1997. First Day Where You Have Been in All the Rooms


Problem Description

The task is to determine the first day on which you have visited all the rooms in a sequence of n rooms. The rooms are labeled from 0 to n-1, and each day is also labeled starting from 0. On the first day (day 0), you visit room 0. The order in which you visit the rooms on subsequent days is determined by two rules and an array nextVisit:

  • If you are visiting a room i for an odd number of times, the next room you will visit is the one with the number nextVisit[i] (where 0 <= nextVisit[i] <= i).
  • If you are visiting room i for an even number of times, the next room you will visit is (i + 1) mod n.

The goal is to return the label of the first day (mod 10^9 + 7) where you have been in all the rooms at least once. The problem guarantees there is always such a day.

Intuition

The solution is based on dynamic programming. To find the first day when all rooms have been visited, an array f is used to store the number of days needed to visit all rooms up to index i.

  • For each room i after the first (since we start in room 0 on day 0), there are a certain number of steps needed to get to room i from room 0.
  • Upon entering room i for the first time, which is always an odd visit, you must follow the nextVisit[i-1] to determine the next room. Then, you will return to room i and go to i+1 on the even visit.
  • The formula f[i] = (f[i - 1] + 1 + f[i - 1] - f[nextVisit[i - 1]] + 1) % mod encapsulates this behavior. It calculates the total days taken to reach room i by adding:
    • The days to reach the previous room, f[i - 1], plus one day to go to nextVisit[i-1] room following the odd visit rule,
    • The days to return from nextVisit[i-1] to room i again, which is f[i - 1] - f[nextVisit[i - 1]] plus one day for the even visit to move to the next room.
  • This process is repeated until the days needed to reach the last room are determined. The value of f[-1] gives the first day when all rooms have been visited, which is taken modulo 10^9 + 7 to keep the number within the integer limits.

The solution leverages the overlapping subproblems characteristic of dynamic programming where the result of visiting previous rooms is reused to determine the number of days needed to reach the current room.

Learn more about Dynamic Programming patterns.

Solution Approach

The reference solution implements a dynamic programming approach to solve the problem. Here's a walk-through of the implementation and the concepts used:

  1. Initialization: We initialize a list f of length n (where n is the total number of rooms), with all values set to 0. This list will hold the minimum number of days required to reach each room for the first time. The room 0 is already visited on day 0, hence f[0] is initially 0.

  2. Modulo Constant: We define mod as 10**9 + 7 which is used for taking modulo after calculations to prevent integer overflow and as required by the problem statement.

  3. Dynamic Programming Iteration: For each room i starting from 1 to n-1 (since room 0 is the starting room and does not require computation), we calculate the minimum number of days to reach this room for the first time, denoted by f[i]. The formula used is:

    1f[i] = (f[i - 1] + 1 + f[i - 1] - f[nextVisit[i - 1]] + 1) % mod

    The terms in the formula have specific meanings:

    • f[i - 1]: Minimum days to reach the previous room (i-1).
    • f[i - 1] - f[nextVisit[i - 1]]: Days spent to visit nextVisit[i - 1] and return to room i. This is the difference of days needed to reach the previous room i-1 and days needed to reach nextVisit[i - 1].
    • The two +1s in the formula account for the days spent for the odd visit (to move to nextVisit[i - 1]) and the even visit (to return to room i and proceed to the next room).
  4. Result: After iterating through all rooms, f[-1] (the last element of f) will contain the first day where all rooms have been visited at least once. This value accounts for all the previous visits and follows the given rules.

  5. Time Complexity: The time complexity of the solution is O(n) since it iterates over the n rooms once to fill out the dynamic programming table f.

  6. Space Complexity: Since only one additional array f of size n is used, the space complexity of the algorithm is also O(n).

By applying this dynamic programming approach, we leverage previous computations to efficiently find the target day, as each step's result depends on the plurality of previous steps but avoids redundant recomputation of those steps.

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Example Walkthrough

Let's consider a small example where n = 4 and the array nextVisit is given by [0, 1, 2, 0]. This means we have 4 rooms in total and the next room you will visit on an odd occasion is determined by the values in nextVisit.

  1. Initialization: We have f = [0, 0, 0, 0] because we have 4 rooms and room 0 is visited on day 0, so f[0] is 0 by default.

