2327. Number of People Aware of a Secret

Description

On day 1, one person discovers a secret.

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 on 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 109 + 7.

Example 1:

Input: n = 6, delay = 2, forget = 4
Output: 5
Explanation:
Day 1: Suppose the first person is named A. (1 person)
Day 2: A is the only person who knows the secret. (1 person)
Day 3: A shares the secret with a new person, B. (2 people)
Day 4: A shares the secret with a new person, C. (3 people)
Day 5: A forgets the secret, and B shares the secret with a new person, D. (3 people)
Day 6: B shares the secret with E, and C shares the secret with F. (5 people)


Example 2:

Input: n = 4, delay = 1, forget = 3
Output: 6
Explanation:
Day 1: The first person is named A. (1 person)
Day 2: A shares the secret with B. (2 people)
Day 3: A and B share the secret with 2 new people, C and D. (4 people)
Day 4: A forgets the secret. B, C, and D share the secret with 3 new people. (6 people)


Constraints:

• 2 <= n <= 1000
• 1 <= delay < forget <= n

Solutions

• class Solution {
private static final int MOD = (int) 1e9 + 7;

public int peopleAwareOfSecret(int n, int delay, int forget) {
int m = (n << 1) + 10;
long[] d = new long[m];
long[] cnt = new long[m];
cnt[1] = 1;
for (int i = 1; i <= n; ++i) {
if (cnt[i] > 0) {
d[i] = (d[i] + cnt[i]) % MOD;
d[i + forget] = (d[i + forget] - cnt[i] + MOD) % MOD;
int nxt = i + delay;
while (nxt < i + forget) {
cnt[nxt] = (cnt[nxt] + cnt[i]) % MOD;
++nxt;
}
}
}
long ans = 0;
for (int i = 1; i <= n; ++i) {
ans = (ans + d[i]) % MOD;
}
return (int) ans;
}
}

• using ll = long long;
const int mod = 1e9 + 7;

class Solution {
public:
int peopleAwareOfSecret(int n, int delay, int forget) {
int m = (n << 1) + 10;
vector<ll> d(m);
vector<ll> cnt(m);
cnt[1] = 1;
for (int i = 1; i <= n; ++i) {
if (!cnt[i]) continue;
d[i] = (d[i] + cnt[i]) % mod;
d[i + forget] = (d[i + forget] - cnt[i] + mod) % mod;
int nxt = i + delay;
while (nxt < i + forget) {
cnt[nxt] = (cnt[nxt] + cnt[i]) % mod;
++nxt;
}
}
int ans = 0;
for (int i = 1; i <= n; ++i) ans = (ans + d[i]) % mod;
return ans;
}
};

• class Solution:
def peopleAwareOfSecret(self, n: int, delay: int, forget: int) -> int:
m = (n << 1) + 10
d = [0] * m
cnt = [0] * m
cnt[1] = 1
for i in range(1, n + 1):
if cnt[i]:
d[i] += cnt[i]
d[i + forget] -= cnt[i]
nxt = i + delay
while nxt < i + forget:
cnt[nxt] += cnt[i]
nxt += 1
mod = 10**9 + 7
return sum(d[: n + 1]) % mod


• func peopleAwareOfSecret(n int, delay int, forget int) int {
m := (n << 1) + 10
d := make([]int, m)
cnt := make([]int, m)
mod := int(1e9) + 7
cnt[1] = 1
for i := 1; i <= n; i++ {
if cnt[i] == 0 {
continue
}
d[i] = (d[i] + cnt[i]) % mod
d[i+forget] = (d[i+forget] - cnt[i] + mod) % mod
nxt := i + delay
for nxt < i+forget {
cnt[nxt] = (cnt[nxt] + cnt[i]) % mod
nxt++
}
}
ans := 0
for i := 1; i <= n; i++ {
ans = (ans + d[i]) % mod
}
return ans
}

• function peopleAwareOfSecret(n: number, delay: number, forget: number): number {
let dp = new Array(n + 1).fill(0n);
dp[1] = 1n;
for (let i = 2; i <= n; i++) {
let pre = 0n;
for (let j = i - forget + 1; j <= i - delay; j++) {
if (j > 0) {
pre += dp[j];
}
}
dp[i] = pre;
}
let pre = 0n;
let i = n + 1;
for (let j = i - forget; j < i; j++) {
if (j > 0) {
pre += dp[j];
}
}
return Number(pre % BigInt(10 ** 9 + 7));
}