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A sequence X_1, X_2, ..., X_n is fibonacci-like if:
n >= 3
X_i + X_{i+1} = X_{i+2} for all i + 2 <= n
Given a strictly increasing array A of positive integers forming a sequence, find the length of the longest fibonacci-like subsequence of A. If one does not exist, return 0.
(Recall that a subsequence is derived from another sequence A by deleting any number of elements (including none) from A, 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: [1,2,3,4,5,6,7,8]
Output: 5
Explanation:
The longest subsequence that is fibonacci-like: [1,2,3,5,8].
Example 2:
Input: [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].
Note:
3 <= A.length <= 1000
1 <= A[0] < A[1] < ... < A[A.length - 1] <= 10^9
(The time limit has been reduced by 50% for submissions in Java, C, and C++.)
|
class Solution:
def lenLongestFibSubseq(self, A: List[int]) -> int:
mp = {}
for v in A:
mp[v] = set()
result = 0
for i in range(len(A)):
for j in range(i + 1, len(A)):
a = A[i]
b = A[j]
l = 2
while True:
if b in mp[a]:
break
mp[a].add(b)
c = a + b
if c not in mp:
break
a = b
b = c
l += 1
if l < 3:
l = 0
result = max(result, l)
return result
|
A sequence X_1, X_2, ..., X_n is fibonacci-like if:
n >= 3
X_i + X_{i+1} = X_{i+2} for all i + 2 <= n
Given a strictly increasing array A of positive integers forming a sequence, find the length of the longest fibonacci-like subsequence of A. If one does not exist, return 0.
(Recall that a subsequence is derived from another sequence A by deleting any number of elements (including none) from A, 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: [1,2,3,4,5,6,7,8]
Output: 5
Explanation:
The longest subsequence that is fibonacci-like: [1,2,3,5,8].
Example 2:
Input: [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].
Note:
3 <= A.length <= 1000
1 <= A[0] < A[1] < ... < A[A.length - 1] <= 10^9
(The time limit has been reduced by 50% for submissions in Java, C, and C++.)
|
class Solution:
def lenLongestFibSubseq(self, A: List[int]) -> int:
unique = set(A)
max_len = 0
for i in range(len(A) - 2):
for j in range(i + 1, len(A) - 1):
f, s = A[i], A[j]
length = 2
while f + s in unique:
f, s = s, f + s
length += 1
max_len = max(max_len, length)
return 0 if max_len == 2 else max_len
|
A sequence X_1, X_2, ..., X_n is fibonacci-like if:
n >= 3
X_i + X_{i+1} = X_{i+2} for all i + 2 <= n
Given a strictly increasing array A of positive integers forming a sequence, find the length of the longest fibonacci-like subsequence of A. If one does not exist, return 0.
(Recall that a subsequence is derived from another sequence A by deleting any number of elements (including none) from A, 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: [1,2,3,4,5,6,7,8]
Output: 5
Explanation:
The longest subsequence that is fibonacci-like: [1,2,3,5,8].
Example 2:
Input: [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].
Note:
3 <= A.length <= 1000
1 <= A[0] < A[1] < ... < A[A.length - 1] <= 10^9
(The time limit has been reduced by 50% for submissions in Java, C, and C++.)
|
class Solution:
def lenLongestFibSubseq(self, A: List[int]) -> int:
# dp = {prev_val, next_val: len}
dp = {}
dp[(A[1], A[0] + A[1])] = 2
for i in range(2, len(A)):
for j in range(i):
if (A[j], A[i]) not in dp:
dp[(A[i], A[j] + A[i])] = 2
else:
dp[(A[i], A[j] + A[i])] = dp[(A[j], A[i])] + 1
del dp[(A[j], A[i])]
#print(A[i], A[j], dp)
longest = max(dp.values())
return longest if longest > 2 else 0
|
A sequence X_1, X_2, ..., X_n is fibonacci-like if:
n >= 3
X_i + X_{i+1} = X_{i+2} for all i + 2 <= n
Given a strictly increasing array A of positive integers forming a sequence, find the length of the longest fibonacci-like subsequence of A. If one does not exist, return 0.
(Recall that a subsequence is derived from another sequence A by deleting any number of elements (including none) from A, 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: [1,2,3,4,5,6,7,8]
Output: 5
Explanation:
The longest subsequence that is fibonacci-like: [1,2,3,5,8].
Example 2:
Input: [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].
Note:
3 <= A.length <= 1000
1 <= A[0] < A[1] < ... < A[A.length - 1] <= 10^9
(The time limit has been reduced by 50% for submissions in Java, C, and C++.)
|
class Solution:
def lenLongestFibSubseq(self, A: List[int]) -> int:
setA=set(A)
r=0
for i in range(len(A)):
start=A[i]
for j in range(i+1,len(A)):
prev_prev=start
prev=A[j]
count=0
while prev+prev_prev in setA:
prev_prev,prev=prev,prev+prev_prev
count+=1
r=max(r,count)
if r!=0:
return r+2
else:
return r
|
A sequence X_1, X_2, ..., X_n is fibonacci-like if:
n >= 3
X_i + X_{i+1} = X_{i+2} for all i + 2 <= n
Given a strictly increasing array A of positive integers forming a sequence, find the length of the longest fibonacci-like subsequence of A. If one does not exist, return 0.
(Recall that a subsequence is derived from another sequence A by deleting any number of elements (including none) from A, 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: [1,2,3,4,5,6,7,8]
Output: 5
Explanation:
The longest subsequence that is fibonacci-like: [1,2,3,5,8].
Example 2:
Input: [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].
Note:
3 <= A.length <= 1000
1 <= A[0] < A[1] < ... < A[A.length - 1] <= 10^9
(The time limit has been reduced by 50% for submissions in Java, C, and C++.)
|
class Solution:
def lenLongestFibSubseq(self, A: List[int]) -> int:
S = set(A)
res = 2
for i in range(len(A)-2):
if A[i] * (res - 1) * (res - 2) > A[-1]:
break
for j in range(i + 1, len(A) - 1):
cnt = 2
a, b = A[j], A[i] + A[j]
while b in S:
cnt += 1
a, b = b , a + b
res = max(res, cnt)
return res if res > 2 else 0
|
A sequence X_1, X_2, ..., X_n is fibonacci-like if:
n >= 3
X_i + X_{i+1} = X_{i+2} for all i + 2 <= n
Given a strictly increasing array A of positive integers forming a sequence, find the length of the longest fibonacci-like subsequence of A. If one does not exist, return 0.
(Recall that a subsequence is derived from another sequence A by deleting any number of elements (including none) from A, 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: [1,2,3,4,5,6,7,8]
Output: 5
Explanation:
The longest subsequence that is fibonacci-like: [1,2,3,5,8].
Example 2:
Input: [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].
Note:
3 <= A.length <= 1000
1 <= A[0] < A[1] < ... < A[A.length - 1] <= 10^9
(The time limit has been reduced by 50% for submissions in Java, C, and C++.)
|
class Solution:
def lenLongestFibSubseq(self, A: List[int]) -> int:
L=len(A)
if L<=2:
return 0
res=dict()
for j in range(1,L):
res[A[0],A[j]]=2
for i in range(1,j):
if (A[j]-A[i], A[i]) in res:
res[A[i],A[j]]=max(res.get((A[i],A[j]), 2), res[A[j]-A[i],A[i]]+1)
else:
res[A[i],A[j]]=2
maxx=max(list(res.values()))
if maxx>=3:
return maxx
else:
return 0
|
A sequence X_1, X_2, ..., X_n is fibonacci-like if:
n >= 3
X_i + X_{i+1} = X_{i+2} for all i + 2 <= n
Given a strictly increasing array A of positive integers forming a sequence, find the length of the longest fibonacci-like subsequence of A. If one does not exist, return 0.
(Recall that a subsequence is derived from another sequence A by deleting any number of elements (including none) from A, 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: [1,2,3,4,5,6,7,8]
Output: 5
Explanation:
The longest subsequence that is fibonacci-like: [1,2,3,5,8].
Example 2:
Input: [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].
Note:
3 <= A.length <= 1000
1 <= A[0] < A[1] < ... < A[A.length - 1] <= 10^9
(The time limit has been reduced by 50% for submissions in Java, C, and C++.)
|
class Solution:
def lenLongestFibSubseq(self, A: List[int]) -> int:
s = set(A)
longest = 2
# brute force type of way
# start with each pair and build the highest fib sequence you can
for i in range(len(A)):
for j in range(i+1, len(A)):
x, y = A[i],A[j]
seqlen = 2
while x+y in s:
x, y = y, x + y
seqlen += 1
longest = max(seqlen, longest)
return longest if longest > 2 else 0
|
A sequence X_1, X_2, ..., X_n is fibonacci-like if:
n >= 3
X_i + X_{i+1} = X_{i+2} for all i + 2 <= n
Given a strictly increasing array A of positive integers forming a sequence, find the length of the longest fibonacci-like subsequence of A. If one does not exist, return 0.
(Recall that a subsequence is derived from another sequence A by deleting any number of elements (including none) from A, 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: [1,2,3,4,5,6,7,8]
Output: 5
Explanation:
The longest subsequence that is fibonacci-like: [1,2,3,5,8].
Example 2:
Input: [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].
Note:
3 <= A.length <= 1000
1 <= A[0] < A[1] < ... < A[A.length - 1] <= 10^9
(The time limit has been reduced by 50% for submissions in Java, C, and C++.)
|
class Solution:
def lenLongestFibSubseq(self, A: List[int]) -> int:
n = len(A)
two_sum = {}
for i in range(n - 1):
for j in range(i + 1, n):
two_sum[(A[i], A[j])] = 2
result = 0
for i in range(n - 1):
for j in range(i + 1, n):
if (A[j] - A[i], A[i]) in two_sum:
two_sum[(A[i], A[j])] = max(two_sum[(A[i], A[j])], two_sum[(A[j] - A[i], A[i])] + 1)
result = max(result, two_sum[(A[i], A[j])])
return result
|
A sequence X_1, X_2, ..., X_n is fibonacci-like if:
n >= 3
X_i + X_{i+1} = X_{i+2} for all i + 2 <= n
Given a strictly increasing array A of positive integers forming a sequence, find the length of the longest fibonacci-like subsequence of A. If one does not exist, return 0.
(Recall that a subsequence is derived from another sequence A by deleting any number of elements (including none) from A, 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: [1,2,3,4,5,6,7,8]
Output: 5
Explanation:
The longest subsequence that is fibonacci-like: [1,2,3,5,8].
Example 2:
Input: [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].
Note:
3 <= A.length <= 1000
1 <= A[0] < A[1] < ... < A[A.length - 1] <= 10^9
(The time limit has been reduced by 50% for submissions in Java, C, and C++.)
|
from functools import lru_cache
class Solution:
def lenLongestFibSubseq(self, A: List[int]) -> int:
n = len(A)
S = set(A)
@lru_cache(None)
def fib(p1, p2):
prevsum = p1 + p2
if prevsum in S:
return 1 + fib(p2, prevsum)
return 0
res = 0
for i in range(n-2):
for j in range(i+1, n-1):
curres = fib(A[i], A[j])
if curres:
res = max(res, 2+curres)
return res
|
A sequence X_1, X_2, ..., X_n is fibonacci-like if:
n >= 3
X_i + X_{i+1} = X_{i+2} for all i + 2 <= n
Given a strictly increasing array A of positive integers forming a sequence, find the length of the longest fibonacci-like subsequence of A. If one does not exist, return 0.
(Recall that a subsequence is derived from another sequence A by deleting any number of elements (including none) from A, 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: [1,2,3,4,5,6,7,8]
Output: 5
Explanation:
The longest subsequence that is fibonacci-like: [1,2,3,5,8].
Example 2:
Input: [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].
Note:
3 <= A.length <= 1000
1 <= A[0] < A[1] < ... < A[A.length - 1] <= 10^9
(The time limit has been reduced by 50% for submissions in Java, C, and C++.)
|
class Solution:
def lenLongestFibSubseq(self, A: List[int]) -> int:
# Create a hashmap to find index from value
val_to_idx = {val: idx for idx, val in enumerate(A)}
def maxLevel(A: List[int], sum_idx: int, var2_idx: int, level: int) -> int:
if var2_idx==0:
return level
#print(f\"In maxLevel: sum={A[sum_idx]} (index={sum_idx}), var2={A[var2_idx]} (index={var2_idx}), level={level}\")
val_to_search = A[sum_idx] - A[var2_idx]
returned_idx = val_to_idx.get(val_to_search, None)
if returned_idx is not None and returned_idx<var2_idx:
#print(f\"Found value at index: {returned_idx}\")
return maxLevel(A, var2_idx, returned_idx, level + 1)
#print(f\"Index not found, returning level: {level}\")
return level
longest = 0
for i in reversed(list(range(len(A)))):
for j in reversed(list(range(i))):
result_length = maxLevel(A, i, j, 2)
longest = max(longest, result_length)
if longest < 3:
return 0
return longest
|
A sequence X_1, X_2, ..., X_n is fibonacci-like if:
n >= 3
X_i + X_{i+1} = X_{i+2} for all i + 2 <= n
Given a strictly increasing array A of positive integers forming a sequence, find the length of the longest fibonacci-like subsequence of A. If one does not exist, return 0.
(Recall that a subsequence is derived from another sequence A by deleting any number of elements (including none) from A, 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: [1,2,3,4,5,6,7,8]
Output: 5
Explanation:
The longest subsequence that is fibonacci-like: [1,2,3,5,8].
Example 2:
Input: [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].
Note:
3 <= A.length <= 1000
1 <= A[0] < A[1] < ... < A[A.length - 1] <= 10^9
(The time limit has been reduced by 50% for submissions in Java, C, and C++.)
|
class Solution:
def lenLongestFibSubseq(self, arr: List[int]) -> int:
fibSeqs = {}
ans = 0
for i3 in range(1, len(arr)):
for i2 in reversed(range(i3)):
v1 = arr[i3] - arr[i2]
cl = fibSeqs.get((v1, arr[i2]))
if cl is None:
fibSeqs[(arr[i2], arr[i3])] = 0
continue
if arr[i2] < v1:
break
fibSeqs[(arr[i2], arr[i3])] = cl+1
ans = max(ans, cl+3)
return ans
|
A sequence X_1, X_2, ..., X_n is fibonacci-like if:
n >= 3
X_i + X_{i+1} = X_{i+2} for all i + 2 <= n
Given a strictly increasing array A of positive integers forming a sequence, find the length of the longest fibonacci-like subsequence of A. If one does not exist, return 0.
(Recall that a subsequence is derived from another sequence A by deleting any number of elements (including none) from A, 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: [1,2,3,4,5,6,7,8]
Output: 5
Explanation:
The longest subsequence that is fibonacci-like: [1,2,3,5,8].
Example 2:
Input: [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].
Note:
3 <= A.length <= 1000
1 <= A[0] < A[1] < ... < A[A.length - 1] <= 10^9
(The time limit has been reduced by 50% for submissions in Java, C, and C++.)
|
class Solution:
def lenLongestFibSubseq(self, A: List[int]) -> int:
dp = []
for i, x in enumerate(A):
nd = collections.defaultdict(lambda: 1)
dp.append(nd)
for j in range(i):
prev = x - A[j]
dp[i][A[j]] = dp[j][prev] + 1
temp = [max(y.values()) for y in dp if y!={}]
if temp:
result = max(temp)
return result if result>=3 else 0
return 0
|
A sequence X_1, X_2, ..., X_n is fibonacci-like if:
n >= 3
X_i + X_{i+1} = X_{i+2} for all i + 2 <= n
Given a strictly increasing array A of positive integers forming a sequence, find the length of the longest fibonacci-like subsequence of A. If one does not exist, return 0.
(Recall that a subsequence is derived from another sequence A by deleting any number of elements (including none) from A, 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: [1,2,3,4,5,6,7,8]
Output: 5
Explanation:
The longest subsequence that is fibonacci-like: [1,2,3,5,8].
Example 2:
Input: [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].
Note:
3 <= A.length <= 1000
1 <= A[0] < A[1] < ... < A[A.length - 1] <= 10^9
(The time limit has been reduced by 50% for submissions in Java, C, and C++.)
|
class Solution:
def lenLongestFibSubseq(self, A: List[int]) -> int:
mem = {}
for idx, x in enumerate(A):
for i2 in range(idx):
v2 = A[i2]
if v2 < x:
mem[(v2, x)] = 1 + mem.get((x - v2, v2), 1)
res = max(mem.values())
return 0 if res == 2 else res
|
A sequence X_1, X_2, ..., X_n is fibonacci-like if:
n >= 3
X_i + X_{i+1} = X_{i+2} for all i + 2 <= n
Given a strictly increasing array A of positive integers forming a sequence, find the length of the longest fibonacci-like subsequence of A. If one does not exist, return 0.