  2. First Calculation (for room 1):

    • We use the formula f[i] = (f[i - 1] + 1 + f[i - 1] - f[nextVisit[i - 1]] + 1) % mod.
    • For room 1, i = 1, nextVisit[0] = 0, so we get:
      • f[1] = (f[0] + 1 + f[0] - f[nextVisit[0]] + 1) % mod = (0 + 1 + 0 - 0 + 1) % mod = 2.
    • This means we will reach room 1 for the first time on day 2.
  3. Second Calculation (for room 2):

    • For room 2, i = 2, nextVisit[1] = 1, we apply the formula:
      • f[2] = (f[1] + 1 + f[1] - f[nextVisit[1]] + 1) % mod = (2 + 1 + 2 - 2 + 1) % mod = 4.
    • This means we will reach room 2 for the first time on day 4.
  4. Third Calculation (for room 3):

    • For room 3, i = 3, nextVisit[2] = 2, we apply the formula:
      • f[3] = (f[2] + 1 + f[2] - f[nextVisit[2]] + 1) % mod = (4 + 1 + 4 - 4 + 1) % mod = 6.
    • This means we will reach room 3 for the first time on day 6.
  5. Final Step: Now, finally, our dynamic array f is [0, 2, 4, 6], and since we are asked to find the first day when we have been in all the rooms at least once, our answer is f[-1], which is 6.

This example clearly demonstrates how the dynamic programming approach builds on the solution to previous rooms to efficiently calculate the arrival at each subsequent room. Here, the array f tracks the day on which each room is first reached using the calculated formula, adhering accurately to the sequence of visits dictated by the nextVisit array.

Solution Implementation

1class Solution:
2    def firstDayBeenInAllRooms(self, nextVisit: List[int]) -> int:
3        # Get the total number of rooms to visit
4        number_of_rooms = len(nextVisit)
5      
6        # Initialize an array to store the days to reach each room for the first time
7        days_to_reach = [0] * number_of_rooms
8      
9        # Define the modulo for large number handling (to prevent integer overflow)
10        mod = 10**9 + 7
11      
12        # Iterate over the rooms starting from the second room, as the first room's day count is zero by default
13        for i in range(1, number_of_rooms):
14            # Calculate number of days to reach this room for the first time
15            # The formula is based on the previous room's day count and the day count at the 
16            # index of the nextVisit for the previous room. We visit the current room after
17            # visiting the previous room twice and once after nextVisit for the previous room.
18            days_to_reach[i] = (days_to_reach[i - 1] + 1 + days_to_reach[i - 1] - days_to_reach[nextVisit[i - 1]] + 1) % mod
19      
20        # Return the number of days to reach the last room for the first time
21        return days_to_reach[-1]
22
1class Solution {
2    public int firstDayBeenInAllRooms(int[] nextVisit) {
3        // Get the number of rooms, which is also the length of the input array.
4        int numRooms = nextVisit.length;
5        // Create an array to store the number of days taken to reach each room for the first time.
6        long[] daysToEnterRoom = new long[numRooms];
7        // Define the modulo constant as per the problem statement (1e9 + 7).
8        final int MODULO = (int) 1e9 + 7;
9      
10        // Iterate over each room starting from the second room (since the first room is the starting point).
11        for (int i = 1; i < numRooms; ++i) {
12            // Calculate the number of days to reach the current room for the first time.
13            // It is based on the days to enter the previous room, plus one day to move to the next current room, plus
14            // the number of days to re-enter the current room after visiting the 'nextVisit' room.
15            // The additional modulo operation ensures that the number stays within the integer range.
16            daysToEnterRoom[i] = (daysToEnterRoom[i - 1] + 1 + daysToEnterRoom[i - 1] - daysToEnterRoom[nextVisit[i - 1]] + 1 + MODULO) % MODULO;
17        }
18      
19        // Return the number of days taken to reach the last room for the first time.
20        return (int) daysToEnterRoom[numRooms - 1];
21    }
22}
23
1class Solution {
2public:
3    int firstDayBeenInAllRooms(vector<int>& nextVisit) {
4        int numRooms = nextVisit.size(); // Store the number of rooms
5        vector<long long> daysToVisit(numRooms); // Create a vector to keep track of the days needed to visit each room
6        const int MOD = 1e9 + 7; // Define the modulo value for large numbers to prevent integer overflow
7
8        // Loop through each room starting from the second room since the first room is always visited on day 0
9        for (int i = 1; i < numRooms; ++i) {
10            // The days to visit current room 'i' is equal to: 
11            // Days to visit the previous room + 1 (for today's visit) +
12            // Days to visit the previous room again after the extra day of waiting, 
13            // plus 2 more days (as you visit nextVisit[i - 1] and then move to room 'i') - 
14            // Days to visit the room pointed to by nextVisit[i - 1].
15            // We also add MOD before taking the modulo to handle any negative numbers.
16            daysToVisit[i] = (daysToVisit[i - 1] + 1 + daysToVisit[i - 1] - daysToVisit[nextVisit[i - 1]] + 2 + MOD) % MOD;
17        }
18
19        // Return the total days needed to visit the last room
20        return daysToVisit[numRooms - 1];
21    }
22};
23
1const MOD = 1e9 + 7;  // Define the modulo constant to prevent integer overflow
2
3function firstDayBeenInAllRooms(nextVisit: number[]): number {
4    const numRooms = nextVisit.length;  // Store the number of rooms
5    let daysToVisit: number[] = new Array(numRooms).fill(0);  // Create an array to keep track of the days needed to visit each room
6
7    // Iterate through each room starting from the second room, since the first room is always visited on day 0
8    for (let i = 1; i < numRooms; i++) {
9        // Calculate the days to visit the current room 'i':
10        // Add one for visiting the previous room, plus the days to reach the previous room again after the revisit, 
11        // plus two more days (for visiting the room at nextVisit[i - 1] and then moving to room 'i'). 
12        // Subtract the days to visit the room pointed to by nextVisit[i - 1].
13        // MOD is added before taking the modulo to handle any negative numbers.
14        daysToVisit[i] = (daysToVisit[i - 1] + 1 + daysToVisit[i - 1] + 2 - daysToVisit[nextVisit[i - 1]] + MOD) % MOD;
15    }
16
17    // Return the total days needed to visit the last room
18    return daysToVisit[numRooms - 1];
19}
20
21// Example usage
22// const nextVisit = [0,0]; // An example input for the function
23// console.log(firstDayBeenInAllRooms(nextVisit)); // Call the function and log the output
24

