From d17d14d45956ea9af8f75e23181d708e4f76979f Mon Sep 17 00:00:00 2001 From: Jack Sangdahl Date: Mon, 8 Sep 2025 21:30:40 -0600 Subject: 2327. number of people aware of a secret --- num_people_aware_secret.c | 32 ++++++++++++++++++++++++++++++++ 1 file changed, 32 insertions(+) create mode 100644 num_people_aware_secret.c (limited to 'num_people_aware_secret.c') 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 +} -- cgit v1.2.3