Track Daily Secret Sharing before Each Person Forgets

### Problem One person learns a secret on day 1. A person who learns it on day `d` tells one new person each day from `d + delay` through `d + forget - 1`. On day `d + forget`, they forget and can no longer share it. Return the number of people who still know the secret at the end of day `n`, modulo `10^9 + 7`. ```cpp class Solution { public: int peopleAwareOfSecret(int n, int delay, int forget); }; ``` ### Examples **Example 1:** ```text Input: n = 3, delay = 1, forget = 2 Output: 2 Explanation: The day-1 learner forgets on day 3. The learners from days 2 and 3 still know the secret. ``` **Example 2:** ```text Input: n = 5, delay = 3, forget = 5 Output: 3 Explanation: The original learner shares on days 4 and 5. Nobody has forgotten by day 5, so three people know it. ``` ### Constraints - `2 <= n <= 1000`. - `1 <= delay < forget <= n`. ### Clarifying Questions - Can someone share on the day they learn? No; `delay >= 1`. - Can someone share on the day they forget? No. - Is the result everyone who ever learned it? No; count only people who still know at the end of day `n`. ```hint Check the boundary Mark the first sharing day and forgetting day for a person who learns on day1. ```