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698A. Vacations

dp, 1400, https://codeforces.com/problemset/problem/698/A

Vasya has n days of vacations! So he decided to improve his IT skills and do sport. Vasya knows the following information about each of this n days: whether that gym opened and whether a contest was carried out in the Internet on that day. For the i-th day there are four options:

  1. on this day the gym is closed and the contest is not carried out;
  2. on this day the gym is closed and the contest is carried out;
  3. on this day the gym is open and the contest is not carried out;
  4. on this day the gym is open and the contest is carried out.

On each of days Vasya can either have a rest or write the contest (if it is carried out on this day), or do sport (if the gym is open on this day).

Find the minimum number of days on which Vasya will have a rest (it means, he will not do sport and write the contest at the same time). The only limitation that Vasya has — he does not want to do the same activity on two consecutive days: it means, he will not do sport on two consecutive days, and write the contest on two consecutive days.

Input

The first line contains a positive integer n (1 ≤ n ≤ 100) — the number of days of Vasya's vacations.

The second line contains the sequence of integers a1, a2, ..., a**n (0 ≤ a**i ≤ 3) separated by space, where:

  • a**i equals 0, if on the i-th day of vacations the gym is closed and the contest is not carried out;
  • a**i equals 1, if on the i-th day of vacations the gym is closed, but the contest is carried out;
  • a**i equals 2, if on the i-th day of vacations the gym is open and the contest is not carried out;
  • a**i equals 3, if on the i-th day of vacations the gym is open and the contest is carried out.

Output

Print the minimum possible number of days on which Vasya will have a rest. Remember that Vasya refuses:

  • to do sport on any two consecutive days,
  • to write the contest on any two consecutive days.

Examples

input

4
1 3 2 0

output

2

input

7
1 3 3 2 1 2 3

output

0

input

2
2 2

output

1

Note

In the first test Vasya can write the contest on the day number 1 and do sport on the day number 3. Thus, he will have a rest for only 2 days.

In the second test Vasya should write contests on days number 1, 3, 5 and 7, in other days do sport. Thus, he will not have a rest for a single day.

In the third test Vasya can do sport either on a day number 1 or number 2. He can not do sport in two days, because it will be contrary to the his limitation. Thus, he will have a rest for only one day.

Plan

  1. Initialize a 2D list dp where dp[i][0] represents the minimum rest days up to day i if Vasya rests on day i, dp[i][1] represents the minimum rest days up to day i if Vasya participates in a contest on day i, and dp[i][2] represents the minimum rest days up to day i if Vasya goes to the gym on day i.
  2. Initialize the first day's values based on the activity available on the first day.
  3. Iterate through each day from the second day to the last day:
    • Update dp[i][0] as the minimum rest days if Vasya rests on day i.
    • Update dp[i][1] as the minimum rest days if Vasya participates in a contest on day i.
    • Update dp[i][2] as the minimum rest days if Vasya goes to the gym on day i.
  4. The result will be the minimum value among dp[n-1][0], dp[n-1][1], and dp[n-1][2].

Code

python
def min_rest_days(n, activities):
    # Initialize dp array
    dp = [[float('inf')] * 3 for _ in range(n)]
    
    # Initialize the first day
    dp[0][0] = 1  # Rest
    if activities[0] == 1 or activities[0] == 3:
        dp[0][1] = 0  # Contest
    if activities[0] == 2 or activities[0] == 3:
        dp[0][2] = 0  # Gym
    
    # Fill the dp array
    for i in range(1, n):
        # Rest on day i
        dp[i][0] = min(dp[i-1]) + 1
        
        # Contest on day i
        if activities[i] == 1 or activities[i] == 3:
            dp[i][1] = min(dp[i-1][0], dp[i-1][2])
        
        # Gym on day i
        if activities[i] == 2 or activities[i] == 3:
            dp[i][2] = min(dp[i-1][0], dp[i-1][1])
    
    # The result is the minimum value on the last day
    return min(dp[n-1])

# Example usage:
n = int(input())
activities = list(map(int, input().split()))
print(min_rest_days(n, activities))
python
n = int(input())
#n = 100
*a, = map(int, input().split())
#b = '3 2 3 3 3 2 3 1 3 2 2 3 2 3 3 3 3 3 3 1 2 2 3 1 3 3 2 2 2 3 1 0 3 3 3 2 3 \
#    3 1 1 3 1 3 3 3 1 3 1 3 0 1 3 2 3 2 1 1 3 2 3 3 3 2 3 1 3 3 3 3 2 2 2 1 3 \
#    1 3 3 3 3 1 3 2 3 3 0 3 3 3 3 3 1 0 2 1 3 3 0 2 3 3'
#*a, = map(int, b.split())
 
dp = [0]*n
if a[0] == 0:
    dp[0] = 1
 
i= 1
while i < n:
    if a[i] == 0:
        dp[i] = 1
        i += 1
        continue
 
    if a[i] == a[i-1] and a[i] != 3:
        a[i] = 0
        dp[i] = 1
        i += 1
        continue
 
    if a[i] == 3 and a[i-1]!=3:
        a[i] = 3 - a[i-1]
 
    i += 1
 
print(sum(dp))