Time and Space Complexity

The provided code defines a function firstDayBeenInAllRooms which calculates the first day a person has been in all rooms given a sequence of nextVisit specifying the next room to visit.

The time complexity of firstDayBeenInAllRooms is O(n), where n is the length of the input list nextVisit. This is because there is a single loop that iterates through each room exactly once. Each iteration involves constant-time arithmetic operations and a modulo operation, which do not depend on the size of the input.

The space complexity of the function is also O(n) because it allocates an array f of size n to store the number of days taken to reach every room until the last one. No other data structures are used that scale with the input size, hence the space complexity is linear with respect to the input size.

In the loop, for each i (room), the following formula is used to calculate f[i], which represents the first day the person has been in all rooms up to room i:

1f[i] = (f[i - 1] + 1 + f[i - 1] - f[nextVisit[i - 1]] + 1) % mod

The modulo operation with mod = 10**9 + 7 ensures that the result stays within the bounds of a 32-bit integer, which is a common practice to avoid overflow in programming contests and problems.

Learn more about how to find time and space complexity quickly using problem constraints.


Fast Track Your Learning with Our Quick Skills Quiz:

What's the output of running the following function using input 56?

1KEYBOARD = {
2    '2': 'abc',
3    '3': 'def',
4    '4': 'ghi',
5    '5': 'jkl',
6    '6': 'mno',
7    '7': 'pqrs',
8    '8': 'tuv',
9    '9': 'wxyz',
10}
11
12def letter_combinations_of_phone_number(digits):
13    def dfs(path, res):
14        if len(path) == len(digits):
15            res.append(''.join(path))
16            return
17
18        next_number = digits[len(path)]
19        for letter in KEYBOARD[next_number]:
20            path.append(letter)
21            dfs(path, res)
22            path.pop()
23
24    res = []
25    dfs([], res)
26    return res
27
1private static final Map<Character, char[]> KEYBOARD = Map.of(
2    '2', "abc".toCharArray(),
3    '3', "def".toCharArray(),
4    '4', "ghi".toCharArray(),
5    '5', "jkl".toCharArray(),
6    '6', "mno".toCharArray(),
7    '7', "pqrs".toCharArray(),
8    '8', "tuv".toCharArray(),
9    '9', "wxyz".toCharArray()
10);
11
12public static List<String> letterCombinationsOfPhoneNumber(String digits) {
13    List<String> res = new ArrayList<>();
14    dfs(new StringBuilder(), res, digits.toCharArray());
15    return res;
16}
17
18private static void dfs(StringBuilder path, List<String> res, char[] digits) {
19    if (path.length() == digits.length) {
20        res.add(path.toString());
21        return;
22    }
23    char next_digit = digits[path.length()];
24    for (char letter : KEYBOARD.get(next_digit)) {
25        path.append(letter);
26        dfs(path, res, digits);
27        path.deleteCharAt(path.length() - 1);
28    }
29}
30
1const KEYBOARD = {
2    '2': 'abc',
3    '3': 'def',
4    '4': 'ghi',
5    '5': 'jkl',
6    '6': 'mno',
7    '7': 'pqrs',
8    '8': 'tuv',
9    '9': 'wxyz',
10}
11
12function letter_combinations_of_phone_number(digits) {
13    let res = [];
14    dfs(digits, [], res);
15    return res;
16}
17
18function dfs(digits, path, res) {
19    if (path.length === digits.length) {
20        res.push(path.join(''));
21        return;
22    }
23    let next_number = digits.charAt(path.length);
24    for (let letter of KEYBOARD[next_number]) {
25        path.push(letter);
26        dfs(digits, path, res);
27        path.pop();
28    }
29}
30

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