(Recall that a subsequence is derived from another sequence A by deleting any number of elements (including none) from A, 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: [1,2,3,4,5,6,7,8]
Output: 5
Explanation:
The longest subsequence that is fibonacci-like: [1,2,3,5,8].
Example 2:
Input: [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].
Note:
3 <= A.length <= 1000
1 <= A[0] < A[1] < ... < A[A.length - 1] <= 10^9
(The time limit has been reduced by 50% for submissions in Java, C, and C++.)
|
class Solution:
def lenLongestFibSubseq(self, A: List[int]) -> int:
res = 2
s = set(A)
for i in range(len(A)):
for j in range(i+1, len(A)):
a, b, l = A[i], A[j], 2
while (a+b) in s:
a,b = b, a+b
l += 1
res = max(res, l)
print(res)
if res > 2:
return res
else:
return 0
|
A sequence X_1, X_2, ..., X_n is fibonacci-like if:
n >= 3
X_i + X_{i+1} = X_{i+2} for all i + 2 <= n
Given a strictly increasing array A of positive integers forming a sequence, find the length of the longest fibonacci-like subsequence of A. If one does not exist, return 0.
(Recall that a subsequence is derived from another sequence A by deleting any number of elements (including none) from A, 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: [1,2,3,4,5,6,7,8]
Output: 5
Explanation:
The longest subsequence that is fibonacci-like: [1,2,3,5,8].
Example 2:
Input: [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].
Note:
3 <= A.length <= 1000
1 <= A[0] < A[1] < ... < A[A.length - 1] <= 10^9
(The time limit has been reduced by 50% for submissions in Java, C, and C++.)
|
class Solution:
def lenLongestFibSubseq(self, A: List[int]) -> int:
dp = [collections.defaultdict(int) for _ in range(len(A))]
longest_seq = 0
s = set(A)
for i in range(2, len(A)):
for j in range(i):
if A[i] - A[j] in s and A[i] - A[j] < A[j]:
two_ago = A[i] - A[j]
one_ago = A[j]
if dp[j][two_ago] > 0:
dp[i][one_ago] = dp[j][two_ago] + 1
else:
dp[i][one_ago] = 3
longest_seq = max(longest_seq, dp[i][one_ago])
return longest_seq
|
A sequence X_1, X_2, ..., X_n is fibonacci-like if:
n >= 3
X_i + X_{i+1} = X_{i+2} for all i + 2 <= n
Given a strictly increasing array A of positive integers forming a sequence, find the length of the longest fibonacci-like subsequence of A. If one does not exist, return 0.
(Recall that a subsequence is derived from another sequence A by deleting any number of elements (including none) from A, 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: [1,2,3,4,5,6,7,8]
Output: 5
Explanation:
The longest subsequence that is fibonacci-like: [1,2,3,5,8].
Example 2:
Input: [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].
Note:
3 <= A.length <= 1000
1 <= A[0] < A[1] < ... < A[A.length - 1] <= 10^9
(The time limit has been reduced by 50% for submissions in Java, C, and C++.)
|
class Solution:
def lenLongestFibSubseq(self, A: List[int]) -> int:
if len(A)<=2:
return 0
N=len(A)
ans=0
dp=dict()
dp[(A[1],A[0])]=2
for i in range(2,N):
for j in range(i):
if (A[j],A[i]-A[j]) in dp:
z=dp[(A[j],A[i]-A[j])]
ans=max(z+1,ans)
dp[(A[i],A[j])]=z+1
else:
dp[(A[i],A[j])]=2
return ans
|
A sequence X_1, X_2, ..., X_n is fibonacci-like if:
n >= 3
X_i + X_{i+1} = X_{i+2} for all i + 2 <= n
Given a strictly increasing array A of positive integers forming a sequence, find the length of the longest fibonacci-like subsequence of A. If one does not exist, return 0.
(Recall that a subsequence is derived from another sequence A by deleting any number of elements (including none) from A, 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: [1,2,3,4,5,6,7,8]
Output: 5
Explanation:
The longest subsequence that is fibonacci-like: [1,2,3,5,8].
Example 2:
Input: [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].
Note:
3 <= A.length <= 1000
1 <= A[0] < A[1] < ... < A[A.length - 1] <= 10^9
(The time limit has been reduced by 50% for submissions in Java, C, and C++.)
|
class Solution:
def lenLongestFibSubseq(self, A: List[int]) -> int:
def getFS(x1, x2):
F = [x1, x2]
while F[-1] <= 1000000000:
F.append(F[-2] + F[-1])
return F
C1 = getFS(1, 0)
C2 = getFS(0, 1)
def getLLFS(start, x1, x2):
max_len = 2
F = [x1, x2]
xi = x1 + x2
for i in range(start, len(A)):
if A[i] == xi:
max_len += 1
F.append(xi)
xi = F[-2] + F[-1]
return max_len
if len(A) < 3:
return len(A)
max_len = 2
for i in range(len(A)):
for j in range(i + 1, len(A)):
x1, x2 = A[i], A[j]
last = x1 * C1[max_len-1] + x2 * C2[max_len-1]
if last > A[-1]:
break
max_len = max(max_len, getLLFS(j + 1, x1, x2))
if max_len < 3:
return 0
return max_len
|
A sequence X_1, X_2, ..., X_n is fibonacci-like if:
n >= 3
X_i + X_{i+1} = X_{i+2} for all i + 2 <= n
Given a strictly increasing array A of positive integers forming a sequence, find the length of the longest fibonacci-like subsequence of A. If one does not exist, return 0.
(Recall that a subsequence is derived from another sequence A by deleting any number of elements (including none) from A, 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: [1,2,3,4,5,6,7,8]
Output: 5
Explanation:
The longest subsequence that is fibonacci-like: [1,2,3,5,8].
Example 2:
Input: [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].
Note:
3 <= A.length <= 1000
1 <= A[0] < A[1] < ... < A[A.length - 1] <= 10^9
(The time limit has been reduced by 50% for submissions in Java, C, and C++.)
|
class Solution:
def lenLongestFibSubseq(self, A: List[int]) -> int:
N = len(A)
indexes = {x:i for i, x in enumerate(A)}
longest = defaultdict(lambda: 1)
for i in range(1, N):
for j in range(i):
k = indexes.get(A[i] - A[j], -1)
longest[(i, j)] = longest[(j, k)] + 1
# print(longest)
longest = max(longest.values())
return longest if longest >= 3 else 0
|
A sequence X_1, X_2, ..., X_n is fibonacci-like if:
n >= 3
X_i + X_{i+1} = X_{i+2} for all i + 2 <= n
Given a strictly increasing array A of positive integers forming a sequence, find the length of the longest fibonacci-like subsequence of A. If one does not exist, return 0.
(Recall that a subsequence is derived from another sequence A by deleting any number of elements (including none) from A, 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: [1,2,3,4,5,6,7,8]
Output: 5
Explanation:
The longest subsequence that is fibonacci-like: [1,2,3,5,8].
Example 2:
Input: [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].
Note:
3 <= A.length <= 1000
1 <= A[0] < A[1] < ... < A[A.length - 1] <= 10^9
(The time limit has been reduced by 50% for submissions in Java, C, and C++.)
|
class Solution:
def lenLongestFibSubseq(self, A: List[int]) -> int:
d = set()
n = len(A)
for c in A:
d.add(c)
max_l = 0
for i in range(n-2):
for j in range(i+1, n-1):
a = A[i]
b = A[j]
count = 0
while (a+b) in d:
count += 1
a,b = b,a+b
max_l = max(max_l, count+2)
if max_l >= 3:
return max_l
return 0
|
A sequence X_1, X_2, ..., X_n is fibonacci-like if:
n >= 3
X_i + X_{i+1} = X_{i+2} for all i + 2 <= n
Given a strictly increasing array A of positive integers forming a sequence, find the length of the longest fibonacci-like subsequence of A. If one does not exist, return 0.
(Recall that a subsequence is derived from another sequence A by deleting any number of elements (including none) from A, 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: [1,2,3,4,5,6,7,8]
Output: 5
Explanation:
The longest subsequence that is fibonacci-like: [1,2,3,5,8].
Example 2:
Input: [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].
Note:
3 <= A.length <= 1000
1 <= A[0] < A[1] < ... < A[A.length - 1] <= 10^9
(The time limit has been reduced by 50% for submissions in Java, C, and C++.)
|
class Solution:
def lenLongestFibSubseq(self, A: List[int]) -> int:
'''
# intuitive, TLE
dp = {i:[[A[i]]] for i in range(len(A))}
res = 0
for i in range(1, len(A)):
for j in range(i):
for sequence in dp[j]:
if len(sequence) <= 1 or\\
sequence[-2]==A[i]-sequence[-1]:
dp[i] += [sequence+[A[i]]]
if len(dp[i][-1]) >= 3:
res = max(res, len(dp[i][-1]))
return res
'''
# kinda of like prefix sum problem
dp = {}
for i in range(len(A)):
b = A[i]
for j in range(i):
a = A[j]
if (b-a, a) not in dp:
dp[(a, b)] = 2
elif b-a < a:
dp[(a, b)] = dp[(b-a, a)]+1
res = max(dp.values())
#print(dp)
return res if res!=2 else 0
|
A sequence X_1, X_2, ..., X_n is fibonacci-like if:
n >= 3
X_i + X_{i+1} = X_{i+2} for all i + 2 <= n
Given a strictly increasing array A of positive integers forming a sequence, find the length of the longest fibonacci-like subsequence of A. If one does not exist, return 0.
(Recall that a subsequence is derived from another sequence A by deleting any number of elements (including none) from A, 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: [1,2,3,4,5,6,7,8]
Output: 5
Explanation:
The longest subsequence that is fibonacci-like: [1,2,3,5,8].
Example 2:
Input: [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].
Note:
3 <= A.length <= 1000
1 <= A[0] < A[1] < ... < A[A.length - 1] <= 10^9
(The time limit has been reduced by 50% for submissions in Java, C, and C++.)
|
class Solution:
def lenLongestFibSubseq(self, A: List[int]) -> int:
n = len(A)
check = set(A)
dp = {}
def helper(i, j):
if (i, j) in dp:
return dp[(i, j)]
total = i + j
if total not in check:
result = 2
else:
val = helper(j, total)
result = val + 1
dp[(i, j)] = result
return result
for i in range(n):
for j in range(i+1, n):
helper(A[i], A[j])
result = max(dp.values())
return result if result > 2 else 0
|
A sequence X_1, X_2, ..., X_n is fibonacci-like if:
n >= 3
X_i + X_{i+1} = X_{i+2} for all i + 2 <= n
Given a strictly increasing array A of positive integers forming a sequence, find the length of the longest fibonacci-like subsequence of A. If one does not exist, return 0.
(Recall that a subsequence is derived from another sequence A by deleting any number of elements (including none) from A, 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: [1,2,3,4,5,6,7,8]
Output: 5
Explanation:
The longest subsequence that is fibonacci-like: [1,2,3,5,8].
Example 2:
Input: [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].
Note:
3 <= A.length <= 1000
1 <= A[0] < A[1] < ... < A[A.length - 1] <= 10^9
(The time limit has been reduced by 50% for submissions in Java, C, and C++.)
|
class Solution:
def lenLongestFibSubseq(self, A: List[int]) -> int:
n = len(A)
dp = [[2]*n for i in range(n)]
map_ = collections.defaultdict(int)
for i,j in enumerate(A):
map_[j] = i+10000
# i:end j:start
for i in range(2, n):
for j in range(1, i):
diff = A[i] - A[j]
if map_[diff] > 0 and map_[diff]-10000 < j:
k = map_[diff]-10000
dp[j][i] = max(dp[j][i], dp[k][j] + 1)
max_ = max(dp[j][i] for i in range(n) for j in range(n))
if max_ == 2:
return 0
else:
return max_
|
A sequence X_1, X_2, ..., X_n is fibonacci-like if:
n >= 3
X_i + X_{i+1} = X_{i+2} for all i + 2 <= n
Given a strictly increasing array A of positive integers forming a sequence, find the length of the longest fibonacci-like subsequence of A. If one does not exist, return 0.
(Recall that a subsequence is derived from another sequence A by deleting any number of elements (including none) from A, 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: [1,2,3,4,5,6,7,8]
Output: 5
Explanation:
The longest subsequence that is fibonacci-like: [1,2,3,5,8].
Example 2:
Input: [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].
Note:
3 <= A.length <= 1000
1 <= A[0] < A[1] < ... < A[A.length - 1] <= 10^9
(The time limit has been reduced by 50% for submissions in Java, C, and C++.)
|
from collections import defaultdict
class Solution:
def lenLongestFibSubseq(self, nums: List[int]) -> int:
num_to_idx = {n: i for i, n in enumerate(nums)}
mat = {}
path_len = 0
for i in range(len(nums)):
for j in range(i + 1, len(nums)):
n = nums[j] - nums[i]
if n < nums[i] and n in num_to_idx:
mat[(i, j)] = mat[(num_to_idx[n], i)] + 1
else:
mat[(i, j)] = 2
path_len = max(path_len, mat[(i, j)])
return path_len if path_len >= 3 else 0
|
A sequence X_1, X_2, ..., X_n is fibonacci-like if:
n >= 3
X_i + X_{i+1} = X_{i+2} for all i + 2 <= n
Given a strictly increasing array A of positive integers forming a sequence, find the length of the longest fibonacci-like subsequence of A. If one does not exist, return 0.
(Recall that a subsequence is derived from another sequence A by deleting any number of elements (including none) from A, 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: [1,2,3,4,5,6,7,8]
Output: 5
Explanation:
The longest subsequence that is fibonacci-like: [1,2,3,5,8].
Example 2:
Input: [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].
Note:
3 <= A.length <= 1000
1 <= A[0] < A[1] < ... < A[A.length - 1] <= 10^9
(The time limit has been reduced by 50% for submissions in Java, C, and C++.)
|
class Solution:
def lenLongestFibSubseq(self, A: List[int]) -> int:
dp={}
dp[(A[0], A[1])]=2
curmax=2
for i in range(2, len(A)):
for j in range(i):
prev=dp[(A[i]-A[j], A[j])] if (A[i]-A[j], A[j]) in dp else 1
dp[(A[j], A[i])]=1+prev
curmax=max(curmax, 1+prev)
if curmax<3:
return 0
return curmax
|
A sequence X_1, X_2, ..., X_n is fibonacci-like if:
n >= 3
X_i + X_{i+1} = X_{i+2} for all i + 2 <= n
Given a strictly increasing array A of positive integers forming a sequence, find the length of the longest fibonacci-like subsequence of A. If one does not exist, return 0.
(Recall that a subsequence is derived from another sequence A by deleting any number of elements (including none) from A, 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: [1,2,3,4,5,6,7,8]
Output: 5
Explanation:
The longest subsequence that is fibonacci-like: [1,2,3,5,8].
Example 2:
Input: [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].
Note:
3 <= A.length <= 1000
1 <= A[0] < A[1] < ... < A[A.length - 1] <= 10^9
(The time limit has been reduced by 50% for submissions in Java, C, and C++.)
|
class Solution:
def lenLongestFibSubseq(self, A: List[int]) -> int:
st = set(A)
lenth = 0
dq = collections.deque()
for j in range(1, len(A) -1):
for i in range(j):
if A[i]+A[j] in st:
lenth = 3
dq.append([3, A[j], A[i] + A[j]])
while dq:
sz = len(dq)
for _ in range(sz):
s = dq.popleft()
if s[1] + s[2] in st:
s[0], s[1], s[2] = s[0] + 1, s[2], s[1] + s[2]
lenth = max(lenth, s[0])
dq.append(s)
return lenth
# class Solution:
# def lenLongestFibSubseq(self, A: List[int]) -> int:
# n = len(A)
# s = set(A)
# res = 0
# for i in range(n):
# for j in range(i + 1, n):
# a, b = A[i], A[j]
# tep = 2
# while a + b in s:
# tep += 1
# a, b = b, a + b
# if tep >= 3: res = max(tep, res)
# # if res >= n - j: break
# # if res >= n - i: return res
# return res
|
A sequence X_1, X_2, ..., X_n is fibonacci-like if:
n >= 3
X_i + X_{i+1} = X_{i+2} for all i + 2 <= n
Given a strictly increasing array A of positive integers forming a sequence, find the length of the longest fibonacci-like subsequence of A. If one does not exist, return 0.
(Recall that a subsequence is derived from another sequence A by deleting any number of elements (including none) from A, 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: [1,2,3,4,5,6,7,8]
Output: 5
Explanation:
The longest subsequence that is fibonacci-like: [1,2,3,5,8].
