {{{0}}};
+ dp[0][0][0] = 1;
+ for (int i = 0; i < 10; i++) {
+ for (int j = 0; j <= even_cnt; j++) {
+ for (int k = 0; k <= target; k++) {
+ if (!dp[i][j][k])
+ continue;
+ int curr = freq[i];
+ for (int l = 0; l <= curr; l++) {
+ if (j + l > even_cnt)
+ break;
+ int ns = k + i * l;
+ if (ns > target)
+ break;
+ long long coef = (ifac[l] * ifac[curr - l]) % mod;
+ dp[i + 1][j + l][ns] =
+ (dp[i + 1][j + l][ns] + dp[i][j][k] * coef) % mod;
+ }
+ }
+ }
+ }
+ long long ans = dp[10][even_cnt][target];
+ ans = (ans * fac[even_cnt]) % mod;
+ ans = (ans * fac[odd_cnt]) % mod;
+ free(fac);
+ free(ifac);
+ return (int)ans;
+}
+
+int main() {
+ char *n1 = "123", *n2 = "112", *n3 = "12345";
+ printf("%d\n", countBalancedPermutations(n1)); // expect: 2
+ printf("%d\n", countBalancedPermutations(n2)); // expect: 1
+ printf("%d\n", countBalancedPermutations(n3)); // expect: 0
+}
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"))