Question
A sequence x1, x2, ..., xn is Fibonacci-like if:
•
n >= 3
•
xi + xi+1 == xi+2 for all i + 2 <= n
Given a strictly increasing array arr of positive integers forming a sequence, return the length of the longest Fibonacci-like subsequence of arr. If one does not exist, return 0.
A subsequence is derived from another sequence arr by deleting any number of elements (including none) from arr, without changing the order of the remaining elements. For example, [3, 5, 8] is a subsequence of [3, 4, 5, 6, 7, 8].
Example 1:
Input: arr = [1,2,3,4,5,6,7,8]
Output: 5
Explanation: The longest subsequence that is fibonacci-like: [1,2,3,5,8].
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Example 2:
Input: arr = [1,3,7,11,12,14,18]
Output: 3
Explanation:The longest subsequence that is fibonacci-like: [1,11,12], [3,11,14] or [7,11,18].
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My Answer
# testcase 26/57
# Fail
class Solution:
def lenLongestFibSubseq(self, arr: List[int]) -> int:
n = len(arr)
max_length = 0
for i in range(n):
for j in range(i+1,n):
x = arr[i]
y = arr[j]
length = 2
while x+y in arr:
next_num = x+y
x = y
y = next_num
length+=1
max_length = max(max_length,length)
return max_length
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Optimized Version
class Solution:
def lenLongestFibSubseq(self, arr: List[int]) -> int:
index = {num: i for i, num in enumerate(arr)} # 각 숫자의 인덱스를 저장하는 딕셔너리
dp = {} # DP 테이블 (i, j) -> subsequence 길이
max_length = 0
n = len(arr)
for j in range(n):
for i in range(j):
x, y = arr[i], arr[j]
prev = x + y
if prev in index and index[prev] > j:
k = index[prev] # 다음 숫자의 인덱스
dp[(j, k)] = dp.get((i, j), 2) + 1 # 기존 subsequence 길이 + 1
max_length = max(max_length, dp[(j, k)])
return max_length if max_length >= 3 else 0 # 길이가 3 이상일 때만 유효
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