Example 2:
Input: [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].
Note:
3 <= A.length <= 1000
1 <= A[0] < A[1] < ... < A[A.length - 1] <= 10^9
(The time limit has been reduced by 50% for submissions in Java, C, and C++.)
|
import bisect
class Solution:
def lenLongestFibSubseq(self, A: List[int]) -> int:
n = len(A)
dp = [0] * n
for i in range(0, n):
dp[i] = [0] * n
record = {elem: idx for idx, elem in enumerate(A)}
for i in range(0, n-1):
for j in range(i+1, n):
target = A[j] - A[i]
if target in record and record[target] < i:
k = record[target]
dp[i][j] = max(dp[i][j], dp[k][i]+1)
ans = 0
for i in range(0, n):
for j in range(0, n):
ans = max(ans, dp[i][j])
if ans > 0:
return ans+2
return 0
|
A sequence X_1, X_2, ..., X_n is fibonacci-like if:
n >= 3
X_i + X_{i+1} = X_{i+2} for all i + 2 <= n
Given a strictly increasing array A of positive integers forming a sequence, find the length of the longest fibonacci-like subsequence of A. If one does not exist, return 0.
(Recall that a subsequence is derived from another sequence A by deleting any number of elements (including none) from A, 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: [1,2,3,4,5,6,7,8]
Output: 5
Explanation:
The longest subsequence that is fibonacci-like: [1,2,3,5,8].
Example 2:
Input: [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].
Note:
3 <= A.length <= 1000
1 <= A[0] < A[1] < ... < A[A.length - 1] <= 10^9
(The time limit has been reduced by 50% for submissions in Java, C, and C++.)
|
from functools import lru_cache
from itertools import combinations
class Solution:
def lenLongestFibSubseq(self, A: List[int]) -> int:
val2idx = {v: i for i, v in enumerate(A)}
@lru_cache(None)
def longest_fib_with_first_second(first, second):
third = val2idx.get(A[first] + A[second], None)
if third:
return 1 + longest_fib_with_first_second(second, third)
return 2
ret = max(
longest_fib_with_first_second(i, j)
for i, j in combinations(list(range(len(A))), 2)
)
return ret if ret >= 3 else 0
|
A sequence X_1, X_2, ..., X_n is fibonacci-like if:
n >= 3
X_i + X_{i+1} = X_{i+2} for all i + 2 <= n
Given a strictly increasing array A of positive integers forming a sequence, find the length of the longest fibonacci-like subsequence of A. If one does not exist, return 0.
(Recall that a subsequence is derived from another sequence A by deleting any number of elements (including none) from A, 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: [1,2,3,4,5,6,7,8]
Output: 5
Explanation:
The longest subsequence that is fibonacci-like: [1,2,3,5,8].
Example 2:
Input: [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].
Note:
3 <= A.length <= 1000
1 <= A[0] < A[1] < ... < A[A.length - 1] <= 10^9
(The time limit has been reduced by 50% for submissions in Java, C, and C++.)
|
class Solution:
def lenLongestFibSubseq(self, A: List[int]) -> int:
set_a = set(A)
seen = set()
n = len(A)
last = A[-1]
g = 0
def dfs(x, y):
seen.add((x, y))
if (z := x + y) in set_a:
return 1 + dfs(y, z)
else:
return 0
for i in range(n - 1):
for j in range(i + 1, n):
if A[i] + A[j] > last:
break
if (A[i], A[j]) in seen:
continue
g = max(g, dfs(A[i], A[j]))
return g + 2 if g else 0
|
A sequence X_1, X_2, ..., X_n is fibonacci-like if:
n >= 3
X_i + X_{i+1} = X_{i+2} for all i + 2 <= n
Given a strictly increasing array A of positive integers forming a sequence, find the length of the longest fibonacci-like subsequence of A. If one does not exist, return 0.
(Recall that a subsequence is derived from another sequence A by deleting any number of elements (including none) from A, 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: [1,2,3,4,5,6,7,8]
Output: 5
Explanation:
The longest subsequence that is fibonacci-like: [1,2,3,5,8].
Example 2:
Input: [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].
Note:
3 <= A.length <= 1000
1 <= A[0] < A[1] < ... < A[A.length - 1] <= 10^9
(The time limit has been reduced by 50% for submissions in Java, C, and C++.)
|
class Solution:
def lenLongestFibSubseq(self, lst: List[int]) -> int:
m = 0
dp ={}
for i in range(len(lst)):
for j in range(i):
j1 = (j,lst[i])
i1 = (i,lst[i]+lst[j])
if j1 in dp:
dp[i1] = 1 + dp[j1]
else:
dp[i1] = 2
m = max(m,dp[i1])
if m <3:
return 0
return m
|
A sequence X_1, X_2, ..., X_n is fibonacci-like if:
n >= 3
X_i + X_{i+1} = X_{i+2} for all i + 2 <= n
Given a strictly increasing array A of positive integers forming a sequence, find the length of the longest fibonacci-like subsequence of A. If one does not exist, return 0.
(Recall that a subsequence is derived from another sequence A by deleting any number of elements (including none) from A, 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: [1,2,3,4,5,6,7,8]
Output: 5
Explanation:
The longest subsequence that is fibonacci-like: [1,2,3,5,8].
Example 2:
Input: [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].
Note:
3 <= A.length <= 1000
1 <= A[0] < A[1] < ... < A[A.length - 1] <= 10^9
(The time limit has been reduced by 50% for submissions in Java, C, and C++.)
|
class Solution:
def lenLongestFibSubseq(self, A: List[int]) -> int:
result = 0
from collections import Counter
d = Counter(A)
for i in range(len(A)-1):
for j in range(i+1,len(A)):
x,y = A[i],A[j]
temp = 0
while d[x+y] != 0:
x,y = y,x+y
temp += 1
result = max(result,temp)
print(result+2)
return (result+2 if result > 0 else 0)
|
A sequence X_1, X_2, ..., X_n is fibonacci-like if:
n >= 3
X_i + X_{i+1} = X_{i+2} for all i + 2 <= n
Given a strictly increasing array A of positive integers forming a sequence, find the length of the longest fibonacci-like subsequence of A. If one does not exist, return 0.
(Recall that a subsequence is derived from another sequence A by deleting any number of elements (including none) from A, 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: [1,2,3,4,5,6,7,8]
Output: 5
Explanation:
The longest subsequence that is fibonacci-like: [1,2,3,5,8].
Example 2:
Input: [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].
Note:
3 <= A.length <= 1000
1 <= A[0] < A[1] < ... < A[A.length - 1] <= 10^9
(The time limit has been reduced by 50% for submissions in Java, C, and C++.)
|
class Solution:
def lenLongestFibSubseq(self, A: List[int]) -> int:
setA=set(A)
overlapping=set()
r=0
for i in range(len(A)):
start=A[i]
for j in range(i+1,len(A)-1):
prev_prev=start
prev=A[j]
count=0
while prev+prev_prev in setA:
overlapping.add(prev)
overlapping.add(prev_prev)
prev_prev,prev=prev,prev+prev_prev
count+=1
r=max(r,count)
if r!=0:
return r+2
else:
return r
|
A sequence X_1, X_2, ..., X_n is fibonacci-like if:
n >= 3
X_i + X_{i+1} = X_{i+2} for all i + 2 <= n
Given a strictly increasing array A of positive integers forming a sequence, find the length of the longest fibonacci-like subsequence of A. If one does not exist, return 0.
(Recall that a subsequence is derived from another sequence A by deleting any number of elements (including none) from A, 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: [1,2,3,4,5,6,7,8]
Output: 5
Explanation:
The longest subsequence that is fibonacci-like: [1,2,3,5,8].
Example 2:
Input: [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].
Note:
3 <= A.length <= 1000
1 <= A[0] < A[1] < ... < A[A.length - 1] <= 10^9
(The time limit has been reduced by 50% for submissions in Java, C, and C++.)
|
from collections import defaultdict
class Solution:
def lenLongestFibSubseq(self, A: List[int]) -> int:
A_indices = {a: i for i, a in enumerate(A)}
memo = {}
self.best = 2
def get_value(i, j):
if (i, j) in memo:
return memo[i, j]
# search for A[i] + A[j] in A
if A[i] + A[j] in A_indices:
memo[i, j] = 1 + get_value(j, A_indices[A[i] + A[j]])
self.best = max(self.best, memo[i, j])
else:
memo[i, j] = 2
return memo[i, j]
for i in range(len(A)):
for j in range(i + 1, len(A)):
get_value(i, j)
return self.best if self.best > 2 else 0
|
A sequence X_1, X_2, ..., X_n is fibonacci-like if:
n >= 3
X_i + X_{i+1} = X_{i+2} for all i + 2 <= n
Given a strictly increasing array A of positive integers forming a sequence, find the length of the longest fibonacci-like subsequence of A. If one does not exist, return 0.
(Recall that a subsequence is derived from another sequence A by deleting any number of elements (including none) from A, 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: [1,2,3,4,5,6,7,8]
Output: 5
Explanation:
The longest subsequence that is fibonacci-like: [1,2,3,5,8].
Example 2:
Input: [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].
Note:
3 <= A.length <= 1000
1 <= A[0] < A[1] < ... < A[A.length - 1] <= 10^9
(The time limit has been reduced by 50% for submissions in Java, C, and C++.)
|
class Solution:
def lenLongestFibSubseq(self, A: List[int]) -> int:
n = len(A)
A_set = set(A)
res_set = []
res = 0
for i in range(n-2):
for j in range(i+1,n):
if A[i] + A[j] in A_set:
res_set.append((A[j],A[i]+A[j]))
if res_set:
res = 3
else:
return 0
while res_set:
length = len(res_set)
while length:
x,y = res_set.pop(0)
if x+y in A_set:
res_set.append((y,x+y))
length -= 1
if res_set:
res += 1
return res
|
A sequence X_1, X_2, ..., X_n is fibonacci-like if:
n >= 3
X_i + X_{i+1} = X_{i+2} for all i + 2 <= n
Given a strictly increasing array A of positive integers forming a sequence, find the length of the longest fibonacci-like subsequence of A. If one does not exist, return 0.
(Recall that a subsequence is derived from another sequence A by deleting any number of elements (including none) from A, 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: [1,2,3,4,5,6,7,8]
Output: 5
Explanation:
The longest subsequence that is fibonacci-like: [1,2,3,5,8].
Example 2:
Input: [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].
Note:
3 <= A.length <= 1000
1 <= A[0] < A[1] < ... < A[A.length - 1] <= 10^9
(The time limit has been reduced by 50% for submissions in Java, C, and C++.)
|
class Solution:
def lenLongestFibSubseq(self, A: List[int]) -> int:
if len(A)<3:
return 0
dic={}
m=max(A)
for i in range(len(A)):
dic[A[i]]=1
m=0
for i in range(len(A)):
for j in range(i+1,len(A)):
c=2
a=A[i]
b=A[j]
try:
while dic[a+b]==1:
temp=b
b=a+b
a=temp
c+=1
except KeyError:
m=max(m,c)
m=max(m,c)
if m<3:
return 0
return m
|
A sequence X_1, X_2, ..., X_n is fibonacci-like if:
n >= 3
X_i + X_{i+1} = X_{i+2} for all i + 2 <= n
Given a strictly increasing array A of positive integers forming a sequence, find the length of the longest fibonacci-like subsequence of A. If one does not exist, return 0.
(Recall that a subsequence is derived from another sequence A by deleting any number of elements (including none) from A, 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: [1,2,3,4,5,6,7,8]
Output: 5
Explanation:
The longest subsequence that is fibonacci-like: [1,2,3,5,8].
Example 2:
Input: [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].
Note:
3 <= A.length <= 1000
1 <= A[0] < A[1] < ... < A[A.length - 1] <= 10^9
(The time limit has been reduced by 50% for submissions in Java, C, and C++.)
|
class Solution:
def lenLongestFibSubseq(self, A: List[int]) -> int:
index = {}
for i in range(len(A)):
index[A[i]] = i
record = {}
def dp(i,j):
if (i,j) in record:
return record[(i,j)]
if A[i] + A[j] in index:
res = 1 + dp(j, index[A[i]+A[j]])
record[(i,j)] = res
return res
else:
return 2
res = 0
for i in range(len(A)-1):
for j in range(i+1, len(A)):
res = max(res, dp(i,j))
if res >= 3:
return res
else:
return 0
|
A sequence X_1, X_2, ..., X_n is fibonacci-like if:
n >= 3
X_i + X_{i+1} = X_{i+2} for all i + 2 <= n
Given a strictly increasing array A of positive integers forming a sequence, find the length of the longest fibonacci-like subsequence of A. If one does not exist, return 0.
(Recall that a subsequence is derived from another sequence A by deleting any number of elements (including none) from A, 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: [1,2,3,4,5,6,7,8]
Output: 5
Explanation:
The longest subsequence that is fibonacci-like: [1,2,3,5,8].
Example 2:
Input: [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].
Note:
3 <= A.length <= 1000
1 <= A[0] < A[1] < ... < A[A.length - 1] <= 10^9
(The time limit has been reduced by 50% for submissions in Java, C, and C++.)
|
class Solution:
def lenLongestFibSubseq(self, x: List[int]) -> int:
def func(a,b):
if (a,b) in dp:
return dp[a,b]
ans=1
if (a+b) in f:
ans=1+func(b,a+b)
dp[a,b]=ans
return ans
l=len(x)
f=set(x)
dp={}
ma=2
for i in range(l):
a=x[i]
for j in range(i+1,l):
b=x[j]
ma=max(1+func(a,b),ma)
return ma if ma>2 else 0
|
A sequence X_1, X_2, ..., X_n is fibonacci-like if:
n >= 3
X_i + X_{i+1} = X_{i+2} for all i + 2 <= n
Given a strictly increasing array A of positive integers forming a sequence, find the length of the longest fibonacci-like subsequence of A. If one does not exist, return 0.
(Recall that a subsequence is derived from another sequence A by deleting any number of elements (including none) from A, 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: [1,2,3,4,5,6,7,8]
Output: 5
Explanation:
The longest subsequence that is fibonacci-like: [1,2,3,5,8].
Example 2:
Input: [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].
Note:
3 <= A.length <= 1000
1 <= A[0] < A[1] < ... < A[A.length - 1] <= 10^9
(The time limit has been reduced by 50% for submissions in Java, C, and C++.)
|
class Solution:
def lenLongestFibSubseq(self, A: List[int]) -> int:
#DP(x,y) = length, expected number for the next, the last number
#how to update?
#DP(x,y) = max(1 + DP(x,y` <y)),
#store values in the
if(len(A)==0):
return 0
mem = {}
dp = [[-1 for x in range(len(A))] for y in range(len(A))]
for idx,a in enumerate(A):
mem[a]=idx
max_l = 1
for i in range(len(A)-1):
for j in range(i+1,len(A)):
max_l = max(2+self.get_valid_pibo(i,j,dp,A,mem),max_l)
if(max_l==2):max_l=0
return max_l
def get_valid_pibo(self,i,j,dp,A,mem):
if(dp[i][j]==-1):
#return dp[i][j]
#else:
next_val = A[i]+A[j]
if(next_val in mem):
length = 1+self.get_valid_pibo(j,mem[next_val],dp,A,mem)
dp[i][j]=length
else:
length = 0
dp[i][j]=0
return dp[i][j]
|
A sequence X_1, X_2, ..., X_n is fibonacci-like if:
n >= 3
X_i + X_{i+1} = X_{i+2} for all i + 2 <= n
Given a strictly increasing array A of positive integers forming a sequence, find the length of the longest fibonacci-like subsequence of A. If one does not exist, return 0.
(Recall that a subsequence is derived from another sequence A by deleting any number of elements (including none) from A, 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: [1,2,3,4,5,6,7,8]
Output: 5
Explanation:
The longest subsequence that is fibonacci-like: [1,2,3,5,8].
Example 2:
Input: [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].
Note:
3 <= A.length <= 1000
1 <= A[0] < A[1] < ... < A[A.length - 1] <= 10^9
(The time limit has been reduced by 50% for submissions in Java, C, and C++.)
|
class Solution:
def lenLongestFibSubseq(self, A: List[int]) -> int:
n = len(A)
set_a = set(A)
seen = set()
res = 0
store = {}
def find(x, y):
if x + y in set_a:
return 1 + find(y, x + y)
else:
return 0
for i in range(n):
for j in range(i + 1, n):
if (A[i], A[j]) not in seen:
res = max(find(A[i], A[j]), res)
if res == 0:
return 0
else:
return res + 2
|
A sequence X_1, X_2, ..., X_n is fibonacci-like if:
n >= 3
X_i + X_{i+1} = X_{i+2} for all i + 2 <= n
Given a strictly increasing array A of positive integers forming a sequence, find the length of the longest fibonacci-like subsequence of A. If one does not exist, return 0.
