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Weekly Contest 257 Q3
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Dynamic Programming

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

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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.

Python3

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]

Java

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];
    }
}

C++

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];
    }
};

Go

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]
}

TypeScript

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];
}

C#

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];
    }
}