LeetCode 2327 - Number of People Aware of a Secret

Table of Contents

• Problem Statement
• Using Dynamic Programming
• Conclusion

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.

Using Dynamic Programming

Explanation

We use Dynamic Programming (DP) to efficiently track secret sharing.

1. DP Array Initialization

   • dp[i] stores the number of people who learn the secret on Day i.

   • Initially, dp[1] = 1 (since only one person knows the secret on Day 1).

2. Simulating Secret Sharing

   • For each day curDay, if people learn the secret on that day, they will:

     • Start sharing from curDay + delay.

     • Stop sharing by curDay + forget - 1.

   • Update dp[nextDay] to track newly aware people.

3. Final Count of Aware People

   • Sum up all dp[i] values for the last forget days (since others have forgotten).
     
     

class Solution {
    public:
        int peopleAwareOfSecret(int n, int delay, int forget) {
            const int MOD = 1e9 + 7;
    
            vector<int> dp(n + 1, 0);
            dp[1] = 1;
    
            for (int curDay = 1; curDay <= n; ++curDay) {
                if (dp[curDay] > 0) {
                    for (int nextDay = curDay + delay; nextDay < curDay + forget && nextDay <= n; ++nextDay) {
                        dp[nextDay] = (dp[nextDay] + dp[curDay]) % MOD;
                    }
                }
            }
            
            int result = 0;
            for (int day = n - forget + 1; day <= n; ++day) {
                if (day >= 1) {
                    result = (result + dp[day]) % MOD;
                }
            }
    
            return result;
        }
    };

Conclusion

• Time Complexity: 𝑂(𝑛 * forget) - Nested loops update future days.

• Space Complexity: 𝑂(𝑛) - Stores DP array.

Source:

2327. Number of People Aware of a Secret