inputs
stringlengths 50
14k
| targets
stringlengths 4
655k
|
|---|---|
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)
index={}
for i in range(N):
index[A[i]]=i
dp=[[0]*N for _ in range(N)]
ans=0
for i in range(N-1,-1,-1):
hi=A[i]
for j in range(i-1,-1,-1):
lo=A[j]
if hi-lo<=0 or hi-lo>=lo:
break
if hi-lo in index:
ind = index[hi-lo]
length = 1+dp[j][i]
dp[ind][j]=length
ans=max(ans,length)
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:
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:
l = 0
else:
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:
index = {x: i for i, x in enumerate(A)}
longest = [[2 for i in range(len(A))] for j in range(len(A))]
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:
longest[j][k] = longest[i][j] + 1
ans = max(ans, longest[j][k])
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)
d = dict()
for i in range(2, len(A)):
for j in range(i):
if A[i] < 2 * A[j] and A[i] - A[j] in s:
d[(A[i], A[j])] = d.get((A[j], A[i]-A[j]), 2) + 1
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:
s = set(A)
n = len(A)
def getlength(a, b):
l = 0
temp = a + b
if temp in s:
l = 2
else: return 0
while temp in s:
temp, b = temp + b, temp
l += 1
return l
mx = 0
for i in range(n):
for j in range(i+1, n):
ll = getlength(A[i], A[j])
if mx < ll:
mx = ll
return mx
|
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)
res = 0
for i in range(len(A)-1):
for j in range(i+1, len(A)):
a, b = A[i], A[j]
temp = 2
while a+b in Aset:
temp += 1
a, b = b, a+b
if temp >= 3:
res = max(res, temp)
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 = [{} for i in range(n)]
result = 0
for i in range(n):
for j in range(i):
prev = A[i] - A[j]
if prev in dp[j]:
l = dp[i][A[j]] = dp[j][prev] + 1
result = max(result, l)
else:
dp[i][A[j]] = 2
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:
L = len(A)
res = 0
if L<3:
return 0
if L==3:
return 3 if A[0]+A[1]==A[2] else 0
s = set(A)
for i in range(2,L):
for j in range(i):
L,R = A[j],A[i]
cur = 2
while R-L in s and R-L<L:
cur += 1
L,R = R-L,L
res = max(res,cur)
#print(A[i],A[j],cur)
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:
set_a = set(A)
res = 0
for i in range(len(A)):
for j in range(i+1, len(A)):
x, y = A[i], A[j]
leng = 2
while x+y in set_a:
x, y = y, x+y
leng += 1
res = max(res, leng)
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:
s = set(A)
res = 2
for i in range(len(A)):
for j in range(i + 1, len(A)):
a = A[i]
b = A[j]
lengthOfSeq = 2
while a + b in s:
a, b, lengthOfSeq = b, a + b, lengthOfSeq + 1
res = max(res,lengthOfSeq)
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)
dp = [[2] * n for i in range(n)]
m = dict()
for i in range(n):
m[A[i]] = i
ans = 0
for i in range(n):
for j in range(i+1, n):
a_k = A[j] - A[i]
if a_k > A[i]:
break
k = m.get(a_k, None)
if k is not None and k < i:
dp[i][j] = dp[k][i] + 1
ans = max(ans, dp[i][j])
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:
n = len(A)
max_ = 0
dp = [[0 for i in range(n)] for j in range(n)]
for i in range(2, n):
l = 0
r = i - 1
while l < r:
sum_ = A[l] + A[r]
if sum_ > A[i]:
r -= 1
elif sum_ < A[i]:
l += 1
else:
dp[r][i] = dp[l][r] + 1
max_ = max(max_, dp[r][i])
r -= 1
l += 1
if max_ == 0:
return 0
else:
return max_ + 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:
n = len(A)
dp = [[2 for _ in range(n)] for _ in range(n)]
indMap = {x: i for i, x in enumerate(A)}
for i in range(n):
for j in range(i+1, n):
diff = A[j] - A[i]
if diff in indMap and indMap[diff] < i:
k = indMap[diff]
dp[i][j] = max(dp[i][j], 1 + dp[k][i])
return max([v for row in dp for v in row if v > 2] 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:
s = set(A)
res = 2
