552. Student Attendance Record II

Given a positive integer n, return the number of all possible attendance records with length n, which will be regarded as rewardable. The answer may be very large, return it after mod 109 + 7.

A student attendance record is a string that only contains the following three characters:

1. 'A' : Absent.
2. 'L' : Late.
3. 'P' : Present.

A record is regarded as rewardable if it doesn't contain more than one 'A' (absent) or more than two continuous 'L' (late).

Example 1:

Input: n = 2
Output: 8
Explanation:
There are 8 records with length 2 will be regarded as rewardable:
"PP" , "AP", "PA", "LP", "PL", "AL", "LA", "LL"
Only "AA" won't be regarded as rewardable owing to more than one absent times.

Note: The value of n won't exceed 100,000.

Approach #1: DP. [C++]

class Solution {
public:
    int checkRecord(int n) {
        int m = 1000000007;
        int *A = new int [n+5];
        int *P = new int [n+5];
        int *L = new int [n+5];

        P[0] = 1;
        L[0] = 1;
        L[1] = 3;
        A[0] = 1;
        A[1] = 2;
        A[2] = 4;

        if(n == 1) return 3;

        for(int i = 1; i < n; i++)
        {
            A[i - 1] %= m;
            P[i - 1] %= m;
            L[i - 1] %= m;

            P[i] = ((A[i - 1] + P[i - 1]) % m + L[i - 1]) % m;

            if(i > 1) L[i] = ((A[i - 1] + P[i - 1]) % m + (A[i - 2] + P[i - 2]) % m) % m;

            if(i > 2) A[i] = ((A[i - 1] + A[i - 2]) % m + A[i - 3]) % m;
        }

        return ((A[n - 1] % m + P[n - 1] % m) % m + L[n - 1] % m) % m;
    }
};

At the first time i initial A, P, L arrays with n, the compiler hint that AddressSanitizer: heap-buffer-overflow on address. After i modified n to n+5, it compiled successfully.

Analysis:

1. Thinking process

1.1 Divide the whole problem into sub-problems:

Before introducing the way to calculate the number of all possible attendance records with length n, we divide the problem into 3 part.

As the attendance records is made by 3 characters('P', 'L', 'A'), the total number can be divided into

Total = ended with P + ended with L + ended with A.

If we define following series

T(n) is the total number of all possible attendance records with length n.

P(n) is the total number of all possible attendance records with length n.

L(n) is the total number of all possible attendance records with length n.

A(n) is the total number of all possible attendance records with length n.

It can be inferred that

T(n) = A(n) + P(n) + L(n), n >= 1

1.2 Solve the sub-problems by dynamic programming

As I use dynamic programming, I need to find out the recurive relaion in 3 sub-problems.

2.2.1 Calculate P(n)

It can be inferred that

If we add a 'P' to an attendance records with length n-1, we will get an attendance records ended with 'P' with length n.

For an attendance record with length n-1.

• If its (n-1)th character is 'P' ---- CAN add 'P'. ('PP')
• If its (n-1)th character is 'A' ---- CAN add 'P'. ('AP')
• If its (n-1)th character is 'L' ---- CAN add 'P'. ('LP')

which means:

P(n) = A(n-1) + P(n-1) + L(n-1), n >= 2.

and we have initial value for the recursive relation

A(1) = P(1) = L(1) = 1.

1.2.2 Calculate L(n)

Similary,

If we add a 'L' to an attendance records with length n-1, we will get an attendance records ended with 'L' with length n.

But the resulting attendance records must be regarded as rewardable!

As the rule is that a record is regarded as rewardable if it doesn't contain.

more than two continuous 'L' (late).

We need to consider the situations when we can add 'L' to an attendance record with length n-1 and it's still regarded as rewardable.

For an attendance record with length n-1.

• If its (n-1)th character is 'P' ---- CAN add 'L'. ('PL')
• If its (n-1)th character is 'A' ---- CAN add 'L'. ('AL')

---

• If its (n-1)th character is 'L':
  • If its (n-2)th character is 'A' ---- CAN add 'L'. ('ALL')
  • If its (n-2)th character is 'P' ---- CAN add 'L'. ('PLL')
  • If its (n-2)th character is 'L' ---- CAN NOT add 'L'. ('LLL' breaks the rule).

which means:

L(n) = A(n-1) + P(n-1) + A(n-2) + P(n-2), n>= 3.

and we have initial value for the recursive relation

A(1) = P(1) = 1

A(2) = 2, P(2) = 3

and

L(1) = 1, L(2) = 3

1.2.3 Calculate A(n).

Similary,

If we add a 'A' to an attendance records with length n-1, we will get an attendance records endeds with 'A' with length n.

But resulting attendance records must be regarded as rewardable!

As the rule is that a record is regarded as rewardable if it doesn't contain

more than one 'A' (absent)

We need to consider the situations when we can add 'A' to an attedance record with length n-1 and it's still regarded as rewardable.

