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Formatted question description: https://leetcode.ca/all/2327.html

# 2327. Number of People Aware of a Secret

• Difficulty: Medium.
• Related Topics: Dynamic Programming, Queue, Simulation.
• Similar Questions: .

## Problem

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

## Solution (Java, C++, Python)

• class Solution {
public int peopleAwareOfSecret(int n, int delay, int forget) {
long[][] dp = new long[n + forget][3];
// 0: people who currently know the secret (includes [1] below)
// 1: people who start sharing the secret on this day
// 2: people who forget on this day
long mod = (long) 1e9 + 7;
dp[0][0] = dp[delay][1] = dp[forget][2] = 1;
for (int i = 1; i < n; i++) {
// dp[i][1] was originally just the i - delay newcomers
dp[i][1] = (dp[i][1] + dp[i - 1][1] - dp[i][2] + mod) % mod;
// these people forget on i + forget day
dp[i + forget][2] = dp[i][1];
// these people start sharing on i + delay day
dp[i + delay][1] = dp[i][1];
// today's total people who know the secret
dp[i][0] = (dp[i - 1][0] + dp[i][1] - dp[i][2] + mod) % mod;
}
return (int) dp[n - 1][0];
}
}

############

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;
}
}

• 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

############

# 2327. Number of People Aware of a Secret
# https://leetcode.com/problems/number-of-people-aware-of-a-secret/

class Solution:
def peopleAwareOfSecret(self, n: int, delay: int, forget: int) -> int:
M = 10 ** 9 + 7
res = 1

canShare = defaultdict(int)

toForget = defaultdict(int)
toForget[forget + 1] = 1

propagate = 1

for day in range(delay + 1, n + 1):
propagate += canShare[day]
propagate -= toForget[day]

res += propagate
res -= toForget[day]
res %= M

canShare[day + delay] += propagate
toForget[day + forget] += propagate

return res % M


• 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;
}
};

• 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));
}



Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).