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-rw-r--r--num_people_aware_secret.c32
-rw-r--r--num_people_aware_secret.py35
2 files changed, 67 insertions, 0 deletions
diff --git a/num_people_aware_secret.c b/num_people_aware_secret.c
new file mode 100644
index 0000000..d9a09d2
--- /dev/null
+++ b/num_people_aware_secret.c
@@ -0,0 +1,32 @@
+// 2327. Number of People Aware of a Secret
+#include "leetcode.h"
+
+/*
+ * on day 1, one person discovers decret. you are given an integer 'delay'
+ * which means that each person will share the secret with a new person every
+ * day starting from 'delay' days after discovering the secret. you are also
+ * given an integer 'forget', which means that each person will forget the
+ * secret 'forget' days after discovering it. a person cannot share the secret
+ * on the same day they forgot it, or any day afterwards. given an integer 'n',
+ * return the number of people who know the secret at the end of day 'n'. since
+ * the answer may be very large, return it modulo 10^9+7.
+ */
+
+int peopleAwareOfSecret(int n, int delay, int forget) {
+ long *dp = (long *)malloc((n + 1) * sizeof(int));
+ dp[1] = 1;
+ int share = 0, mod = 1e9 + 7, ans = 0;
+ for (int i = 2; i <= n; ++i)
+ dp[i] = share = (share + dp[(int)fmax(i - delay, 0)] -
+ dp[(int)fmax(i - forget, 0)] + mod) %
+ mod;
+ for (int i = n - forget + 1; i <= n; ++i)
+ ans = (ans + dp[i]) % mod;
+ free(dp);
+ return ans;
+}
+
+int main() {
+ printf("%d\n", peopleAwareOfSecret(6, 2, 4)); // expect: 5
+ printf("%d\n", peopleAwareOfSecret(4, 1, 3)); // expect: 6
+}
diff --git a/num_people_aware_secret.py b/num_people_aware_secret.py
new file mode 100644
index 0000000..8b13fea
--- /dev/null
+++ b/num_people_aware_secret.py
@@ -0,0 +1,35 @@
+# 2327. Number of People Aware of a Secret
+
+"""
+on day 1, one person discovers decret. you are given an integer 'delay'
+which means that each person will share the secret with a new person every
+day starting from 'delay' days after discovering the secret. you are also
+given an integer 'forget', which means that each person will forget the
+secret 'forget' days after discovering it. a person cannot share the secret
+on the same day they forgot it, or any day afterwards. given an integer 'n',
+return the number of people who know the secret at the end of day 'n'. since
+the answer may be very large, return it modulo 10^9+7.
+"""
+
+
+class Solution(object):
+ def peopleAwareOfSecret(self, n, delay, forget):
+ """
+ :type n: int
+ :type delay: int
+ :type forget: int
+ :rtype: int
+ """
+ dp = [1] + [0] * forget
+ share, mod = 0, 10**9 + 7
+ for i in range(1, n):
+ dp[i % forget] = share = (
+ share + dp[(i - delay) % forget] - dp[i % forget]
+ ) % forget
+ return sum(dp) % mod
+
+
+if __name__ == "__main__":
+ obj = Solution()
+ print(obj.peopleAwareOfSecret(6, 2, 4))
+ print(obj.peopleAwareOfSecret(4, 1, 3))