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 <= 10001 <= delay < forget <= n
Solutions
Solution 1: Difference Array
We use a difference array d[i] to record the change in the number of people who know the secret on day i, and an array cnt[i] to record the number of people who newly learn the secret on day i.
For the cnt[i] people who newly learn the secret on day i, they can share the secret with another cnt[i] people each day during the interval [i+delay, i+forget).
The answer is ∑_{i=1}n d[i].
The time complexity is O(n2), and the space complexity is O(n), where n is the given integer.
PythonJavaC++GoTypeScriptRust
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(code-box)
class Solution {
public int peopleAwareOfSecret(int n, int delay, int forget) {
final int mod = (int) 1e9 + 7;
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;
}
}(code-box)
class Solution {
public:
int peopleAwareOfSecret(int n, int delay, int forget) {
const int mod = 1e9 + 7;
int m = (n << 1) + 10;
vector<long long> d(m), cnt(m);
cnt[1] = 1;
for (int i = 1; i <= n; i++) {
if (cnt[i]) {
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 long ans = 0;
for (int i = 0; i <= n; i++) {
ans += d[i];
}
return ans % mod;
}
};(code-box)
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
}(code-box)
function peopleAwareOfSecret(n: number, delay: number, forget: number): number {
const mod = 1e9 + 7;
const m = (n << 1) + 10;
const d: number[] = Array(m).fill(0);
const cnt: number[] = Array(m).fill(0);
cnt[1] = 1;
for (let 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;
let nxt = i + delay;
while (nxt < i + forget) {
cnt[nxt] = (cnt[nxt] + cnt[i]) % mod;
++nxt;
}
}
}
let ans = 0;
for (let i = 1; i <= n; ++i) {
ans = (ans + d[i]) % mod;
}
return ans;
}(code-box)
impl Solution {
pub fn people_aware_of_secret(n: i32, delay: i32, forget: i32) -> i32 {
let n = n as usize;
let delay = delay as usize;
let forget = forget as usize;
let m = (n << 1) + 10;
let modulo: i64 = 1_000_000_007;
let mut d = vec![0i64; m];
let mut cnt = vec![0i64; m];
cnt[1] = 1;
for i in 1..=n {
if cnt[i] > 0 {
d[i] = (d[i] + cnt[i]) % modulo;
d[i + forget] = (d[i + forget] - cnt[i] + modulo) % modulo;
let mut nxt = i + delay;
while nxt < i + forget {
cnt[nxt] = (cnt[nxt] + cnt[i]) % modulo;
nxt += 1;
}
}
}
let mut ans: i64 = 0;
for i in 1..=n {
ans = (ans + d[i]) % modulo;
}
ans as i32
}
}(code-box)