For an attendance record with length n-1.

• If its (n-1)th character is 'A' ---- CAN NOT add 'A'. ("AA" break the rule)
• If its (n-1)th character is 'P' and has no 'A' ---- CAN add 'A'.
• If its (n-1)th character is 'L' and has no 'A' ---- CAN add 'A'.

If we define series

noAP(n) is the total number of all possible attendance records ended with 'P' with length n and with no 'A'.

noAL(n) is the total number of all possible attendance records ended with 'L' with length n and with no 'A'.

It can be inferred that

A(n) = noAP(n-1) + noAL(n-1), n>= 2.

and we have initial value for the recursive relation

A(1) = 1

noAP(1) = noAL(1) = 1

1.2.4 Calculate noAP(n) and noAL(n)

In 2.2.3, 2 new series noAP(n) and noAL(n) is introduced. Now we focus on the recursive relation in noAP(n) and noAL(n).

For noAP(n), we need to consider the situation when we can add 'P' to an attendance record with length n-1 and no 'A' and it's still regarded as rewardable.

Since noAP(n) has no 'A', we don't need to consider the situation when its (n-1)th character is 'A'.

For an attendance record with length n-1, we can get only 2 situations

• If its (n-1)th character is 'P' and has no 'A' ---- CAN add 'P'.
• If its (n-1)th character is 'L' and has no 'A' ---- CAN add 'P'.

which means

noAP(n) = noAP(n-1) + noAP(n-1), n>= 2.

and we have initial value for the recursive relation

noAP(1) = noAL(1) = 1.

For noAL(n), we need to consider the situations when we can add 'L' to an attendance record with length n-1 and no 'A' ans it's still regraded as rewardable.

Since noAL(n) has no 'A', we don'r need to consider the situation when its (n-1)th character is 'A'.

For an attendance record with length n-1, we can get

• If its (n-1)th character is 'P' and has no 'A' ---- CAN add 'L'. ("PL")
• If its (n-1)th character is ‘L' and has no 'A'.
  • If its (n-2)th character is 'P' and has no 'A' ---- CAN add 'L'.
  • If its (n-2)th character is 'L' and has no 'A' ---- CAN NOT  add 'L'.

which mean

noAL(n) = noAP(n-1) + noAP(n-2), n>= 3.

and we have initial value for the recursive relation

noAP(1) = noAL(1) = 1.

and

noAL(2) = 2

1.3 Recursive relationship summarizaion.

The answer to the whole problem is T(n), and

T(n) = A(n) + P(n) + L(n), n >= 1.

Recursive formula:

P(n) = A(n - 1) + P(n - 1) + L(n - 1), n ≥ 2.

A(n) = noAP(n - 1) + noAL(n - 1), n ≥ 2.

noAP(n) = noAP(n - 1) + noAL(n - 1), n ≥ 2.

L(n) = A(n - 1) + P(n - 1) + A(n - 2) + P(n - 2), n ≥ 3.

noAL(n) = noAP(n - 1) + noAP(n - 2), n ≥ 3.

with initial value

A(1) = P(1) = L(1) = 1.

noAP(1) = noAL(1) = 1.

L(2) = 3.

noAL(2) = 2.

1.4 Simplifying.

When n >= 4, the 3 formulas

A(n) = noAP(n - 1) + noAL(n - 1), n ≥ 2.

noAP(n) = noAP(n - 1) + noAL(n - 1), n ≥ 2.

noAL(n) = noAP(n - 1) + noAP(n - 2), n ≥ 3.

can be simplified to

A(n) = A(n - 1) + A(n - 2) + A(n - 3), n ≥ 4.

Finally, the recursive formula group becomes

P(n) = A(n - 1) + P(n - 1) + L(n - 1), n ≥ 2.

L(n) = A(n - 1) + P(n - 1) + A(n - 2) + P(n - 2), n ≥ 3.

A(n) = A(n - 1) + A(n - 2) + A(n - 3), n ≥ 4.

Here, noAP(n) and noAL(n) disappeared.

with initial value

P(1) = 1.

L(1) = 1, L(2) = 3.

A(1) = 1, A(2) = 2, A(3) = 4.

1.5 Do modulus

The result need to be return after mod 10^9 + 7.

Since the result is generated by adding a lot of middle results together. in order to make sure that every middle result and the final result won't exceed INT_MAX, we need to do modulus for every middle result. and for every 2-middle-result-addition.

2. Complexity analysis

2.1 Time complexity

Since the algorithm is one-pass from 1 to n.

The time complexity is O(n).

2.2 Space Complexity

Since 3 array are used P(n), L(n), A(n), the total size is 3n.

The space complexity is O(n).

Reference:

https://leetcode.com/problems/student-attendance-record-ii/discuss/101643/Share-my-O(n)-C%2B%2B-DP-solution-with-thinking-process-and-explanation