Is Number of People Aware of a Secret easy or hard?

Number of People Aware of a Secret is classified as Medium difficulty. The challenge lies in modeling when people start sharing and when they forget, then maintaining the correct window of active sharers. Once the lifecycle is understood, the dynamic programming transition becomes straightforward.

Number of People Aware of a Secret Python/Java solution

Both Python and Java implementations follow the same DP recurrence. Maintain an array for people who learn the secret each day, update the active sharer count using delay and forget offsets, and compute values modulo 1e9+7. The algorithm runs in O(n) time and works easily across Python, Java, C++, and other languages.

How to solve Number of People Aware of a Secret in O(n)?

Maintain dp[i] as the number of people who learn the secret on day i. Track a running count of people allowed to share by adding dp[i - delay] when they become eligible and subtracting dp[i - forget] when they forget. The number of new learners each day equals the active sharer count, producing an O(n) dynamic programming solution.

What is the best approach for Number of People Aware of a Secret?

The most practical approach uses dynamic programming with a running count of active sharers. Track how many people learn the secret each day and update a sliding window of those who can share between delay and forget days. This gives an O(n) time solution with O(n) space while directly modeling the lifecycle of each group.

Is Number of People Aware of a Secret asked at Google/Amazon/Meta?

This problem represents a common interview pattern involving lifecycle simulation and dynamic programming. Variants of spreading or expiring information frequently appear in interviews at companies like Amazon, Google, and Meta because they test window management and state transitions.

What data structure is used in Number of People Aware of a Secret?

Most solutions rely on a dynamic programming array combined with a sliding window or running sum. Some implementations conceptually treat the active sharers as a queue of groups entering and leaving the sharing window. These structures help efficiently track who can share on each day.

What is the time complexity of Number of People Aware of a Secret?

The optimal solution runs in O(n) time because each day from 1 to n is processed once. Updates only involve adding or removing groups entering or leaving the sharing window. Space complexity is O(n) with a DP array or O(1) with the optimized rolling-window version.

Description Hints 3 Solution Video 45:15

2327. Number of People Aware of a Secret

Medium60.9% acceptance6 companies

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

Asked at 6 companies

Arcesium NCR

Google Bloomberg Amazon Microsoft

Topics

Hints open one at a time

Hint 1

Let dp[i][j] be the number of people who have known the secret for exactly j + 1 days, at day i.

Hint 2

If j > 0, dp[i][j] = dp[i – 1][j – 1].

Hint 3

dp[i][0] = sum(dp[i – 1][j]) for j in [delay – 1, forget – 2].

45:15

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1class Solution:
2    def peopleAwareOfSecret(self, n: int, delay: int, forget: int) -> int:
3        

Starter codePython 3Spaces: 4

Examples

Case 1Case 2

Input

6
2
4

Input

4
1
3

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