Leetcode daily question (1997. first day where you have been in all the rooms)

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

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This problem seems to have been done before , It may just be similar , But I have no impression at all

The questions I have done recently are cheap , The difficulty is not in solving problems , All in various details

hypothesis dp[i][odd] For an odd number of times into the i Steps required for a room , dp[i][even] Enter the... For an even number of times i Steps required for a room , Then we can push down the following two ;

dp[i][odd] = dp[i-1][even] + 1
dp[i][even] = dp[i][odd] + (dp[i][odd]-dp[k][odd]) + 1

among k = next_visit[i]
The first of these two is easy to understand , Only after Ou enters a room several times can he push back into a room . The second actual expression is , Enter the third even number of times i The premise of a room must be an odd number of times , Because it is an odd number of entries, we need to return to a certain point in front , Be careful , The title says to return to next_visit[i] Any point before , But what we want is the minimum distance , According to the rules of the topic , The farther the point we return from the current point , The number of steps we need to go back to the current point must be more , So don't worry , The return point we choose must be next_visit[i], So we walked again from next_visit[i] To i Distance of , So is dp[i][odd] - dp[k][odd] + 1. The whole calculation process needs attention , Because the results are done mod Operational , So there may be dp[i][odd] < dp[k][odd] The situation of , At this time, the result is negative , To avoid this situation, you can dp[i][odd] + 10.pow(9) + 7

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use std::collections::HashMap;

impl Solution {
    
    fn dp(
        next_visit: &Vec<i32>,
        i: usize,
        is_odd: bool,
        cache: &mut HashMap<(usize, bool), i128>,
    ) -> i128 {
    
        let m = 10i128.pow(9) + 7;
        if is_odd {
    
            let next = if let Some(c) = cache.get(&(i - 1, false)) {
    
                *c
            } else {
    
                Solution::dp(next_visit, i - 1, false, cache)
            };
            let ans = (next + 1) % m;
            cache.insert((i, is_odd), ans);
            return ans;
        }
        let odd = if let Some(c) = cache.get(&(i, true)) {
    
            *c
        } else {
    
            Solution::dp(next_visit, i, true, cache)
        };
        let prev = if let Some(c) = cache.get(&(next_visit[i] as usize, true)) {
    
            *c
        } else {
    
            Solution::dp(next_visit, next_visit[i] as usize, true, cache)
        };
        let ans = if 2 * odd < prev {
    
            (2 * odd + m - prev + 1) % m
        } else {
    
            (2 * odd - prev + 1) % m
        };
        cache.insert((i, is_odd), ans);
        ans
    }
    pub fn first_day_been_in_all_rooms(next_visit: Vec<i32>) -> i32 {
    
        let mut cache = HashMap::new();
        cache.insert((0, true), 0);
        cache.insert((0, false), 1);
        Solution::dp(&next_visit, next_visit.len() - 1, true, &mut cache) as i32
    }
}