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, l = b, a + b, l + 1
res = max(res, l)
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:
S = set(A)
i = 0
max_len = 0
while i < len(A)-1:
j = i + 1
while j < len(A):
subSeq = [A[i], A[j]]
while subSeq[-1] + subSeq[-2] in S:
subSeq.append(subSeq[-1]+subSeq[-2])
# print(subSeq)
if len(subSeq) > max_len:
max_len = len(subSeq)
j += 1
i += 1
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)
s = set(A)
res = 2
for i in range(n):
for j in range(i+1, n):
a, b, l = A[i], A[j], 2
while a+b in s:
a,b,l = b,a+b,l+1
res = max(res, l)
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)
dp = [[2 for _ in range(n)] for _ in range(n)]
indMap = {x: i for i, x in enumerate(A)}
for i in range(n):
for j in range(i+1, n):
diff = A[j] - A[i]
if diff in indMap and indMap[diff] < i:
k = indMap[diff]
dp[i][j] = max(dp[i][j], 1 + dp[k][i])
res = 0
for row in dp:
for v in row:
if v > 2:
res = max(res, v)
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:
d={}
for i in range(len(A)):
d[A[i]]=i
ans=0
for i in range(0,len(A)):
for j in range(i+1,len(A)):
y=A[i]+A[j]
x=A[j]
l=2
while y in d and d[y]>j:
# print(y,l)
z=x+y
x=y
y=z
l+=1
ans=max(ans,l)
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:
sA = set(A)
B = Counter()
ans = 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] = B[b, c] + 1 # if recording B[b,c], everything is counted 1. We are tracking back like 3+5, 2+3, 1+2. So we want the 1+2 to be counted the most.
ans = max(ans, B[a, b] + 2) # [a, b, c] = [1 ,2, 3], B[b, c] = 1
if c > A[-1]:
break # the inner loop needn't continue
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:
res = 0
A_dict = {}
for x in A:
A_dict[x] = A_dict.get(x,0) + 1
for i in range(len(A)-1):
for j in range(i+1, len(A)):
a, b = A[i], A[j]
c = a + b
length = 0
while c in A_dict:
length += 1
a, b = b, c
c = a + b
res = max(res, length)
return res + 2 if res > 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:
ans = 0
setA = set(A)
for i in range(len(A)):
for j in range(i+1, len(A)):
temp = 2
x = A[j]
y = A[i] + A[j]
while y in setA:
temp += 1
x, y = y, x+y
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:
''' correct but too slow lol :'(
l = len(A)
def loop(one: int, two: int, ind: int, length: int) -> int:
if (ind == l):
return length
elif (one + two == A[ind]):
#print(\"one, two are \" + str(one) + \" \" + str(two))
return loop(two, A[ind], ind + 1, length + 1)
else:
return loop(one, two, ind + 1, length)
maxLen = 0
for i in range(l):
for j in range(i+1, l):
c = loop(A[i], A[j], j+1, 0)
maxLen = max(maxLen, c)
return (maxLen + 2) if maxLen else maxLen
'''
mySet = set(A)
maxCount = 0
l = len(A)
for i in range(l):
for j in range(i+1, l):
one = A[i]
two = A[j]
count = 0
while(one + two) in mySet:
temp = two
two = one + two
one = temp
count += 1
maxCount = max(maxCount, count)
return maxCount + 2 if maxCount 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:
max_length = 0
S = set(A)
for i in range(len(A)):
for j in range(i+1, len(A)):
x, y = A[i], A[j]
expected = x+y
length = 2
while expected in S:
x=y
y=expected
expected = x+y
length += 1
max_length = max(max_length, length)
return max_length if max_length >= 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)):
for j in range(i + 1,len(A)):
a = A[i]
b = A[j]
l = 2
while a + b in s:
a,b,l = b,a + b,l + 1
res = max(res,l) % (10 ** 9 + 7)
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:
res = 0
A_dict = {}
for x in A:
A_dict[x] = A_dict.get(x,0) + 1
for i in range(len(A)-1):
for j in range(i+1, len(A)):
a, b = A[i], A[j] # loop all initial candidates
c = a + b
length = 0
while c in A_dict:
length += 1
a, b = b, c
c = a + b
res = max(res, length)