(Recall that a subsequence is derived from another sequence A by deleting any number of elements (including none) from A, 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: [1,2,3,4,5,6,7,8]
Output: 5
Explanation:
The longest subsequence that is fibonacci-like: [1,2,3,5,8].
Example 2:
Input: [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].
Note:
3 <= A.length <= 1000
1 <= A[0] < A[1] < ... < A[A.length - 1] <= 10^9
(The time limit has been reduced by 50% for submissions in Java, C, and C++.)
|
class Solution:
def lenLongestFibSubseq(self, A: List[int]) -> int:
num_set = set(A)
used = set()
max_len = 2
for x in range(len(A)):
a = A[x]
for y in range(x+1, len(A)):
i, j = a, A[y]
if (i, j) not in used:
used.add((i, j))
count = 2
while i+j in num_set:
count += 1
i, j = j, i+j
used.add((i, j))
max_len = max(count, max_len)
return 0 if max_len == 2 else max_len
|
A sequence X_1, X_2, ..., X_n is fibonacci-like if:
n >= 3
X_i + X_{i+1} = X_{i+2} for all i + 2 <= n
Given a strictly increasing array A of positive integers forming a sequence, find the length of the longest fibonacci-like subsequence of A. If one does not exist, return 0.
(Recall that a subsequence is derived from another sequence A by deleting any number of elements (including none) from A, 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: [1,2,3,4,5,6,7,8]
Output: 5
Explanation:
The longest subsequence that is fibonacci-like: [1,2,3,5,8].
Example 2:
Input: [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].
Note:
3 <= A.length <= 1000
1 <= A[0] < A[1] < ... < A[A.length - 1] <= 10^9
(The time limit has been reduced by 50% for submissions in Java, C, and C++.)
|
class Solution:
def lenLongestFibSubseq(self, A: List[int]) -> int:
dp = dict()
res = 1
for i in range(len(A)):
for j in range(i):
if (A[j], A[i]-A[j]) in dp:
dp[(A[i], A[j])] = dp[(A[j], A[i]-A[j])] + 1
else:
dp[(A[i], A[j])] = 2
res = max(res, dp[(A[i], A[j])])
return res if res > 2 else 0
|
A sequence X_1, X_2, ..., X_n is fibonacci-like if:
n >= 3
X_i + X_{i+1} = X_{i+2} for all i + 2 <= n
Given a strictly increasing array A of positive integers forming a sequence, find the length of the longest fibonacci-like subsequence of A. If one does not exist, return 0.
(Recall that a subsequence is derived from another sequence A by deleting any number of elements (including none) from A, 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: [1,2,3,4,5,6,7,8]
Output: 5
Explanation:
The longest subsequence that is fibonacci-like: [1,2,3,5,8].
Example 2:
Input: [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].
Note:
3 <= A.length <= 1000
1 <= A[0] < A[1] < ... < A[A.length - 1] <= 10^9
(The time limit has been reduced by 50% for submissions in Java, C, and C++.)
|
class Solution:
def lenLongestFibSubseq(self, A: List[int]) -> int:
dp = dict()
res = 2
for i in range(len(A)):
for j in range(i):
if (A[j], A[i]-A[j]) in dp:
dp[A[i], A[j]] = dp[A[j], A[i]-A[j]] + 1
else:
dp[A[i], A[j]] = 2
res = max(res, dp[A[i], A[j]])
return res if res > 2 else 0
|
A sequence X_1, X_2, ..., X_n is fibonacci-like if:
n >= 3
X_i + X_{i+1} = X_{i+2} for all i + 2 <= n
Given a strictly increasing array A of positive integers forming a sequence, find the length of the longest fibonacci-like subsequence of A. If one does not exist, return 0.
(Recall that a subsequence is derived from another sequence A by deleting any number of elements (including none) from A, 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: [1,2,3,4,5,6,7,8]
Output: 5
Explanation:
The longest subsequence that is fibonacci-like: [1,2,3,5,8].
Example 2:
Input: [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].
Note:
3 <= A.length <= 1000
1 <= A[0] < A[1] < ... < A[A.length - 1] <= 10^9
(The time limit has been reduced by 50% for submissions in Java, C, and C++.)
|
class Solution:
def lenLongestFibSubseq(self, x: List[int]) -> int:
def func(a,b):
if (a,b) in dp:return dp[a,b]
ans=1
if (a+b) in f:ans=1+func(b,a+b)
dp[a,b]=ans
return ans
l=len(x)
f=set(x)
dp={}
ma=2
for i in range(l):
a=x[i]
for j in range(i+1,l):
b=x[j]
ma=max(1+func(a,b),ma)
return ma if ma>2 else 0
|
A sequence X_1, X_2, ..., X_n is fibonacci-like if:
n >= 3
X_i + X_{i+1} = X_{i+2} for all i + 2 <= n
Given a strictly increasing array A of positive integers forming a sequence, find the length of the longest fibonacci-like subsequence of A. If one does not exist, return 0.
(Recall that a subsequence is derived from another sequence A by deleting any number of elements (including none) from A, 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: [1,2,3,4,5,6,7,8]
Output: 5
Explanation:
The longest subsequence that is fibonacci-like: [1,2,3,5,8].
Example 2:
Input: [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].
Note:
3 <= A.length <= 1000
1 <= A[0] < A[1] < ... < A[A.length - 1] <= 10^9
(The time limit has been reduced by 50% for submissions in Java, C, and C++.)
|
class Solution:
def lenLongestFibSubseq(self, A: List[int]) -> int:
index = {x:i for i, x in enumerate(A)}
m = collections.defaultdict(lambda:2)
res = 0
for i2 in range(2, len(A)):
for i0 in range(i2):
x1 = A[i2] - A[i0]
if x1 <= A[i0]:
break
if x1 in index:
i1 = index[x1]
m[(i1, i2)] = m[(i0, i1)] + 1
res = max(res, m[(i1, i2)])
return res
|
A sequence X_1, X_2, ..., X_n is fibonacci-like if:
n >= 3
X_i + X_{i+1} = X_{i+2} for all i + 2 <= n
Given a strictly increasing array A of positive integers forming a sequence, find the length of the longest fibonacci-like subsequence of A. If one does not exist, return 0.
(Recall that a subsequence is derived from another sequence A by deleting any number of elements (including none) from A, 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: [1,2,3,4,5,6,7,8]
Output: 5
Explanation:
The longest subsequence that is fibonacci-like: [1,2,3,5,8].
Example 2:
Input: [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].
Note:
3 <= A.length <= 1000
1 <= A[0] < A[1] < ... < A[A.length - 1] <= 10^9
(The time limit has been reduced by 50% for submissions in Java, C, and C++.)
|
class Solution:
def lenLongestFibSubseq(self, A: List[int]) -> int:
fibHash = {}
ans = 0
n = len(A)
for i in range(1, n):
for j in reversed(range(i)):
f1 = A[i] - A[j]
size = fibHash[(A[j], A[i])] = fibHash.get((f1, A[j]), 1) + 1
ans = max(ans, size)
return ans if ans > 2 else 0
|
A sequence X_1, X_2, ..., X_n is fibonacci-like if:
n >= 3
X_i + X_{i+1} = X_{i+2} for all i + 2 <= n
Given a strictly increasing array A of positive integers forming a sequence, find the length of the longest fibonacci-like subsequence of A. If one does not exist, return 0.
(Recall that a subsequence is derived from another sequence A by deleting any number of elements (including none) from A, 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: [1,2,3,4,5,6,7,8]
Output: 5
Explanation:
The longest subsequence that is fibonacci-like: [1,2,3,5,8].
Example 2:
Input: [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].
Note:
3 <= A.length <= 1000
1 <= A[0] < A[1] < ... < A[A.length - 1] <= 10^9
(The time limit has been reduced by 50% for submissions in Java, C, and C++.)
|
class Solution:
def lenLongestFibSubseq(self, A: List[int]) -> int:
n = len(A)
mem = collections.defaultdict(Counter)
ans = 0
for i in range(0, n):
for j in range(i-1, -1, -1):
dij = A[i] - A[j]
djk = A[j] - dij
if djk in mem[A[j]]:
mem[A[i]][dij] = 1 + mem[A[j]][djk]
else:
mem[A[i]][dij] = 2
ans = max(ans, mem[A[i]][dij])
# print(mem)
if ans < 3:
return 0
return ans
|
A sequence X_1, X_2, ..., X_n is fibonacci-like if:
n >= 3
X_i + X_{i+1} = X_{i+2} for all i + 2 <= n
Given a strictly increasing array A of positive integers forming a sequence, find the length of the longest fibonacci-like subsequence of A. If one does not exist, return 0.
(Recall that a subsequence is derived from another sequence A by deleting any number of elements (including none) from A, 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: [1,2,3,4,5,6,7,8]
Output: 5
Explanation:
The longest subsequence that is fibonacci-like: [1,2,3,5,8].
Example 2:
Input: [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].
Note:
3 <= A.length <= 1000
1 <= A[0] < A[1] < ... < A[A.length - 1] <= 10^9
(The time limit has been reduced by 50% for submissions in Java, C, and C++.)
|
class Solution:
def lenLongestFibSubseq(self, A: List[int]) -> int:
def memo(f):
dic = {}
def fa(*a):
if a not in dic:
dic[a] = f(*a)
return dic[a]
return fa
cache = {val:idx for idx,val in enumerate(A)}
@memo
def leng(i,j):
if A[i]+A[j] not in cache:
return 2
return 1 + leng(j, cache[A[i]+A[j]])
rec = 0
for j in range(len(A)):
for i in range(j):
rec = max(rec, leng(i,j))
return rec if rec>2 else 0
|
A sequence X_1, X_2, ..., X_n is fibonacci-like if:
n >= 3
X_i + X_{i+1} = X_{i+2} for all i + 2 <= n
Given a strictly increasing array A of positive integers forming a sequence, find the length of the longest fibonacci-like subsequence of A. If one does not exist, return 0.
(Recall that a subsequence is derived from another sequence A by deleting any number of elements (including none) from A, 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: [1,2,3,4,5,6,7,8]
Output: 5
Explanation:
The longest subsequence that is fibonacci-like: [1,2,3,5,8].
Example 2:
Input: [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].
Note:
3 <= A.length <= 1000
1 <= A[0] < A[1] < ... < A[A.length - 1] <= 10^9
(The time limit has been reduced by 50% for submissions in Java, C, and C++.)
|
class Solution:
def lenLongestFibSubseq(self, A: List[int]) -> int:
setA = set(A)
answer = 0
max_ = A[-1]
visited = set()
for i in range(len(A)-1):
curr1 = A[i]
for j in range(i+1, len(A)):
curr2 = A[j]
if (curr1, curr2) in visited:
continue
visited.add((curr1, curr2))
sum_ = curr1 + curr2
l = 2 if sum_ in setA else 0
while sum_ in setA and sum_ <= max_:
curr1, curr2 = curr2, sum_
sum_ = curr1 + curr2
visited.add((curr1, curr2))
l += 1
answer = max(answer, l)
curr1 = A[i]
return answer
|
A sequence X_1, X_2, ..., X_n is fibonacci-like if:
n >= 3
X_i + X_{i+1} = X_{i+2} for all i + 2 <= n
Given a strictly increasing array A of positive integers forming a sequence, find the length of the longest fibonacci-like subsequence of A. If one does not exist, return 0.
(Recall that a subsequence is derived from another sequence A by deleting any number of elements (including none) from A, 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: [1,2,3,4,5,6,7,8]
Output: 5
Explanation:
The longest subsequence that is fibonacci-like: [1,2,3,5,8].
Example 2:
Input: [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].
Note:
3 <= A.length <= 1000
1 <= A[0] < A[1] < ... < A[A.length - 1] <= 10^9
(The time limit has been reduced by 50% for submissions in Java, C, and C++.)
|
class Solution:
def lenLongestFibSubseq(self, A: List[int]) -> int:
indexDict = {a:index for index,a in enumerate(A)}
indexToLength = {}
aLength = len(A)
def computeMaxFibonacci(i,j):##gives the value of the longest sequence that starts with A[i] and A[j]. We assume that i < j. We are told it is strictly increasing so we are going to get monotonically decreasing sequences.
if (i,j) in indexToLength:
return indexToLength[(i,j)]
else:
first = A[i]
second = A[j]
second += first
first = second-first
if second in indexDict:
answer = 1+computeMaxFibonacci(j,indexDict[second])
else:
answer = 2
indexToLength[(i,j)] = answer
return answer
maxFib = 0
for i in range(aLength):
for j in range(i+1,aLength):##we have to ensure that i < j.
maxFib = max(maxFib,computeMaxFibonacci(i,j))
##print(indexToLength)
return maxFib if maxFib >= 3 else 0##however it isn't as simple as returning maxFib because if it happens to be 2, then we need to return 0 per conditions of problem. Need to always consider test cases that would give variety of outputs.
|
A sequence X_1, X_2, ..., X_n is fibonacci-like if:
n >= 3
X_i + X_{i+1} = X_{i+2} for all i + 2 <= n
Given a strictly increasing array A of positive integers forming a sequence, find the length of the longest fibonacci-like subsequence of A. If one does not exist, return 0.
(Recall that a subsequence is derived from another sequence A by deleting any number of elements (including none) from A, 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: [1,2,3,4,5,6,7,8]
Output: 5
Explanation:
The longest subsequence that is fibonacci-like: [1,2,3,5,8].
Example 2:
Input: [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].
Note:
3 <= A.length <= 1000
1 <= A[0] < A[1] < ... < A[A.length - 1] <= 10^9
(The time limit has been reduced by 50% for submissions in Java, C, and C++.)
|
class Solution:
def lenLongestFibSubseq(self, A: List[int]) -> int:
dp = collections.defaultdict(lambda: 2)
ret, idx = 0, {A[0]:0}
for j in range(1, len(A) -1):
idx[A[j]] = j
for i in range(j+1, len(A)):
diff = A[i] - A[j]
if diff >= A[j]:
break
elif diff not in idx:
continue
dp[j,i] = dp[idx[diff], j] +1
ret = max(ret, dp[j,i])
return ret
|
A sequence X_1, X_2, ..., X_n is fibonacci-like if:
n >= 3
X_i + X_{i+1} = X_{i+2} for all i + 2 <= n
Given a strictly increasing array A of positive integers forming a sequence, find the length of the longest fibonacci-like subsequence of A. If one does not exist, return 0.
(Recall that a subsequence is derived from another sequence A by deleting any number of elements (including none) from A, 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: [1,2,3,4,5,6,7,8]
Output: 5
Explanation:
The longest subsequence that is fibonacci-like: [1,2,3,5,8].
Example 2:
Input: [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].
Note:
3 <= A.length <= 1000
1 <= A[0] < A[1] < ... < A[A.length - 1] <= 10^9
(The time limit has been reduced by 50% for submissions in Java, C, and C++.)
|
class Solution:
def lenLongestFibSubseq(self, A: List[int]) -> int:
setA = set(A)
answer = 0
max_ = A[-1]
visited = set()
queue = collections.deque()
for i in range(len(A)-1):
curr1 = A[i]
for j in range(i+1, len(A)):
curr2 = A[j]
if curr1 + curr2 in setA:
queue.append((curr1, curr2))
while queue:
num1, num2 = queue.popleft()
if (num1, num2) in visited:
continue
visited.add((num1, num2))
sum_ = num1 + num2
l = 2
while sum_ in setA and sum_ <= max_:
num1, num2 = num2, sum_
sum_ = num1 + num2
visited.add((num1, num2))
l += 1
answer = max(answer, l)
return answer
|
A sequence X_1, X_2, ..., X_n is fibonacci-like if:
n >= 3
X_i + X_{i+1} = X_{i+2} for all i + 2 <= n
Given a strictly increasing array A of positive integers forming a sequence, find the length of the longest fibonacci-like subsequence of A. If one does not exist, return 0.
(Recall that a subsequence is derived from another sequence A by deleting any number of elements (including none) from A, 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: [1,2,3,4,5,6,7,8]
Output: 5
Explanation:
The longest subsequence that is fibonacci-like: [1,2,3,5,8].
Example 2:
Input: [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].
