LeetCode 1997. First Day Where You Have Been in All the Rooms Solution in Java, C++, Python & More | Explanation + Code

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 ≤ nextVisit[i] ≤ 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 10⁹ + 7, and to prevent negative numbers, we need to add 10⁹ + 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.

PythonJavaC++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](code-box)

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]; } }(code-box)

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]; } };(code-box)

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] }(code-box)

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]; }(code-box)

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]; } }(code-box)