N-th tribonacci number [Fibonacci Number]¶
Time: O(LogN); Space: O(1); easy
The Tribonacci sequence Tn is defined as follows:
T0 = 0, T1 = 1, T2 = 1, and
Tn+3 = Tn + Tn+1 + Tn+2 for n >= 0.
Given n, return the value of Tn.
Example 1:
Input: n = 4
Output: 4
Explanation:
T_3 = 0 + 1 + 1 = 2
T_4 = 1 + 1 + 2 = 4
Example 2:
Input: n = 25
Output: 1389537
Constraints:
0 <= n <= 37
The answer is guaranteed to fit within a 32-bit integer, ie. answer <= 2^31 - 1.
Hints:
Make an array F of length 38, and set F[0] = 0, F[1] = F[2] = 1.
Now write a loop where you set F[n+3] = F[n] + F[n+1] + F[n+2], and return F[n].
1. Dynamic programming [O(LogN), O(1)]¶
[5]:
class Solution1(object):
"""
Time: O(LogN)
Space: O(1)
"""
def tribonacci(self, n):
"""
:type n: int
:rtype: int
"""
def matrix_mult(m1, m2):
return [[sum(a * b for a, b in zip(m1_row, m2_col)) for m2_col in zip(*m2)] for m1_row in m1]
def identity(size):
size = range(size)
return [[(i == j) * 1 for i in size] for j in size]
def matrix_exp(m, pow):
assert pow >= 0 and int(pow) == pow, "Only non-negative, integer powers allowed"
accumulator = identity(len(m))
for i in range(pow):
accumulator = matrix_mult(accumulator, m)
return accumulator
def print_table(data):
for row in data:
print(' '.join('%-5s' % ('%s' % cell) for cell in row))
T = [[1, 1, 0],
[1, 0, 1],
[1, 0, 0]]
m_exp = matrix_exp(T, n)
res = matrix_mult([[1, 0, 0]], m_exp)
return res[0][1]
[6]:
s = Solution1()
n = 4
assert s.tribonacci(n) == 4
n = 25
assert s.tribonacci(n) == 1389537
2. Simplest¶
[7]:
class Solution2(object):
def tribonacci(self, n):
"""
:type n: int
:rtype: int
"""
a, b, c = 0, 1, 1
for _ in range(n):
a, b, c = b, c, a+b+c
return a
[8]:
s = Solution2()
n = 4
assert s.tribonacci(n) == 4
n = 25
assert s.tribonacci(n) == 1389537