Note:
3 <= A.length <= 1000
1 <= A[0] < A[1] < ... < A[A.length - 1] <= 10^9
(The time limit has been reduced by 50% for submissions in Java, C, and C++.)
|
from typing import List
from collections import defaultdict
class Solution:
def lenLongestFibSubseq(self, A: List[int]) -> int:
result = 0
starterAtIdx = [] # starterAtIdx[i] = {numbers}, where each next num is A[i] + A[j] where j < i
fullSeqAtIdx = [] # fullSeqAtIdx[i] = {num: length}, where num is next number in seq of length
for idx, num in enumerate(A):
starter = set()
fullSeq = defaultdict(int)
for j in range(0, idx):
prevNum = A[j]
if num in starterAtIdx[j]:
fullSeq[prevNum + num] = max(fullSeq[prevNum + num], 3)
if num in fullSeqAtIdx[j]:
fullSeq[prevNum + num] = max(fullSeq[prevNum + num], 1 + fullSeqAtIdx[j][num])
result = max(result, fullSeq[prevNum + num])
starter.add(num + prevNum)
starterAtIdx.append(starter)
fullSeqAtIdx.append(fullSeq)
return result
|
A sequence X_1, X_2, ..., X_n is fibonacci-like if:
n >= 3
X_i + X_{i+1} = X_{i+2} for all i + 2 <= n
Given a strictly increasing array A of positive integers forming a sequence, find the length of the longest fibonacci-like subsequence of A. If one does not exist, return 0.
(Recall that a subsequence is derived from another sequence A by deleting any number of elements (including none) from A, 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: [1,2,3,4,5,6,7,8]
Output: 5
Explanation:
The longest subsequence that is fibonacci-like: [1,2,3,5,8].
Example 2:
Input: [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].
Note:
3 <= A.length <= 1000
1 <= A[0] < A[1] < ... < A[A.length - 1] <= 10^9
(The time limit has been reduced by 50% for submissions in Java, C, and C++.)
|
class Solution:
def lenLongestFibSubseq(self, A: List[int]) -> int:
seen,n={},len(A)
A_set=set(A)
def find_lens(ind,prev,curr):
if ind==n:
return 0
key=(prev,curr)
if key in seen:
return seen[key]
ans=0
if prev+curr in A_set:
for i in range(ind,n):
if curr+prev==A[i]:
ans=max(ans,find_lens(i+1,curr,A[i])+1)
seen[key]=ans
return ans
ans=0
for i in range(n-3):
for j in range(i+1,n-2):
ans=max(ans,find_lens(j+1,A[i],A[j]))
return ans+2 if ans!=0 else 0
|
A sequence X_1, X_2, ..., X_n is fibonacci-like if:
n >= 3
X_i + X_{i+1} = X_{i+2} for all i + 2 <= n
Given a strictly increasing array A of positive integers forming a sequence, find the length of the longest fibonacci-like subsequence of A. If one does not exist, return 0.
(Recall that a subsequence is derived from another sequence A by deleting any number of elements (including none) from A, 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: [1,2,3,4,5,6,7,8]
Output: 5
Explanation:
The longest subsequence that is fibonacci-like: [1,2,3,5,8].
Example 2:
Input: [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].
Note:
3 <= A.length <= 1000
1 <= A[0] < A[1] < ... < A[A.length - 1] <= 10^9
(The time limit has been reduced by 50% for submissions in Java, C, and C++.)
|
class Solution:
def lenLongestFibSubseq(self, A: List[int]) -> int:
a = set(A)
cache = {}
ans = 0
for i in range(len(A)):
for j in range(i+1, len(A)):
ans = max(ans, self.helper(a, A[i], A[j], cache))
# print(cache)
return ans if ans > 2 else 0
def helper(self, a, n1, n2, cache):
if (n1, n2) not in cache:
if n1+n2 not in a:
cache[(n1, n2)] = 2
else:
cache[(n1, n2)] = self.helper(a, n2, n1+n2, cache) + 1
return cache[(n1, n2)]
|
A sequence X_1, X_2, ..., X_n is fibonacci-like if:
n >= 3
X_i + X_{i+1} = X_{i+2} for all i + 2 <= n
Given a strictly increasing array A of positive integers forming a sequence, find the length of the longest fibonacci-like subsequence of A. If one does not exist, return 0.
(Recall that a subsequence is derived from another sequence A by deleting any number of elements (including none) from A, 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: [1,2,3,4,5,6,7,8]
Output: 5
Explanation:
The longest subsequence that is fibonacci-like: [1,2,3,5,8].
Example 2:
Input: [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].
Note:
3 <= A.length <= 1000
1 <= A[0] < A[1] < ... < A[A.length - 1] <= 10^9
(The time limit has been reduced by 50% for submissions in Java, C, and C++.)
|
class Solution:
def lenLongestFibSubseq(self, A: List[int]) -> int:
record = {}
for i, a in enumerate(A):
record[a] = i
adj = {}
for i in range(len(A)-2):
for j in range(i+1, len(A)-1):
if (A[i] + A[j]) in record:
adj[(i,j)] = (j, record[A[i]+A[j]])
mem = {}
def helper(i,j):
if (i,j) in mem:
return mem[(i,j)]
else:
if (i,j) in adj:
res = 1 + helper(adj[(i,j)][0],adj[(i,j)][1])
mem[(i,j)] = res
return res
else:
mem[(i,j)] = 2
return 2
res = 0
for i in range(len(A)-2):
for j in range(i+1, len(A)-1):
res = max(res, helper(i,j))
if res >= 3:
return res
else:
return 0
|
A sequence X_1, X_2, ..., X_n is fibonacci-like if:
n >= 3
X_i + X_{i+1} = X_{i+2} for all i + 2 <= n
Given a strictly increasing array A of positive integers forming a sequence, find the length of the longest fibonacci-like subsequence of A. If one does not exist, return 0.
(Recall that a subsequence is derived from another sequence A by deleting any number of elements (including none) from A, 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: [1,2,3,4,5,6,7,8]
Output: 5
Explanation:
The longest subsequence that is fibonacci-like: [1,2,3,5,8].
Example 2:
Input: [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].
Note:
3 <= A.length <= 1000
1 <= A[0] < A[1] < ... < A[A.length - 1] <= 10^9
(The time limit has been reduced by 50% for submissions in Java, C, and C++.)
|
class Solution:
def lenLongestFibSubseq(self, A: List[int]) -> int:
max_l = 0
list_poss = [{} for i in range(len(A))]
for i in range(len(A) - 1):
for j in range(i+1):
if not (A[i+1] + A[j]) in list_poss[i+1]:
list_poss[i+1][A[i+1] + A[j]] = 2
if A[i+1] in list_poss[j]:
if not(A[i+1] + A[j]) in list_poss[i+1]:
list_poss[i+1][A[i+1] + A[j]] = list_poss[j][A[i+1]] + 1
else:
if list_poss[j][A[i+1]] + 1 > list_poss[i+1][A[i+1] + A[j]]:
list_poss[i+1][A[i+1] + A[j]] = list_poss[j][A[i+1]] + 1
if list_poss[i+1][A[i+1] + A[j]] > max_l:
max_l = list_poss[i+1][A[i+1] + A[j]]
# print(list_poss[i+1])
return max_l
|
A sequence X_1, X_2, ..., X_n is fibonacci-like if:
n >= 3
X_i + X_{i+1} = X_{i+2} for all i + 2 <= n
Given a strictly increasing array A of positive integers forming a sequence, find the length of the longest fibonacci-like subsequence of A. If one does not exist, return 0.
(Recall that a subsequence is derived from another sequence A by deleting any number of elements (including none) from A, 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: [1,2,3,4,5,6,7,8]
Output: 5
Explanation:
The longest subsequence that is fibonacci-like: [1,2,3,5,8].
Example 2:
Input: [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].
Note:
3 <= A.length <= 1000
1 <= A[0] < A[1] < ... < A[A.length - 1] <= 10^9
(The time limit has been reduced by 50% for submissions in Java, C, and C++.)
|
from functools import lru_cache
from bisect import bisect_left
class Solution:
def lenLongestFibSubseq(self, A: List[int]) -> int:
n = len(A)
@lru_cache(None)
def fib(p1, p2, i):
prevsum = p1 + p2
i = bisect_left(A, prevsum)
if i < n and A[i] == prevsum:
return 1 + fib(p2, A[i], i+1)
return 0
res = 0
for i in range(n-2):
for j in range(i+1, n-1):
curres = fib(A[i], A[j], j+1)
if curres:
res = max(res, 2+curres)
return res
|
A sequence X_1, X_2, ..., X_n is fibonacci-like if:
n >= 3
X_i + X_{i+1} = X_{i+2} for all i + 2 <= n
Given a strictly increasing array A of positive integers forming a sequence, find the length of the longest fibonacci-like subsequence of A. If one does not exist, return 0.
(Recall that a subsequence is derived from another sequence A by deleting any number of elements (including none) from A, 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: [1,2,3,4,5,6,7,8]
Output: 5
Explanation:
The longest subsequence that is fibonacci-like: [1,2,3,5,8].
Example 2:
Input: [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].
Note:
3 <= A.length <= 1000
1 <= A[0] < A[1] < ... < A[A.length - 1] <= 10^9
(The time limit has been reduced by 50% for submissions in Java, C, and C++.)
|
class Solution:
def lenLongestFibSubseq(self, A: List[int]) -> int:
dp = {}
for i in range(1, len(A)):
for j in range(i):
dp[(A[j], A[i])] = max(dp.get((A[j], A[i]), 2), dp.get((A[i]-A[j], A[j]), 1)+1)
res = max(dp.values())
return res if res>2 else 0
|
A sequence X_1, X_2, ..., X_n is fibonacci-like if:
n >= 3
X_i + X_{i+1} = X_{i+2} for all i + 2 <= n
Given a strictly increasing array A of positive integers forming a sequence, find the length of the longest fibonacci-like subsequence of A. If one does not exist, return 0.
(Recall that a subsequence is derived from another sequence A by deleting any number of elements (including none) from A, 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: [1,2,3,4,5,6,7,8]
Output: 5
Explanation:
The longest subsequence that is fibonacci-like: [1,2,3,5,8].
Example 2:
Input: [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].
Note:
3 <= A.length <= 1000
1 <= A[0] < A[1] < ... < A[A.length - 1] <= 10^9
(The time limit has been reduced by 50% for submissions in Java, C, and C++.)
|
from collections import defaultdict
class Solution:
def lenLongestFibSubseq(self, A: List[int]) -> int:
maps = {num:i for i, num in enumerate(A)}
n = len(A)
dp = defaultdict(lambda: 2)
max_len = 0
for i in range(2, n):
for j in range(i):
diff = A[i] - A[j]
if diff in maps and diff < A[j]:
idx = maps[diff]
dp[j, i] = max(dp[j, i], dp[idx, j] + 1)
max_len = max(dp[j, i], max_len)
return max_len if max_len > 2 else 0
|
A sequence X_1, X_2, ..., X_n is fibonacci-like if:
n >= 3
X_i + X_{i+1} = X_{i+2} for all i + 2 <= n
Given a strictly increasing array A of positive integers forming a sequence, find the length of the longest fibonacci-like subsequence of A. If one does not exist, return 0.
(Recall that a subsequence is derived from another sequence A by deleting any number of elements (including none) from A, 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: [1,2,3,4,5,6,7,8]
Output: 5
Explanation:
The longest subsequence that is fibonacci-like: [1,2,3,5,8].
Example 2:
Input: [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].
Note:
3 <= A.length <= 1000
1 <= A[0] < A[1] < ... < A[A.length - 1] <= 10^9
(The time limit has been reduced by 50% for submissions in Java, C, and C++.)
|
class Solution:
def lenLongestFibSubseq(self, A: List[int]) -> int:
nums = {}
for i, num in enumerate(A):
nums[num] = i
def dfs(prev, cur, visit):
if (prev,cur) in visit:
return visit[(prev, cur)]
if prev == -1:
res = 0
for i in range(len(A)):
for j in range(i+1, len(A)):
if A[i]+A[j] in nums:
res = max(res, 1+dfs(i,j,visit))
return res
if A[prev] + A[cur] in nums:
visit[(prev,cur)] = 1 + dfs(cur, nums[A[prev]+A[cur]], visit)
else:
visit[(prev,cur)] = 1
return visit[(prev,cur)]
return dfs(-1,-1,{})
|
A sequence X_1, X_2, ..., X_n is fibonacci-like if:
n >= 3
X_i + X_{i+1} = X_{i+2} for all i + 2 <= n
Given a strictly increasing array A of positive integers forming a sequence, find the length of the longest fibonacci-like subsequence of A. If one does not exist, return 0.
(Recall that a subsequence is derived from another sequence A by deleting any number of elements (including none) from A, 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: [1,2,3,4,5,6,7,8]
Output: 5
Explanation:
The longest subsequence that is fibonacci-like: [1,2,3,5,8].
Example 2:
Input: [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].
Note:
3 <= A.length <= 1000
1 <= A[0] < A[1] < ... < A[A.length - 1] <= 10^9
(The time limit has been reduced by 50% for submissions in Java, C, and C++.)
|
class Solution:
def lenLongestFibSubseq(self, A: List[int]) -> int:
n = len(A)
d = {}
s = set(A)
for j in range(n):
for i in range(j):
if (A[j] - A[i]) in s and (A[j] - A[i]) < A[i]:
d[(A[i], A[j])] = d.get((A[j] - A[i], A[i]), 2) + 1
#print(d)
return max(d.values() or [0])
|
A sequence X_1, X_2, ..., X_n is fibonacci-like if:
n >= 3
X_i + X_{i+1} = X_{i+2} for all i + 2 <= n
Given a strictly increasing array A of positive integers forming a sequence, find the length of the longest fibonacci-like subsequence of A. If one does not exist, return 0.
(Recall that a subsequence is derived from another sequence A by deleting any number of elements (including none) from A, 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: [1,2,3,4,5,6,7,8]
Output: 5
Explanation:
The longest subsequence that is fibonacci-like: [1,2,3,5,8].
Example 2:
Input: [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].
Note:
3 <= A.length <= 1000
1 <= A[0] < A[1] < ... < A[A.length - 1] <= 10^9
(The time limit has been reduced by 50% for submissions in Java, C, and C++.)
|
class Solution:
def lenLongestFibSubseq(self, A: List[int]) -> int:
nums = {}
for i, num in enumerate(A):
nums[num] = i
def dfs(prev, cur, visit):
if (prev,cur) in visit:
return visit[(prev, cur)]
if A[prev] + A[cur] in nums:
visit[(prev,cur)] = 1 + dfs(cur, nums[A[prev]+A[cur]], visit)
else:
visit[(prev,cur)] = 1
return visit[(prev,cur)]
visit = {}
res = 0
for i in range(len(A)):
for j in range(i+1, len(A)):
if A[i] + A[j] in nums:
res = max(res, 1+dfs(i, j, visit))
return res
|
A sequence X_1, X_2, ..., X_n is fibonacci-like if:
n >= 3
X_i + X_{i+1} = X_{i+2} for all i + 2 <= n
Given a strictly increasing array A of positive integers forming a sequence, find the length of the longest fibonacci-like subsequence of A. If one does not exist, return 0.
(Recall that a subsequence is derived from another sequence A by deleting any number of elements (including none) from A, 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: [1,2,3,4,5,6,7,8]
Output: 5
Explanation:
The longest subsequence that is fibonacci-like: [1,2,3,5,8].
Example 2:
Input: [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].
Note:
3 <= A.length <= 1000
1 <= A[0] < A[1] < ... < A[A.length - 1] <= 10^9
(The time limit has been reduced by 50% for submissions in Java, C, and C++.)
|
class Solution:
def lenLongestFibSubseq(self, A: List[int]) -> int:
'''
dp(i, j): Maximum length of Fib sequence ending at A[i] & A[j]
dp(i, j) = dp(k, j) + 1
'''
cache = set(A)
dp = collections.defaultdict(int)
for i in range(len(A)):
for j in range(i + 1, len(A)):
if A[j] - A[i] in cache and A[j] - A[i] < A[i]:
dp[(A[i], A[j])] = dp.get((A[j] - A[i], A[i]), 2) + 1
return max(list(dp.values()) + [0])
|
A sequence X_1, X_2, ..., X_n is fibonacci-like if:
n >= 3
X_i + X_{i+1} = X_{i+2} for all i + 2 <= n
Given a strictly increasing array A of positive integers forming a sequence, find the length of the longest fibonacci-like subsequence of A. If one does not exist, return 0.
(Recall that a subsequence is derived from another sequence A by deleting any number of elements (including none) from A, 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: [1,2,3,4,5,6,7,8]
Output: 5
Explanation:
The longest subsequence that is fibonacci-like: [1,2,3,5,8].
Example 2:
Input: [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].