return res + 2 if res > 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:
def getLength(orgprev,orgcurr,prev,curr):
nxt = curr+prev
if nxt in Aset:
return 1+getLength(orgprev,orgcurr,curr,nxt)
else:
return 2 if orgprev != prev and orgcurr != curr else 0
Aset = set(A)
return max(max([getLength(a,b,a,b) for a in A[:i]],default=float('-inf')) for i,b in enumerate(A))
|
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 = defaultdict(int)
ans = 0
for i in range(len(A)):
for j in range(i+1, len(A)):
prev, curr, l = A[i], A[j], 0
while prev + curr in s:
l += 1
dp[prev+curr] = l
prev, curr = curr, prev + curr
ans = max(ans, l)
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):
s = set(A)
res = 2
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, l = b, a + b, l + 1
res = max(res, l)
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:
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], float('inf'))
if i < j:
longest[j, k] = longest[i, j] + 1
if longest[j, k] > result:
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(object):
def lenLongestFibSubseq(self, A):
S = 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 S:
x, y = y, x + y
length += 1
ans = max(ans, length)
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 new_lenLongestFibSubseq(self, A: List[int]) -> int:
exists = set(A)
max_length = 2
for i in range(len(A)):
first = A[i]
for j in range(i):
second = A[j]
start = second
curr_length = 2
while (first + second) in exists:
curr_length += 1
curr = first + second
second = first
first = curr
max_length = max(max_length, curr_length)
if max_length < 3: return 0
return max_length
def lenLongestFibSubseq(self, A: List[int]) -> int:
exists = set(A)
max_length = 2
for i in range(len(A)):
for j in range(i):
b = A[i]
a = A[j]
curr_length = 2
while (a + b) in exists:
curr_length += 1
curr = a + b
a = b
b = curr
max_length = max(max_length, curr_length)
if max_length <= 2: return 0
return max_length
# [1,2,3,4,5,6,7,8]
#
# **** START == 1 ****
#
# second = 1
# first = 2
# curr_length = 3
# curr = 3
# cache[1] = max(cache[1]+1, 3)
# cache[1] = max(0+1, 3)
# cache[1] = 3
#
# second = 2
# first = 3
# curr_length = 4
# curr = 5
#
# cache[1] = max(cache[1], curr_length)
# cache[1] = max(3, 3)
# cache[1] = 4
#
# second = 3
# first = 5
# curr_length = 5
# curr = 8
#
# cache[1] = max(cache[1], 5)
# cache[1] = 5
#
# second = 5
# first = 8
# curr_length = 6
# curr = 13
#
# 13 not in A, next pair
#
#
# **** START == 1 ****
#
# second = 1
# first = 3
# curr_length = 3
# curr = 4
#
# cache[1] = max(5, 3)
# cache[1] = 5
#
# second = 3
# first = 4
# curr = 7
# curr_length = 4
#
# cache[1] = max(5, 4)
# cache[1] = 5
#
# second = 4
# first = 7
# curr = 11
#
# 11 not in A, exit
#
|
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=0
d={x:i for i,x in enumerate(A)}
n=len(A)
dp=[[2 for j in range(n)] for i in range(n)]
for i in range(2,n):
for j in range(i):
z=A[i]
x=A[j]
y=z-x
k=d.get(y,-1)
#x<y<z
if x<y and k!=-1:
dp[i][k]=1+dp[k][j]
res=max(res,dp[i][k])
if res>2:
return res
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)
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 S:
x, y = y, x + y
length += 1
ans = max(ans, length)
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:
if len(A) < 3:
return 0
S = set(A)
ans = 0
for i in range(len(A) - 1):
for j in range(i + 1, len(A)):
l = 2
x, y = A[j], A[i] + A[j]
while y in S:
l += 1
x, y = y, x + y
ans = max(ans, l)
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:
exists = set(A)
max_length = 2
for i in range(len(A)):
for j in range(i+1, len(A)):
a = A[i]
b = A[j]
curr_length = 2
while (a + b) in exists:
curr_length += 1
#curr = a + b
#a = b
#b = curr
a, b = b, a + b
