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: Dynamic Programming

• Time complexity: O(n)
• Space complexity: O(n)

 

JavaScript

/**
 * @param {number[]} nextVisit
 * @return {number}
 */
const firstDayBeenInAllRooms = function (nextVisit) {
  const MODULO = 10 ** 9 + 7;
  const size = nextVisit.length;
  const visitDay = new Array(size).fill(0);

  for (let index = 1; index < size; index++) {
    const previousRoom = index - 1;
    const day = visitDay[previousRoom];
    const oddTimes = day - visitDay[nextVisit[previousRoom]] + 1;
    const evenTimes = day + 1;

    visitDay[index] = (oddTimes + evenTimes + MODULO) % MODULO;
  }
  return visitDay[size - 1];
};