Note:
3 <= A.length <= 1000
1 <= A[0] < A[1] < ... < A[A.length - 1] <= 10^9
(The time limit has been reduced by 50% for submissions in Java, C, and C++.)
|
from collections import defaultdict
class Solution:
def lenLongestFibSubseq(self, A: List[int]) -> int:
A_indices = {a: i for i, a in enumerate(A)}
lengths = defaultdict(lambda: 2)
def get_value(i, j):
if (i, j) in memo:
return memo[i, j]
# search for A[i] + A[j] in A
if A[i] + A[j] in A_indices:
memo[i, j] = 1 + get_value(j, A_indices[A[i] + A[j]])
self.best = max(self.best, memo[i, j])
else:
memo[i, j] = 2
return memo[i, j]
best = 2
for i in range(len(A) - 3, -1, -1):
for j in range(len(A) - 2, i, -1):
if A[i] + A[j] in A_indices:
lengths[i, j] = 1 + lengths[j, A_indices[A[i] + A[j]]]
best = max(best, lengths[i, j])
return best if best > 2 else 0
|
A sequence X_1, X_2, ..., X_n is fibonacci-like if:
n >= 3
X_i + X_{i+1} = X_{i+2} for all i + 2 <= n
Given a strictly increasing array A of positive integers forming a sequence, find the length of the longest fibonacci-like subsequence of A. If one does not exist, return 0.
(Recall that a subsequence is derived from another sequence A by deleting any number of elements (including none) from A, 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: [1,2,3,4,5,6,7,8]
Output: 5
Explanation:
The longest subsequence that is fibonacci-like: [1,2,3,5,8].
Example 2:
Input: [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].
Note:
3 <= A.length <= 1000
1 <= A[0] < A[1] < ... < A[A.length - 1] <= 10^9
(The time limit has been reduced by 50% for submissions in Java, C, and C++.)
|
class Solution:
def lenLongestFibSubseq(self, A: List[int]) -> int:
indices = {n: i for i, n in enumerate(A)}
d = defaultdict(lambda: 2)
ans = 0
for k, n in enumerate(A):
for j in range(k-1, 0, -1):
m = n - A[j]
if m not in indices:
continue
i = indices[m]
if i >= j:
break
d[k, j] = d[j, i] + 1
ans = max(ans, d[k, j])
return ans
|
A sequence X_1, X_2, ..., X_n is fibonacci-like if:
n >= 3
X_i + X_{i+1} = X_{i+2} for all i + 2 <= n
Given a strictly increasing array A of positive integers forming a sequence, find the length of the longest fibonacci-like subsequence of A. If one does not exist, return 0.
(Recall that a subsequence is derived from another sequence A by deleting any number of elements (including none) from A, 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: [1,2,3,4,5,6,7,8]
Output: 5
Explanation:
The longest subsequence that is fibonacci-like: [1,2,3,5,8].
Example 2:
Input: [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].
Note:
3 <= A.length <= 1000
1 <= A[0] < A[1] < ... < A[A.length - 1] <= 10^9
(The time limit has been reduced by 50% for submissions in Java, C, and C++.)
|
class Solution:
def lenLongestFibSubseq(self, arr: List[int]) -> int:
fibSeqs = {}
values = set(arr)
for i in range(len(arr)):
for j in range(i-1, -1, -1):
a = arr[i] - arr[j]
if a >= arr[j]:
break
if a in values:
fibSeqs[arr[j], arr[i]] = fibSeqs.get((a, arr[j]), 2) + 1
return max(fibSeqs.values() or [0])
|
A sequence X_1, X_2, ..., X_n is fibonacci-like if:
n >= 3
X_i + X_{i+1} = X_{i+2} for all i + 2 <= n
Given a strictly increasing array A of positive integers forming a sequence, find the length of the longest fibonacci-like subsequence of A. If one does not exist, return 0.
(Recall that a subsequence is derived from another sequence A by deleting any number of elements (including none) from A, 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: [1,2,3,4,5,6,7,8]
Output: 5
Explanation:
The longest subsequence that is fibonacci-like: [1,2,3,5,8].
Example 2:
Input: [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].
Note:
3 <= A.length <= 1000
1 <= A[0] < A[1] < ... < A[A.length - 1] <= 10^9
(The time limit has been reduced by 50% for submissions in Java, C, and C++.)
|
class Solution:
def lenLongestFibSubseq(self, A: List[int]) -> int:
# index = {a:i for i, a in enumerate(A)}
candidates = set(A)
dp = {}
N = len(A)
ans = 0
for r in range(N):
for l in range(r - 1):
if A[l] > A[r] // 2:
break
if A[r] - A[l] in candidates:
val = dp[(A[r] - A[l], A[r])] = dp.get((A[l], A[r] - A[l]), 2) + 1
ans = max(ans, val)
return ans if ans >= 3 else 0
|
A sequence X_1, X_2, ..., X_n is fibonacci-like if:
n >= 3
X_i + X_{i+1} = X_{i+2} for all i + 2 <= n
Given a strictly increasing array A of positive integers forming a sequence, find the length of the longest fibonacci-like subsequence of A. If one does not exist, return 0.
(Recall that a subsequence is derived from another sequence A by deleting any number of elements (including none) from A, 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: [1,2,3,4,5,6,7,8]
Output: 5
Explanation:
The longest subsequence that is fibonacci-like: [1,2,3,5,8].
Example 2:
Input: [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].
Note:
3 <= A.length <= 1000
1 <= A[0] < A[1] < ... < A[A.length - 1] <= 10^9
(The time limit has been reduced by 50% for submissions in Java, C, and C++.)
|
from collections import defaultdict
class Solution:
def lenLongestFibSubseq(self, A: List[int]) -> int:
d = defaultdict(lambda: 1)
set_A = set(A)
for first in range(len(A)):
for second in range(first + 1, len(A)):
if A[second] - A[first] in set_A and A[first] > A[second] - A[first]:
d[A[first], A[second]] = d[A[second] - A[first], A[first]] + 1
if len(d.values()) == 0:
return 0
ret = max(d.values()) + 1
return ret if ret >= 3 else 0
|
A sequence X_1, X_2, ..., X_n is fibonacci-like if:
n >= 3
X_i + X_{i+1} = X_{i+2} for all i + 2 <= n
Given a strictly increasing array A of positive integers forming a sequence, find the length of the longest fibonacci-like subsequence of A. If one does not exist, return 0.
(Recall that a subsequence is derived from another sequence A by deleting any number of elements (including none) from A, 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: [1,2,3,4,5,6,7,8]
Output: 5
Explanation:
The longest subsequence that is fibonacci-like: [1,2,3,5,8].
Example 2:
Input: [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].
Note:
3 <= A.length <= 1000
1 <= A[0] < A[1] < ... < A[A.length - 1] <= 10^9
(The time limit has been reduced by 50% for submissions in Java, C, and C++.)
|
class Solution:
def lenLongestFibSubseq(self, A: List[int]) -> int:
# # straight forward N^2log(M)
# n = len(A)
# s = set()
# for i in A:
# s.add(i)
# res = 0
# ll = 0
# for i in range(n):
# for j in range(i+1,n):
# f1 = A[i]
# f2 = A[j]
# if f1+f2 in s:
# ll = 2
# while f1+f2 in s:
# ll += 1
# tmp = f2
# f2 = f1+f2
# f1 = tmp
# res = max(res,ll)
# return res
# # DP N^2 time N^2 space using std. dp array, if using two key dict as dp
# n = len(A)
# d = {}
# for i in range(n):
# d[A[i]] = i
# dp = [[0]*n for i in range(n)]
# res = 0
# for i in range(n):
# for j in range(i+1,n):
# tar = A[i]+A[j]
# if tar in d:
# k = d[tar]
# dp[j][k] = max(dp[j][k], dp[i][j]+1, 3)
# res = max(res,dp[j][k])
# # print(res,i,j,dp)
# return res
# DP using using two key dict as dp , N^2 time NlogM space
n = len(A)
d = {}
for i in range(n):
d[A[i]] = i
dp = {}
res = 0
for i in range(n):
for j in range(i+1,n):
tar = A[i]+A[j]
if tar in d:
k = d[tar]
dp[(j,k)] = max(dp.get((i,j),2)+1, dp.get((j,k),2))
res = max(res,dp[(j,k)])
# print(res,i,j,dp)
return res
|
A sequence X_1, X_2, ..., X_n is fibonacci-like if:
n >= 3
X_i + X_{i+1} = X_{i+2} for all i + 2 <= n
Given a strictly increasing array A of positive integers forming a sequence, find the length of the longest fibonacci-like subsequence of A. If one does not exist, return 0.
(Recall that a subsequence is derived from another sequence A by deleting any number of elements (including none) from A, 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: [1,2,3,4,5,6,7,8]
Output: 5
Explanation:
The longest subsequence that is fibonacci-like: [1,2,3,5,8].
Example 2:
Input: [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].
Note:
3 <= A.length <= 1000
1 <= A[0] < A[1] < ... < A[A.length - 1] <= 10^9
(The time limit has been reduced by 50% for submissions in Java, C, and C++.)
|
class Solution:
def lenLongestFibSubseq(self, A: List[int]) -> int:
n = len(A)
dp = [[2] * n for _ in range(n)]
idxs = {num: i for i, num in enumerate(A)}
result = 0
for j in range(2, n):
for i in range(1, j):
target = A[j] - A[i]
if target in idxs and target < A[i]:
# A[j] - A[i] 在 A 中且 在 A[i] 的左方
dp[i][j] = dp[idxs[target]][i] + 1
result = max(result, dp[i][j])
return result
|
A sequence X_1, X_2, ..., X_n is fibonacci-like if:
n >= 3
X_i + X_{i+1} = X_{i+2} for all i + 2 <= n
Given a strictly increasing array A of positive integers forming a sequence, find the length of the longest fibonacci-like subsequence of A. If one does not exist, return 0.
(Recall that a subsequence is derived from another sequence A by deleting any number of elements (including none) from A, 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: [1,2,3,4,5,6,7,8]
Output: 5
Explanation:
The longest subsequence that is fibonacci-like: [1,2,3,5,8].
Example 2:
Input: [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].
Note:
3 <= A.length <= 1000
1 <= A[0] < A[1] < ... < A[A.length - 1] <= 10^9
(The time limit has been reduced by 50% for submissions in Java, C, and C++.)
|
class Solution:
def lenLongestFibSubseq(self, A: List[int]) -> int:
n=len(A)
Aset=set(A)
ans=0
dp={}
for i in range(1,n):
for j in range(i):
diff=A[i]-A[j]
if diff < A[j] and diff in Aset:
dp[(A[j],A[i])]=dp.get((diff, A[j]),2)+1
ans=max(ans, dp[(A[j],A[i])])
return ans
|
A sequence X_1, X_2, ..., X_n is fibonacci-like if:
n >= 3
X_i + X_{i+1} = X_{i+2} for all i + 2 <= n
Given a strictly increasing array A of positive integers forming a sequence, find the length of the longest fibonacci-like subsequence of A. If one does not exist, return 0.
(Recall that a subsequence is derived from another sequence A by deleting any number of elements (including none) from A, 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: [1,2,3,4,5,6,7,8]
Output: 5
Explanation:
The longest subsequence that is fibonacci-like: [1,2,3,5,8].
Example 2:
Input: [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].
Note:
3 <= A.length <= 1000
1 <= A[0] < A[1] < ... < A[A.length - 1] <= 10^9
(The time limit has been reduced by 50% for submissions in Java, C, and C++.)
|
class Solution:
def lenLongestFibSubseq(self, A: List[int]) -> int:
dp = [[2]*len(A) for _ in range(len(A))]
d = {}
for i, num in enumerate(A):
d[num] = i
res = 2
for i in range(len(A)):
for j in range(i+1, len(A)):
diff = A[j] - A[i]
if diff in d and d[diff] < i:
dp[i][j] = dp[d[diff]][i] + 1
res = max(res, dp[i][j])
return res if res>=3 else 0
|
A sequence X_1, X_2, ..., X_n is fibonacci-like if:
n >= 3
X_i + X_{i+1} = X_{i+2} for all i + 2 <= n
Given a strictly increasing array A of positive integers forming a sequence, find the length of the longest fibonacci-like subsequence of A. If one does not exist, return 0.
(Recall that a subsequence is derived from another sequence A by deleting any number of elements (including none) from A, 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: [1,2,3,4,5,6,7,8]
Output: 5
Explanation:
The longest subsequence that is fibonacci-like: [1,2,3,5,8].
Example 2:
Input: [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].
Note:
3 <= A.length <= 1000
1 <= A[0] < A[1] < ... < A[A.length - 1] <= 10^9
(The time limit has been reduced by 50% for submissions in Java, C, and C++.)
|
# class Solution:
# def lenLongestFibSubseq(self, A: List[int]) -> int:
# aSet = set(A)
# ans = 0
# for i in range(len(A)):
# for j in range(i+1, len(A)):
# x, y = A[j], A[i] + A[j]
# length = 2
# while y in aSet:
# x, y = y, x + y
# length += 1
# ans = max(ans, length)
# return ans if ans >= 3 else 0
class Solution:
def lenLongestFibSubseq(self, A: List[int]) -> int:
aSet = set(A)
dp = collections.defaultdict(int)
ans = 0
for i in range(len(A)):
for j in range(i):
if A[i] - A[j] < A[j] and A[i] - A[j] in aSet:
dp[(A[j], A[i])] = dp.get((A[i] - A[j], A[j]), 2) + 1
return max(dp.values() or [0])
|
A sequence X_1, X_2, ..., X_n is fibonacci-like if:
n >= 3
X_i + X_{i+1} = X_{i+2} for all i + 2 <= n
Given a strictly increasing array A of positive integers forming a sequence, find the length of the longest fibonacci-like subsequence of A. If one does not exist, return 0.
(Recall that a subsequence is derived from another sequence A by deleting any number of elements (including none) from A, 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: [1,2,3,4,5,6,7,8]
Output: 5
Explanation:
The longest subsequence that is fibonacci-like: [1,2,3,5,8].
Example 2:
Input: [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].
Note:
3 <= A.length <= 1000
1 <= A[0] < A[1] < ... < A[A.length - 1] <= 10^9
(The time limit has been reduced by 50% for submissions in Java, C, and C++.)
|
class Solution:
def lenLongestFibSubseq(self, A: List[int]) -> int:
dp = collections.defaultdict(int)
s = set(A)
for j in range(len(A)):
for i in range(j):
if A[j]-A[i] < A[i] and A[j]-A[i] in s:
dp[(A[i], A[j])] = dp.get((A[j]-A[i], A[i]), 2) + 1
return max(dp.values() or [0])
|
A sequence X_1, X_2, ..., X_n is fibonacci-like if:
n >= 3
X_i + X_{i+1} = X_{i+2} for all i + 2 <= n
Given a strictly increasing array A of positive integers forming a sequence, find the length of the longest fibonacci-like subsequence of A. If one does not exist, return 0.
(Recall that a subsequence is derived from another sequence A by deleting any number of elements (including none) from A, 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: [1,2,3,4,5,6,7,8]
Output: 5
Explanation:
The longest subsequence that is fibonacci-like: [1,2,3,5,8].
Example 2:
Input: [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].
Note:
3 <= A.length <= 1000
1 <= A[0] < A[1] < ... < A[A.length - 1] <= 10^9
(The time limit has been reduced by 50% for submissions in Java, C, and C++.)
|
class Solution:
# dp, time O(n^2), space O(n^2)
'''dp[a, b] represents the length of fibo sequence ends up with (a, b)
Then we have dp[a, b] = (dp[b - a, a] + 1) or 2
'''
def lenLongestFibSubseq(self, A: List[int]) -> int:
n = len(A)
dp = collections.defaultdict(int)
st = set(A)
for i in range(n):
for j in range(i + 1, n):
if A[j] - A[i] < A[i] and A[j] - A[i] in st:
dp[A[i], A[j]] = dp.get((A[j] - A[i], A[i]), 2) + 1
return max(dp.values() or [0])
|
A sequence X_1, X_2, ..., X_n is fibonacci-like if:
n >= 3
X_i + X_{i+1} = X_{i+2} for all i + 2 <= n
Given a strictly increasing array A of positive integers forming a sequence, find the length of the longest fibonacci-like subsequence of A. If one does not exist, return 0.
(Recall that a subsequence is derived from another sequence A by deleting any number of elements (including none) from A, 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: [1,2,3,4,5,6,7,8]
Output: 5
Explanation:
The longest subsequence that is fibonacci-like: [1,2,3,5,8].
Example 2:
Input: [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].
