1137. N-th Tribonacci Number
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Difficulty: easy Acceptance: 64% Topics: Math, Dynamic Programming, Memoization
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Reference solution (spoiler · python)
# Time: O(logn)
# Space: O(1)
import itertools
class Solution(object):
def tribonacci(self, n):
"""
:type n: int
:rtype: int
"""
def matrix_expo(A, K):
result = [[int(i==j) for j in xrange(len(A))] \
for i in xrange(len(A))]
while K:
if K % 2:
result = matrix_mult(result, A)
A = matrix_mult(A, A)
K /= 2
return result
def matrix_mult(A, B):
ZB = zip(*B)
return [[sum(a*b for a, b in itertools.izip(row, col)) \
for col in ZB] for row in A]
T = [[1, 1, 0],
[1, 0, 1],
[1, 0, 0]]
return matrix_mult([[1, 0, 0]], matrix_expo(T, n))[0][1] # [a1, a0, a(-1)] * T^n
# Time: O(n)
# Space: O(1)
class Solution2(object):
def tribonacci(self, n):
"""
:type n: int
:rtype: int
"""
a, b, c = 0, 1, 1
for _ in xrange(n):
a, b, c = b, c, a+b+c
return a
Solution from kamyu104/LeetCode-Solutions · MIT
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