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LeetCode 2327: Number of People Aware of a Secret Solution

Master LeetCode problem 2327 (Number of People Aware of a Secret), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

2327. Number of People Aware of a Secret

Medium
Dynamic ProgrammingQueueSimulation

Problem Explanation

Explanation:

To solve this problem, we can simulate the process day by day. We can maintain a set of people who know the secret on each day. On each day, we iterate through the set of people who know the secret and update it based on the rules given in the problem description. We also keep track of the total number of people who know the secret at the end of each day.

Algorithm:

  1. Initialize a set known with one person who knows the secret on day 1.
  2. Iterate from day 2 to day n:
    • Create a new set new_known to store people who will know the secret on the current day.
    • For each person in the known set:
      • If the person has not forgotten the secret, add them to the new_known set.
      • If the person can share the secret on the current day, add a new person to the new_known set.
    • Update known set to be equal to new_known.
    • Update the total number of people who know the secret.
  3. Return the total number of people who know the secret modulo 10^9 + 7.

Time Complexity:

The time complexity of this algorithm is O(n) where n is the given integer representing the number of days.

Space Complexity:

The space complexity is also O(n) as we are storing a set of people who know the secret on each day.

:

Solution Code

class Solution {
    public int findSecret(int n, int delay, int forget) {
        int MOD = 1000000007;
        Set<Integer> known = new HashSet<>();
        known.add(1);
        
        for (int day = 2; day <= n; day++) {
            Set<Integer> newKnown = new HashSet<>();
            for (int person : known) {
                if (day - person <= forget) {
                    newKnown.add(person);
                }
                if (day - person >= delay && day - person <= forget) {
                    newKnown.add(day);
                }
            }
            known = newKnown;
        }
        
        return known.size() % MOD;
    }
}

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