Leetcode 1997: First Day Where You Have Been in All the Rooms
You are given a list of rooms and their visiting order for each day based on a set of rules. On each day, you either revisit a room based on the number of visits or proceed to the next room in a cyclic manner. Your task is to find the first day when all rooms have been visited.
Problem
Approach
Steps
Complexity
Input: The input consists of a list 'nextVisit' of length n (2 <= n <= 100,000), where each element is an integer indicating the next room to visit based on the rules provided.
Example: nextVisit = [0, 0]
Constraints:
• 2 <= n <= 100,000
• 0 <= nextVisit[i] <= i
Output: Return the first day when all rooms have been visited. The answer should be returned modulo 10^9 + 7.
Example: 2
Constraints:
Goal: We need to track how many times each room is visited and determine when all rooms have been visited at least once.
Steps:
• 1. Use dynamic programming (dp) to track the total number of visits up to each day.
• 2. For each room, depending on whether it has been visited an even or odd number of times, decide the next room to visit.
• 3. Continue until all rooms have been visited at least once, then return the day modulo 10^9 + 7.
Goal: The input list must meet the following constraints to ensure valid input and efficient processing.
Steps:
• The list 'nextVisit' has length between 2 and 100,000.
• Each element in 'nextVisit' is between 0 and the index of that element.
Assumptions:
• The input list will always contain valid values.
• Input: nextVisit = [0, 0]
• Explanation: On day 0, you visit room 0. On day 1, you visit room 0 again, and on day 2, you visit room 1, completing the visit to all rooms.
Approach: The solution uses dynamic programming to track the visits and determine when all rooms are visited.
Observations:
• We can simulate each day by using a dp array to store the cumulative steps taken and track which room is visited.
• Since the number of rooms is large (up to 100,000), we need an efficient approach to calculate the first day all rooms are visited.
Steps:
• 1. Initialize a dp array of size n to track the number of visits for each room.
• 2. For each day, calculate which room to visit based on the current number of visits for each room and update the dp array.
• 3. Continue this until all rooms are visited at least once and return the day modulo 10^9 + 7.
Empty Inputs:
• The problem guarantees a minimum input length of 2.
Large Inputs:
• Ensure that the solution can handle the largest input sizes efficiently within time limits.
Special Values:
• The modulo operation (10^9 + 7) ensures the result remains within valid range.
Constraints:
• The input will always satisfy the constraints.
int n;
int firstDayBeenInAllRooms(vector<int>& nxt) {
n = nxt.size();
int mod = (int) 1e9 + 7;
vector<long long> dp(n, 0); // step
for(int i = 1; i < n; i++)
dp[i] = (dp[i - 1] + 1 + (dp[i - 1] - dp[nxt[i - 1]]) + 1 + mod) % mod;
return dp[n - 1];
}
1 : Variable Declaration
int n;
Declares the variable `n` to store the size of the `nxt` vector, representing the number of rooms.
2 : Function Definition
int firstDayBeenInAllRooms(vector<int>& nxt) {
Defines the function `firstDayBeenInAllRooms`, which takes a reference to a vector `nxt` that represents the connections between rooms, and returns the minimum number of steps modulo 1e9+7.
3 : Array Size Initialization
n = nxt.size();
Initializes `n` with the size of the `nxt` vector, which represents the total number of rooms.
4 : Modulo Constant Declaration
int mod = (int) 1e9 + 7;
Declares and initializes the constant `mod` with the value 1e9 + 7, used for modulo operations to prevent overflow.
5 : Dynamic Programming Initialization
vector<long long> dp(n, 0); // step
Declares a vector `dp` of type `long long` initialized with 0s, which will store the steps for each room. The vector size is `n`.
6 : Loop Start
for(int i = 1; i < n; i++)
Starts a loop from the second room (i=1) to the last room (i<n), calculating the minimum number of steps for each room.
7 : Dynamic Programming Update
dp[i] = (dp[i - 1] + 1 + (dp[i - 1] - dp[nxt[i - 1]]) + 1 + mod) % mod;
Updates the `dp` value for the current room `i` based on the steps from the previous room, adjusted by the modulo to handle overflow.
8 : Return Statement
return dp[n - 1];
Returns the value stored in `dp[n-1]`, representing the minimum number of steps required to visit all rooms.
Best Case: O(n)
Average Case: O(n)
Worst Case: O(n)
Description: The time complexity is O(n) since we only iterate through the rooms once, and the calculations for each room are constant time operations.
Best Case: O(n)
Worst Case: O(n)
Description: The space complexity is O(n) due to the space needed for the dp array to track visits.
| LeetCode Solutions Library / DSA Sheets / Course Catalog |
|---|
comments powered by Disqus