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

Description

There are n rooms you need to visit, labeled from 0 to n - 1. Each day is labeled, starting from 0. You will go in and visit one room a day.

Initially on day 0, you visit room 0. The order you visit the rooms for the coming days is determined by the following rules and a given 0-indexed array nextVisit of length n:

• Assuming that on a day, you visit room i,
• if you have been in room i an odd number of times (including the current visit), on the next day you will visit a room with a lower or equal room number specified by nextVisit[i] where 0 <= nextVisit[i] <= i;
• if you have been in room i an even number of times (including the current visit), on the next day you will visit room (i + 1) mod n.

Return the label of the first day where you have been in all the rooms. It can be shown that such a day exists. Since the answer may be very large, return it modulo 109 + 7.

 

Example 1:

Input: nextVisit = [0,0]
Output: 2
Explanation:
- On day 0, you visit room 0. The total times you have been in room 0 is 1, which is odd.
  On the next day you will visit room nextVisit[0] = 0
- On day 1, you visit room 0, The total times you have been in room 0 is 2, which is even.
  On the next day you will visit room (0 + 1) mod 2 = 1
- On day 2, you visit room 1. This is the first day where you have been in all the rooms.

Example 2:

Input: nextVisit = [0,0,2]
Output: 6
Explanation:
Your room visiting order for each day is: [0,0,1,0,0,1,2,...].
Day 6 is the first day where you have been in all the rooms.

Example 3:

Input: nextVisit = [0,1,2,0]
Output: 6
Explanation:
Your room visiting order for each day is: [0,0,1,1,2,2,3,...].
Day 6 is the first day where you have been in all the rooms.

 

Constraints:

• n == nextVisit.length
• 2 <= n <= 105
• 0 <= nextVisit[i] <= i

Solutions

Solution 1: Dynamic Programming

We define \(f[i]\) as the date number of the first visit to the \(i\)-th room, so the answer is \(f[n - 1]\).

Consider the date number of the first arrival at the \((i-1)\)-th room, denoted as \(f[i-1]\). At this time, it takes one day to return to the \(nextVisit[i-1]\)-th room. Why return? Because the problem restricts \(0 \leq nextVisit[i] \leq i\).

After returning, the \(nextVisit[i-1]\)-th room is visited an odd number of times, and the rooms from \(nextVisit[i-1]+1\) to \(i-1\) are visited an even number of times. At this time, we go to the \((i-1)\)-th room again from the \(nextVisit[i-1]\)-th room, which takes \(f[i-1] - f[nextVisit[i-1]]\) days, and then it takes one more day to reach the \(i\)-th room. Therefore, \(f[i] = f[i-1] + 1 + f[i-1] - f[nextVisit[i-1]] + 1\). Since \(f[i]\) may be very large, we need to take the remainder of \(10^9 + 7\), and to prevent negative numbers, we need to add \(10^9 + 7\).

Finally, return \(f[n-1]\).

The time complexity is \(O(n)\), and the space complexity is \(O(n)\), where \(n\) is the number of rooms.

Python3JavaC++GoTypeScriptC#

class Solution:
    def firstDayBeenInAllRooms(self, nextVisit: List[int]) -> int:
        n = len(nextVisit)
        f = [0] * n
        mod = 10**9 + 7
        for i in range(1, n):
            f[i] = (f[i - 1] + 1 + f[i - 1] - f[nextVisit[i - 1]] + 1) % mod
        return f[-1]

class Solution {
    public int firstDayBeenInAllRooms(int[] nextVisit) {
        int n = nextVisit.length;
        long[] f = new long[n];
        final int mod = (int) 1e9 + 7;
        for (int i = 1; i < n; ++i) {
            f[i] = (f[i - 1] + 1 + f[i - 1] - f[nextVisit[i - 1]] + 1 + mod) % mod;
        }
        return (int) f[n - 1];
    }
}

class Solution {
public:
    int firstDayBeenInAllRooms(vector<int>& nextVisit) {
        int n = nextVisit.size();
        vector<long long> f(n);
        const int mod = 1e9 + 7;
        for (int i = 1; i < n; ++i) {
            f[i] = (f[i - 1] + 1 + f[i - 1] - f[nextVisit[i - 1]] + 1 + mod) % mod;
        }
        return f[n - 1];
    }
};

func firstDayBeenInAllRooms(nextVisit []int) int {
    n := len(nextVisit)
    f := make([]int, n)
    const mod = 1e9 + 7
    for i := 1; i < n; i++ {
        f[i] = (f[i-1] + 1 + f[i-1] - f[nextVisit[i-1]] + 1 + mod) % mod
    }
    return f[n-1]
}

function firstDayBeenInAllRooms(nextVisit: number[]): number {
    const n = nextVisit.length;
    const mod = 1e9 + 7;
    const f: number[] = new Array<number>(n).fill(0);
    for (let i = 1; i < n; ++i) {
        f[i] = (f[i - 1] + 1 + f[i - 1] - f[nextVisit[i - 1]] + 1 + mod) % mod;
    }
    return f[n - 1];
}

public class Solution {
    public int FirstDayBeenInAllRooms(int[] nextVisit) {
        int n = nextVisit.Length;
        long[] f = new long[n];
        int mod = (int)1e9 + 7;
        for (int i = 1; i < n; ++i) {
            f[i] = (f[i - 1] + 1 + f[i - 1] - f[nextVisit[i - 1]] + 1 + mod) % mod;
        }
        return (int)f[n - 1];
    }
}

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