Introduction

In this blog post, we will explore how to solve the LeetCode problem "2327" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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

Explanation

Here's a solution that efficiently calculates the number of people who know the secret at the end of day n, considering the delay and forget parameters.

• Dynamic Programming: Use dynamic programming to track the number of people who start sharing the secret on each day. This avoids recalculating information repeatedly.

• Iterative Calculation: Iterate through each day from 1 to n. For each day, determine how many new people learn the secret based on how many people were sharing the secret within their active "sharing" window (delay to forget).

• Modulo Arithmetic: Apply the modulo operator (% (10**9 + 7)) at each step to prevent integer overflow.

• Time and Space Complexity: O(n), O(n)

Code

    def people_aware_of_secret(n: int, delay: int, forget: int) -> int:
    """
    Calculates the number of people who know the secret at the end of day n.

    Args:
        n: The number of days.
        delay: The number of days after discovering the secret before a person starts sharing.
        forget: The number of days after discovering the secret when a person forgets it.

    Returns:
        The number of people who know the secret at the end of day n, modulo 10^9 + 7.
    """

    MOD = 10**9 + 7
    shares = [0] * (n + 1)  # shares[i] is the number of people who *start* sharing on day i
    shares[1] = 1  # One person knows the secret on day 1

    for i in range(1, n + 1):
        for j in range(i + delay, min(i + forget, n + 1)):
            shares[j] = (shares[j] + shares[i]) % MOD

    total_aware = 0
    for i in range(n - forget + 1, n + 1):
        total_aware = (total_aware + shares[i]) % MOD

    return total_aware