Note:
3 <= A.length <= 1000
1 <= A[0] < A[1] < ... < A[A.length - 1] <= 10^9
(The time limit has been reduced by 50% for submissions in Java, C, and C++.)
|
import collections
class Solution:
def lenLongestFibSubseq(self, A: List[int]) -> int:
dict1 = collections.defaultdict(lambda :2)
L = len(A)
set_A = set(A)
if L<3:
return 0
d = 2
for i,A_i in enumerate(A):
for j in range(i,0,-1):
if A_i-A[j]>A[j]:
break
elif A_i-A[j] in set_A:
dd = dict1[A_i-A[j],A[j]]+1
dict1[A[j],A_i] = dd
d = max(d,dd)
return d if d>2 else 0
|
A sequence X_1, X_2, ..., X_n is fibonacci-like if:
n >= 3
X_i + X_{i+1} = X_{i+2} for all i + 2 <= n
Given a strictly increasing array A of positive integers forming a sequence, find the length of the longest fibonacci-like subsequence of A. If one does not exist, return 0.
(Recall that a subsequence is derived from another sequence A by deleting any number of elements (including none) from A, 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: [1,2,3,4,5,6,7,8]
Output: 5
Explanation:
The longest subsequence that is fibonacci-like: [1,2,3,5,8].
Example 2:
Input: [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].
Note:
3 <= A.length <= 1000
1 <= A[0] < A[1] < ... < A[A.length - 1] <= 10^9
(The time limit has been reduced by 50% for submissions in Java, C, and C++.)
|
class Solution:
def lenLongestFibSubseq(self, A: List[int]) -> int:
# https://blog.csdn.net/fuxuemingzhu/article/details/82715323
n = len(A)
m = dict()
for i, a in enumerate(A):
m[a] = i
res = 0
# dp[i][j] := max len of seq ends with A[i], A[j]
dp = [[2 for i in range(n)] for j in range(n)]
for j in range(n):
for k in range(j + 1, n):
a_i = A[k] - A[j]
if a_i >= A[j]:
break
if a_i in m:
i = m[a_i]
dp[j][k] = dp[i][j] + 1
res = max(res, dp[j][k])
return res
|
A sequence X_1, X_2, ..., X_n is fibonacci-like if:
n >= 3
X_i + X_{i+1} = X_{i+2} for all i + 2 <= n
Given a strictly increasing array A of positive integers forming a sequence, find the length of the longest fibonacci-like subsequence of A. If one does not exist, return 0.
(Recall that a subsequence is derived from another sequence A by deleting any number of elements (including none) from A, 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: [1,2,3,4,5,6,7,8]
Output: 5
Explanation:
The longest subsequence that is fibonacci-like: [1,2,3,5,8].
Example 2:
Input: [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].
Note:
3 <= A.length <= 1000
1 <= A[0] < A[1] < ... < A[A.length - 1] <= 10^9
(The time limit has been reduced by 50% for submissions in Java, C, and C++.)
|
class Solution:
def lenLongestFibSubseq(self, A: List[int]) -> int:
s = set(A)
dp = collections.defaultdict(int)
for j in range(len(A)):
for i in range(j):
if(A[j]-A[i]<A[i] and (A[j]-A[i]) in s):
dp[(A[i], A[j])] = dp.get((A[j]-A[i], A[i]), 2)+1
return max(dp.values() or [0])
|
A sequence X_1, X_2, ..., X_n is fibonacci-like if:
n >= 3
X_i + X_{i+1} = X_{i+2} for all i + 2 <= n
Given a strictly increasing array A of positive integers forming a sequence, find the length of the longest fibonacci-like subsequence of A. If one does not exist, return 0.
(Recall that a subsequence is derived from another sequence A by deleting any number of elements (including none) from A, 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: [1,2,3,4,5,6,7,8]
Output: 5
Explanation:
The longest subsequence that is fibonacci-like: [1,2,3,5,8].
Example 2:
Input: [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].
Note:
3 <= A.length <= 1000
1 <= A[0] < A[1] < ... < A[A.length - 1] <= 10^9
(The time limit has been reduced by 50% for submissions in Java, C, and C++.)
|
from collections import defaultdict
class Solution:
def lenLongestFibSubseq(self, A: List[int]) -> int:
dp = defaultdict(int)
s = set(A)
for i in range(len(A)):
for j in range(i):
if A[i] - A[j] < A[j] and A[i] - A[j] in s:
dp [(A[i], A[j])] = dp.get((A[j],A[i]-A[j]),2) + 1
return max(dp.values()) if dp else 0
|
A sequence X_1, X_2, ..., X_n is fibonacci-like if:
n >= 3
X_i + X_{i+1} = X_{i+2} for all i + 2 <= n
Given a strictly increasing array A of positive integers forming a sequence, find the length of the longest fibonacci-like subsequence of A. If one does not exist, return 0.
(Recall that a subsequence is derived from another sequence A by deleting any number of elements (including none) from A, 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: [1,2,3,4,5,6,7,8]
Output: 5
Explanation:
The longest subsequence that is fibonacci-like: [1,2,3,5,8].
Example 2:
Input: [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].
Note:
3 <= A.length <= 1000
1 <= A[0] < A[1] < ... < A[A.length - 1] <= 10^9
(The time limit has been reduced by 50% for submissions in Java, C, and C++.)
|
class Solution:
def lenLongestFibSubseq(self, A: List[int]) -> int:
indexes = {n: i for i, n in enumerate(A)}
dp, ans = collections.defaultdict(lambda: 2), 0
for i, n in enumerate(A):
for j in range(i):
idx = indexes.get(n - A[j], None)
if idx is not None and idx < j:
cand = dp[j, i] = dp[idx, j] + 1
ans = max(ans, cand)
return ans if ans >= 3 else 0
|
A sequence X_1, X_2, ..., X_n is fibonacci-like if:
n >= 3
X_i + X_{i+1} = X_{i+2} for all i + 2 <= n
Given a strictly increasing array A of positive integers forming a sequence, find the length of the longest fibonacci-like subsequence of A. If one does not exist, return 0.
(Recall that a subsequence is derived from another sequence A by deleting any number of elements (including none) from A, 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: [1,2,3,4,5,6,7,8]
Output: 5
Explanation:
The longest subsequence that is fibonacci-like: [1,2,3,5,8].
Example 2:
Input: [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].
Note:
3 <= A.length <= 1000
1 <= A[0] < A[1] < ... < A[A.length - 1] <= 10^9
(The time limit has been reduced by 50% for submissions in Java, C, and C++.)
|
class Solution:
def lenLongestFibSubseq(self, A: List[int]) -> int:
s=set(A)
n = len(A)
result = 0
for i in range(n-1):
for j in range(i+1,n):
a,b = A[i],A[j]
count = 2
while a+b in s:
a,b= b,a+b
count+=1
result=max(count,result)
return result if result>2 else 0
|
A sequence X_1, X_2, ..., X_n is fibonacci-like if:
n >= 3
X_i + X_{i+1} = X_{i+2} for all i + 2 <= n
Given a strictly increasing array A of positive integers forming a sequence, find the length of the longest fibonacci-like subsequence of A. If one does not exist, return 0.
(Recall that a subsequence is derived from another sequence A by deleting any number of elements (including none) from A, 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: [1,2,3,4,5,6,7,8]
Output: 5
Explanation:
The longest subsequence that is fibonacci-like: [1,2,3,5,8].
Example 2:
Input: [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].
Note:
3 <= A.length <= 1000
1 <= A[0] < A[1] < ... < A[A.length - 1] <= 10^9
(The time limit has been reduced by 50% for submissions in Java, C, and C++.)
|
class Solution:
def lenLongestFibSubseq(self, A: List[int]) -> int:
s = set(A)
n = len(A)
result = 0
for i in range(n-1):
for j in range(i+1, n):
a, b = A[i], A[j]
count = 2
while a+b in s:
a, b = b, a+b
count += 1
result = max(result, count)
return result if result > 2 else 0
|
A sequence X_1, X_2, ..., X_n is fibonacci-like if:
n >= 3
X_i + X_{i+1} = X_{i+2} for all i + 2 <= n
Given a strictly increasing array A of positive integers forming a sequence, find the length of the longest fibonacci-like subsequence of A. If one does not exist, return 0.
(Recall that a subsequence is derived from another sequence A by deleting any number of elements (including none) from A, 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: [1,2,3,4,5,6,7,8]
Output: 5
Explanation:
The longest subsequence that is fibonacci-like: [1,2,3,5,8].
Example 2:
Input: [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].
Note:
3 <= A.length <= 1000
1 <= A[0] < A[1] < ... < A[A.length - 1] <= 10^9
(The time limit has been reduced by 50% for submissions in Java, C, and C++.)
|
class Solution:
def lenLongestFibSubseq(self, A: List[int]) -> int:
s = set(A)
res = 0
for i in range(len(A) - 2):
for j in range(i+1, len(A) - 1):
a, b = A[i], A[j]
cnt = 2
while a+b in s:
a, b = b, a+b
cnt += 1
res = max(res, cnt)
return res
|
A sequence X_1, X_2, ..., X_n is fibonacci-like if:
n >= 3
X_i + X_{i+1} = X_{i+2} for all i + 2 <= n
Given a strictly increasing array A of positive integers forming a sequence, find the length of the longest fibonacci-like subsequence of A. If one does not exist, return 0.
(Recall that a subsequence is derived from another sequence A by deleting any number of elements (including none) from A, 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: [1,2,3,4,5,6,7,8]
Output: 5
Explanation:
The longest subsequence that is fibonacci-like: [1,2,3,5,8].
Example 2:
Input: [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].
Note:
3 <= A.length <= 1000
1 <= A[0] < A[1] < ... < A[A.length - 1] <= 10^9
(The time limit has been reduced by 50% for submissions in Java, C, and C++.)
|
class Solution:
def lenLongestFibSubseq(self, A: List[int]) -> int:
# 1. Brutal Force
# l = ans = 0
# mapA = set(a)
# for i in range(len(A)):
# for j in range(i+1, len(A)):
# x, y = A[j], A[i] + A[j]
# l = 2
# while y in mapA:
# x, y = y, x + y
# l += 1
# ans = max(ans, l)
# return ans if ans >= 3 else 0
# 2. DP
from collections import defaultdict
ans = 0
mapA = {n: i for i, n in enumerate(A)}
dp = defaultdict(lambda: 2)
for k, z in enumerate(A):
for j in range(k):
i = mapA.get(z-A[j])
if i is not None and i < j:
temp = dp[(j, k)] = dp[(i, j)] + 1
ans = max(ans, temp)
return ans if ans >= 3 else 0
|
A sequence X_1, X_2, ..., X_n is fibonacci-like if:
n >= 3
X_i + X_{i+1} = X_{i+2} for all i + 2 <= n
Given a strictly increasing array A of positive integers forming a sequence, find the length of the longest fibonacci-like subsequence of A. If one does not exist, return 0.
(Recall that a subsequence is derived from another sequence A by deleting any number of elements (including none) from A, 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: [1,2,3,4,5,6,7,8]
Output: 5
Explanation:
The longest subsequence that is fibonacci-like: [1,2,3,5,8].
Example 2:
Input: [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].
Note:
3 <= A.length <= 1000
1 <= A[0] < A[1] < ... < A[A.length - 1] <= 10^9
(The time limit has been reduced by 50% for submissions in Java, C, and C++.)
|
class Solution:
def lenLongestFibSubseq(self, A: List[int]) -> int:
# dp = {prev_val, next_val: len}
n = len(A)
dp = [[2] * n for _ in range(n)]
loc = {A[0]: 0, A[1]: 1}
longest = 0
for k in range(2, n):
loc[A[k]] = k
for j in range(k):
target = A[k] - A[j]
if target in loc and loc[target] < j:
dp[j][k] = dp[loc[target]][j] + 1
longest = max(longest, dp[j][k])
return longest
'''dp = {}
dp[(A[1], A[0] + A[1])] = 2
for i in range(2, len(A)):
for j in range(i):
if (A[j], A[i]) not in dp:
dp[(A[i], A[j] + A[i])] = 2
else:
dp[(A[i], A[j] + A[i])] = dp[(A[j], A[i])] + 1
del dp[(A[j], A[i])]
longest = max(dp.values())
return longest if longest > 2 else 0'''
|
A sequence X_1, X_2, ..., X_n is fibonacci-like if:
n >= 3
X_i + X_{i+1} = X_{i+2} for all i + 2 <= n
Given a strictly increasing array A of positive integers forming a sequence, find the length of the longest fibonacci-like subsequence of A. If one does not exist, return 0.
(Recall that a subsequence is derived from another sequence A by deleting any number of elements (including none) from A, 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: [1,2,3,4,5,6,7,8]
Output: 5
Explanation:
The longest subsequence that is fibonacci-like: [1,2,3,5,8].
Example 2:
Input: [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].
Note:
3 <= A.length <= 1000
1 <= A[0] < A[1] < ... < A[A.length - 1] <= 10^9
(The time limit has been reduced by 50% for submissions in Java, C, and C++.)
|
class Solution:
def lenLongestFibSubseq(self, A: List[int]) -> int:
m={}
for i in range(len(A)):
m[A[i]]=i
n=len(A)
dp=[[2]*n for _ in range(n)]
res=0
for j in range(n):
for k in range(j+1,n):
ai=A[k]-A[j]
if ai>=A[j]:
break
if ai not in m:
continue
i=m[ai]
dp[j][k]=dp[i][j]+1
res=max(res,dp[j][k])
return res
|
A sequence X_1, X_2, ..., X_n is fibonacci-like if:
n >= 3
X_i + X_{i+1} = X_{i+2} for all i + 2 <= n
Given a strictly increasing array A of positive integers forming a sequence, find the length of the longest fibonacci-like subsequence of A. If one does not exist, return 0.
(Recall that a subsequence is derived from another sequence A by deleting any number of elements (including none) from A, 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: [1,2,3,4,5,6,7,8]
Output: 5
Explanation:
The longest subsequence that is fibonacci-like: [1,2,3,5,8].
Example 2:
Input: [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].
Note:
3 <= A.length <= 1000
1 <= A[0] < A[1] < ... < A[A.length - 1] <= 10^9
(The time limit has been reduced by 50% for submissions in Java, C, and C++.)
|
class Solution:
def lenLongestFibSubseq(self, A: List[int]) -> int:
index = {x:i for i, x in enumerate(A)}
dp = collections.defaultdict(lambda:2)
ans = 0
for k, z in enumerate(A):
for j in range(k):
i = index.get(z-A[j], None)
if i is not None and i<j:
cand = dp[j, k] = dp[i, j]+1
ans = max(ans, cand)
return ans if ans>=3 else 0
|
A sequence X_1, X_2, ..., X_n is fibonacci-like if:
n >= 3
X_i + X_{i+1} = X_{i+2} for all i + 2 <= n
Given a strictly increasing array A of positive integers forming a sequence, find the length of the longest fibonacci-like subsequence of A. If one does not exist, return 0.
(Recall that a subsequence is derived from another sequence A by deleting any number of elements (including none) from A, 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: [1,2,3,4,5,6,7,8]
Output: 5
Explanation:
The longest subsequence that is fibonacci-like: [1,2,3,5,8].
Example 2:
Input: [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].
Note:
3 <= A.length <= 1000
1 <= A[0] < A[1] < ... < A[A.length - 1] <= 10^9
(The time limit has been reduced by 50% for submissions in Java, C, and C++.)
|
class Solution(object):
def lenLongestFibSubseq(self, A):
index = {x: i for i, x in enumerate(A)}
longest = collections.defaultdict(lambda: 2)
ans = 0
for k, z in enumerate(A):
for j in range(k):
i = index.get(z - A[j], None)
if i is not None and i < j:
cand = longest[j, k] = longest[i, j] + 1
ans = max(ans, cand)
return ans if ans >= 3 else 0
|
A sequence X_1, X_2, ..., X_n is fibonacci-like if:
n >= 3
X_i + X_{i+1} = X_{i+2} for all i + 2 <= n
Given a strictly increasing array A of positive integers forming a sequence, find the length of the longest fibonacci-like subsequence of A. If one does not exist, return 0.
(Recall that a subsequence is derived from another sequence A by deleting any number of elements (including none) from A, 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: [1,2,3,4,5,6,7,8]
Output: 5
Explanation:
The longest subsequence that is fibonacci-like: [1,2,3,5,8].
Example 2:
Input: [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].
Note:
3 <= A.length <= 1000
1 <= A[0] < A[1] < ... < A[A.length - 1] <= 10^9
(The time limit has been reduced by 50% for submissions in Java, C, and C++.)
|
class Solution:
def lenLongestFibSubseq(self, A: List[int]) -> int:
s=set(A)
n=len(A)
result=0
for i in range(n-1):
for j in range(i+1,n):
a,b=A[i],A[j]
count=2
while a+b in s:
a,b=b,a+b
count+=1
result=max(result,count)
return result if result>2 else 0
|
A sequence X_1, X_2, ..., X_n is fibonacci-like if:
n >= 3
X_i + X_{i+1} = X_{i+2} for all i + 2 <= n
Given a strictly increasing array A of positive integers forming a sequence, find the length of the longest fibonacci-like subsequence of A. If one does not exist, return 0.
