2408. Number Of People Aware Of A Secret¶

Difficulty: Medium

2408. Number of People Aware of a Secret

Medium

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¶

class Solution {
    private int dp[];
    private final int mod = (int)(1e9 + 7);
    public int peopleAwareOfSecret(int n, int delay, int forget) {
        dp = new int[n + 1];
        dp[1] = 1;
        for (int day = 1; day <= n; day++) {
            for (int prevDay = day - forget + 1; prevDay <= day - delay; prevDay++) {
                if (prevDay >= 1) 
                    dp[day] = (dp[day] + dp[prevDay]) % mod;
            }
        }
        long res = 0;
        for (int day = n - forget + 1; day <= n; day++) {
            if (day >= 1) 
                res = (res + dp[day]) % mod;
        }
        return (int)(res);
    }
}

Complexity Analysis¶

Time Complexity: O(?)
Space Complexity: O(?)

Approach¶

Detailed explanation of the approach will be added here