(Recall that a subsequence is derived from another sequence A by deleting any number of elements (including none) from A, 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: [1,2,3,4,5,6,7,8]
Output: 5
Explanation:
The longest subsequence that is fibonacci-like: [1,2,3,5,8].
Example 2:
Input: [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].
Note:
3 <= A.length <= 1000
1 <= A[0] < A[1] < ... < A[A.length - 1] <= 10^9
(The time limit has been reduced by 50% for submissions in Java, C, and C++.)
|
class Solution:
def lenLongestFibSubseq1(self, A: List[int]) -> int:
mp = {}
for v in A:
mp[v] = set()
result = 0
for i in range(len(A)):
for j in range(i + 1, len(A)):
a = A[i]
b = A[j]
l = 2
while True:
if b in mp[a]:
break
if l != 2:
mp[a].add(b)
c = a + b
if c not in mp:
break
a, b = b, c
l += 1
if l >= 3:
result = max(result, l)
return result
def lenLongestFibSubseq(self, A):
index = {x: i for i, x in enumerate(A)}
longest = collections.defaultdict(lambda: 2)
ans = 0
for k, z in enumerate(A):
for j in range(k):
i = index.get(z - A[j], None)
if i is not None and i < j:
cand = longest[j, k] = longest[i, j] + 1
ans = max(ans, cand)
return ans if ans >= 3 else 0
|
A sequence X_1, X_2, ..., X_n is fibonacci-like if:
n >= 3
X_i + X_{i+1} = X_{i+2} for all i + 2 <= n
Given a strictly increasing array A of positive integers forming a sequence, find the length of the longest fibonacci-like subsequence of A. If one does not exist, return 0.
(Recall that a subsequence is derived from another sequence A by deleting any number of elements (including none) from A, 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: [1,2,3,4,5,6,7,8]
Output: 5
Explanation:
The longest subsequence that is fibonacci-like: [1,2,3,5,8].
Example 2:
Input: [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].
Note:
3 <= A.length <= 1000
1 <= A[0] < A[1] < ... < A[A.length - 1] <= 10^9
(The time limit has been reduced by 50% for submissions in Java, C, and C++.)
|
class Solution:
def lenLongestFibSubseq(self, A: List[int]) -> int:
index = {x: i for i, x in enumerate(A)}
longest = collections.defaultdict(lambda: 2)
result = 0
for k, z in enumerate(A):
for j in range(k):
i = index.get(z - A[j])
if i is not None and i < j:
longest[(j, k)] = longest[(i, j)] + 1
result = max(result, longest[(j, k)])
return result if result >= 3 else 0
|
A sequence X_1, X_2, ..., X_n is fibonacci-like if:
n >= 3
X_i + X_{i+1} = X_{i+2} for all i + 2 <= n
Given a strictly increasing array A of positive integers forming a sequence, find the length of the longest fibonacci-like subsequence of A. If one does not exist, return 0.
(Recall that a subsequence is derived from another sequence A by deleting any number of elements (including none) from A, 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: [1,2,3,4,5,6,7,8]
Output: 5
Explanation:
The longest subsequence that is fibonacci-like: [1,2,3,5,8].
Example 2:
Input: [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].
Note:
3 <= A.length <= 1000
1 <= A[0] < A[1] < ... < A[A.length - 1] <= 10^9
(The time limit has been reduced by 50% for submissions in Java, C, and C++.)
|
class Solution:
def lenLongestFibSubseq(self, A: List[int]) -> int:
seen,n={},len(A)
A_set=set(A)
def find_lens(ind,prev,curr):
if ind==n:
return 0
key=(prev,curr)
if key in seen:
return seen[key]
ans=0
if prev+curr in A_set:
for i in range(ind,n):
if curr+prev==A[i]:
ans=max(ans,find_lens(i+1,curr,A[i])+1)
seen[key]=ans
return ans
ans=0
for i in range(n-3):
for j in range(i+1,n-2):
if A[i]+A[j] not in A_set: continue
ans=max(ans,find_lens(j+1,A[i],A[j]))
return ans+2 if ans!=0 else 0
|
A sequence X_1, X_2, ..., X_n is fibonacci-like if:
n >= 3
X_i + X_{i+1} = X_{i+2} for all i + 2 <= n
Given a strictly increasing array A of positive integers forming a sequence, find the length of the longest fibonacci-like subsequence of A. If one does not exist, return 0.
(Recall that a subsequence is derived from another sequence A by deleting any number of elements (including none) from A, 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: [1,2,3,4,5,6,7,8]
Output: 5
Explanation:
The longest subsequence that is fibonacci-like: [1,2,3,5,8].
Example 2:
Input: [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].
Note:
3 <= A.length <= 1000
1 <= A[0] < A[1] < ... < A[A.length - 1] <= 10^9
(The time limit has been reduced by 50% for submissions in Java, C, and C++.)
|
class Solution:
def lenLongestFibSubseq(self, A: List[int]) -> int:
# d = set()
# n = len(A)
# for c in A:
# d.add(c)
# max_l = 0
# for i in range(n-2):
# for j in range(i+1, n-1):
# a = A[i]
# b = A[j]
# count = 0
# while (a+b) in d:
# count += 1
# a,b = b,a+b
# max_l = max(max_l, count+2)
# if max_l >= 3:
# return max_l
# return 0
dp = collections.defaultdict(int)
s = set(A)
for j in range(len(A)):
for i in range(j):
if A[j] - A[i] < A[i] and A[j] - A[i] in s:
dp[A[i], A[j]] = dp.get((A[j] - A[i], A[i]), 2) + 1
return max(dp.values() or [0])
|
A sequence X_1, X_2, ..., X_n is fibonacci-like if:
n >= 3
X_i + X_{i+1} = X_{i+2} for all i + 2 <= n
Given a strictly increasing array A of positive integers forming a sequence, find the length of the longest fibonacci-like subsequence of A. If one does not exist, return 0.
(Recall that a subsequence is derived from another sequence A by deleting any number of elements (including none) from A, 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: [1,2,3,4,5,6,7,8]
Output: 5
Explanation:
The longest subsequence that is fibonacci-like: [1,2,3,5,8].
Example 2:
Input: [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].
Note:
3 <= A.length <= 1000
1 <= A[0] < A[1] < ... < A[A.length - 1] <= 10^9
(The time limit has been reduced by 50% for submissions in Java, C, and C++.)
|
'''
----BRUTE (but is also smart)----
KEY: use a set to hold all elements of input array.
So if you find a+b in the set, you can keep moving forward and updating 1 \"jump\" forward for a-->b, b--> a+b.
Then continue. So change the if-check to a while loop.
get every pair (i,j)
then try to find their next value in set. If found then update
holdj = j
j = i+j
i = holdj
len++
res = max(res,len)
return res if res>2
----big-O----
O((N^2) * logM)
N^2 to get each pair.
logM to get to the end of each fibonacci \"chain\"
----dp----
dp[a,b] = len of fib sequence ending with a,b
dp[a,b] = dp[b-a,a]+1 , if val[b]-val[a] <val[a] and (val[b]-val[a]) exists in set
= 2 , if (val[b]-val[a],val[a]) doesnt exist in dp
return the max of all dp cells, or [0]
---big-O--
O(N^2) for nested loop.
'''
class Solution:
def lenLongestFibSubseq(self, A: List[int]) -> int:
res = 0
if len(A)<=2:
return 0
find = set(A)
for i in range(len(A)-1):
for j in range(i+1,len(A)):
currLen = 2
a = A[i]
b = A[j]
while (a+b) in find:
currLen+=1
holdb = b #you're not moving along i and j, but the actual values themselves. Because you already have the values you need in the set.
b = a+b
a = holdb
res = max(currLen,res)
if res>2:
return res
else:
return 0
|
A sequence X_1, X_2, ..., X_n is fibonacci-like if:
n >= 3
X_i + X_{i+1} = X_{i+2} for all i + 2 <= n
Given a strictly increasing array A of positive integers forming a sequence, find the length of the longest fibonacci-like subsequence of A. If one does not exist, return 0.
(Recall that a subsequence is derived from another sequence A by deleting any number of elements (including none) from A, 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: [1,2,3,4,5,6,7,8]
Output: 5
Explanation:
The longest subsequence that is fibonacci-like: [1,2,3,5,8].
Example 2:
Input: [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].
Note:
3 <= A.length <= 1000
1 <= A[0] < A[1] < ... < A[A.length - 1] <= 10^9
(The time limit has been reduced by 50% for submissions in Java, C, and C++.)
|
class Solution:
def lenLongestFibSubseq(self, A: List[int]) -> int:
sA = set(A)
last = A[-1]
B = Counter()
best = 0
for i in reversed(range(len(A))):
a = A[i]
for b in A[i+1:]:
c = a+b
if c in sA:
B[a,b] = 1 + B[b,c]
best = max(best , B[a,b]+2)
elif c>last:
break
return best
|
A sequence X_1, X_2, ..., X_n is fibonacci-like if:
n >= 3
X_i + X_{i+1} = X_{i+2} for all i + 2 <= n
Given a strictly increasing array A of positive integers forming a sequence, find the length of the longest fibonacci-like subsequence of A. If one does not exist, return 0.
(Recall that a subsequence is derived from another sequence A by deleting any number of elements (including none) from A, 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: [1,2,3,4,5,6,7,8]
Output: 5
Explanation:
The longest subsequence that is fibonacci-like: [1,2,3,5,8].
Example 2:
Input: [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].
Note:
3 <= A.length <= 1000
1 <= A[0] < A[1] < ... < A[A.length - 1] <= 10^9
(The time limit has been reduced by 50% for submissions in Java, C, and C++.)
|
class Solution:
def lenLongestFibSubseq(self, A: List[int]) -> int:
s =set(A)
res = 2
for i in range(len(A) - 2):
for j in range(i+1, len(A) - 1):
a, b = A[i], A[j]
cnt = 2
while a+b in s:
cnt += 1
res = max(res, cnt)
a, b = b, a+b
return res if res > 2 else 0
|
A sequence X_1, X_2, ..., X_n is fibonacci-like if:
n >= 3
X_i + X_{i+1} = X_{i+2} for all i + 2 <= n
Given a strictly increasing array A of positive integers forming a sequence, find the length of the longest fibonacci-like subsequence of A. If one does not exist, return 0.
(Recall that a subsequence is derived from another sequence A by deleting any number of elements (including none) from A, 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: [1,2,3,4,5,6,7,8]
Output: 5
Explanation:
The longest subsequence that is fibonacci-like: [1,2,3,5,8].
Example 2:
Input: [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].
Note:
3 <= A.length <= 1000
1 <= A[0] < A[1] < ... < A[A.length - 1] <= 10^9
(The time limit has been reduced by 50% for submissions in Java, C, and C++.)
|
from collections import defaultdict
class Solution:
def lenLongestFibSubseq(self, A: List[int]) -> int:
# Brute Force with Set
# Time complexity: O(N^2 x logM), where N is the length of A,
# and M is the maximum value of A.
# Space complexity: O(N)
# S, ans = set(A), 0
# for i in range(len(A)):
# for j in range(i + 1, len(A)):
# x, y, l = A[j], A[i] + A[j], 2
# while y in S:
# x, y = y, x + y
# l += 1
# ans = max(ans, l)
# return ans if ans >= 3 else 0
# Dynamic Programming
# Time complexity: O(N^2)
# Space complexity: O(NlogM), where M is the largest element of A.
index = {x: i for i, x in enumerate(A)}
longest = defaultdict(lambda: 2)
ans = 0
for k, z in enumerate(A):
for j in range(k):
i = index.get(z - A[j], None)
if i is not None and i < j:
cand = longest[j, k] = longest[i, j] + 1
ans = max(ans, cand)
return ans if ans >= 3 else 0
|
A sequence X_1, X_2, ..., X_n is fibonacci-like if:
n >= 3
X_i + X_{i+1} = X_{i+2} for all i + 2 <= n
Given a strictly increasing array A of positive integers forming a sequence, find the length of the longest fibonacci-like subsequence of A. If one does not exist, return 0.
(Recall that a subsequence is derived from another sequence A by deleting any number of elements (including none) from A, 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: [1,2,3,4,5,6,7,8]
Output: 5
Explanation:
The longest subsequence that is fibonacci-like: [1,2,3,5,8].
Example 2:
Input: [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].
Note:
3 <= A.length <= 1000
1 <= A[0] < A[1] < ... < A[A.length - 1] <= 10^9
(The time limit has been reduced by 50% for submissions in Java, C, and C++.)
|
class Solution:
def lenLongestFibSubseq(self, A: List[int]) -> int:
S = set(A)
res = 2
for i in range(len(A) - 2):
if A[i] * (res - 1) * (res - 2) > A[-1]:
break
for j in range(i + 1, len(A) - 1):
cnt = 2
a, b = A[j], A[i] + A[j]
while b in S:
cnt += 1
a, b = b, a + b
res = max(res, cnt)
return res if res > 2 else 0
|
A sequence X_1, X_2, ..., X_n is fibonacci-like if:
n >= 3
X_i + X_{i+1} = X_{i+2} for all i + 2 <= n
Given a strictly increasing array A of positive integers forming a sequence, find the length of the longest fibonacci-like subsequence of A. If one does not exist, return 0.
(Recall that a subsequence is derived from another sequence A by deleting any number of elements (including none) from A, 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: [1,2,3,4,5,6,7,8]
Output: 5
Explanation:
The longest subsequence that is fibonacci-like: [1,2,3,5,8].
Example 2:
Input: [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].
Note:
3 <= A.length <= 1000
1 <= A[0] < A[1] < ... < A[A.length - 1] <= 10^9
(The time limit has been reduced by 50% for submissions in Java, C, and C++.)
|
class Solution:
def lenLongestFibSubseq(self, A: List[int]) -> int:
n = len(A)
ret = 0
a_set = set(A)
for i in range(n):
for j in range(i + 1, n):
num_1 = A[i]
num_2 = A[j]
length = 2
while num_1 + num_2 in a_set:
num_3 = num_1 + num_2
num_1 = num_2
num_2 = num_3
length += 1
if length > 2:
ret = max(ret, length)
return ret
|
A sequence X_1, X_2, ..., X_n is fibonacci-like if:
n >= 3
X_i + X_{i+1} = X_{i+2} for all i + 2 <= n
Given a strictly increasing array A of positive integers forming a sequence, find the length of the longest fibonacci-like subsequence of A. If one does not exist, return 0.
(Recall that a subsequence is derived from another sequence A by deleting any number of elements (including none) from A, 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: [1,2,3,4,5,6,7,8]
Output: 5
Explanation:
The longest subsequence that is fibonacci-like: [1,2,3,5,8].
Example 2:
Input: [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].
Note:
3 <= A.length <= 1000
1 <= A[0] < A[1] < ... < A[A.length - 1] <= 10^9
(The time limit has been reduced by 50% for submissions in Java, C, and C++.)
|
class Solution:
def lenLongestFibSubseq(self, A: List[int]) -> int:
N = len(A)
dp = [[2]*N for _ in range(N)]
a = {val:i for i, val in enumerate(A)}
for i in range(1, len(A)):
for j in range(i):
diff = A[i] - A[j]
if diff in a and a[diff] < j:
dp[i][j] = max(dp[i][j], dp[j][a[diff]] + 1)
res = max([max(i) for i in dp])
return res if res > 2 else 0
|
A sequence X_1, X_2, ..., X_n is fibonacci-like if:
n >= 3
X_i + X_{i+1} = X_{i+2} for all i + 2 <= n
Given a strictly increasing array A of positive integers forming a sequence, find the length of the longest fibonacci-like subsequence of A. If one does not exist, return 0.
(Recall that a subsequence is derived from another sequence A by deleting any number of elements (including none) from A, 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: [1,2,3,4,5,6,7,8]
Output: 5
Explanation:
The longest subsequence that is fibonacci-like: [1,2,3,5,8].
Example 2:
Input: [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].
Note:
3 <= A.length <= 1000
1 <= A[0] < A[1] < ... < A[A.length - 1] <= 10^9
(The time limit has been reduced by 50% for submissions in Java, C, and C++.)
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class Solution:
def lenLongestFibSubseq(self, A: List[int]) -> int:
N = len(A)
D = { num:idx for idx, num in enumerate(A) }
dp = [ [2] * N for _ in range(N) ]
for j in range(1, N):
for i in range(j):
if A[j] - A[i] in D and D[A[j] - A[i]] < i:
#print(i, j)
dp[i][j] = dp[ D[A[j] - A[i]] ][i] + 1
res = max(map(max, dp))
return res if res > 2 else 0
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