max_length = max(max_length, curr_length)
if max_length <= 2: return 0
return max_length
# [1,2,3,4,5,6,7,8]
#
# **** START == 1 ****
#
# second = 1
# first = 2
# curr_length = 3
# curr = 3
# cache[1] = max(cache[1]+1, 3)
# cache[1] = max(0+1, 3)
# cache[1] = 3
#
# second = 2
# first = 3
# curr_length = 4
# curr = 5
#
# cache[1] = max(cache[1], curr_length)
# cache[1] = max(3, 3)
# cache[1] = 4
#
# second = 3
# first = 5
# curr_length = 5
# curr = 8
#
# cache[1] = max(cache[1], 5)
# cache[1] = 5
#
# second = 5
# first = 8
# curr_length = 6
# curr = 13
#
# 13 not in A, next pair
#
#
# **** START == 1 ****
#
# second = 1
# first = 3
# curr_length = 3
# curr = 4
#
# cache[1] = max(5, 3)
# cache[1] = 5
#
# second = 3
# first = 4
# curr = 7
# curr_length = 4
#
# cache[1] = max(5, 4)
# cache[1] = 5
#
# second = 4
# first = 7
# curr = 11
#
# 11 not in A, exit
#
|
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)
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 S:
x, y = y, x + y
length += 1
ans = max(ans, length)
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:
indexes={A[i]:i for i in range(len(A))}
dp=[[2 for i in range(len(A))] for j in range(len(A))]
z=0
for i in range(1,len(A)):
for j in range(0,i):
idx=indexes.get(A[i]+A[j],-1)
if idx==-1:
continue
else:
dp[i][idx]=dp[j][i]+1
z=max(max(dp[i]),z)
if z<3:
return 0
return z
|
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_val = set(A)
max_len = 0
for i in range(len(A)):
for j in range(i+1, len(A)):
p2, p1 = A[i],A[j]
cnt = 0
while p2 + p1 in set_val:
z = p2 + p1
p2 = p1
p1 = z
cnt += 1
max_len = max(max_len, cnt + 2 if cnt > 0 else cnt)
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:
if len(A) == 0: return A
result = 0
aref = set(A)
for i in range(len(A)):
for j in range(i+1, len(A)):
x, y = A[j], A[i] + A[j]
seq_length = 0
while y in aref:
if seq_length == 0: seq_length = 2
seq_length += 1
x, y = y, x+y
result = max(result,seq_length)
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:
s = set(A)
res = 2
for i in range(len(A)):
for j in range(i+1,len(A)):
a = A[i]
b = A[j]
l = 2
while a + b in s:
a,b,l = b,a + b,l + 1
res = max(res,l) % (10 ** 9 + 7)
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:
mapA = {n: i for i, n in enumerate(A)}
# 1. Brutal Force
l = ans = 0
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
|
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:
# [1,2,3,4,5,6,7,8]
# [0,0,3,]
memo = defaultdict(int)
visited = dict()
visited[A[0]] = True
visited[A[1]] = True
ret = 0
for i in range(2, len(A)):
num = A[i]
visited[num] = True
for j in range(i - 1, -1, -1):
if num - A[j] in visited and num - A[j] < A[j]:
old_key = (num - A[j], A[j])
val = memo[old_key] + 1
ret = max(ret, val)
key = (A[j], num)
# print(old_key)
# print(key, val)
# print('-----')
memo[key] = val
if ret:
ret += 2
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:
longest = 0
coll = set(A)
n = len(A)
lookup = [[None for _ in range(n)] for _ in range(n)]
for i in range(n - 2):
for j in range(i + 1, n - 1):
target = A[i] + A[j]
prev = A[j]
while target in coll:
if lookup[i][j]:
lookup[i][j] += 1
else:
lookup[i][j] = 3
temp = target
target += prev
prev = temp
longest = max(longest, lookup[i][j])
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, 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)
n = len(A)
res = 0
for i in range(n):
for j in range(i+1,n):
x, y = A[j],A[i] + A[j]
ans = 2
while y in S:
x, y = y, x + y
ans += 1
res = max(res, ans)
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:
s = set(A)
res = 2
for i in range(len(A)):
for j in range(i + 1,len(A)):
a = A[i]
b = A[j]
l = 2
while a + b in s:
a,b,l = b,a + b,l + 1
res = max(res,l) % (10 ** 9 + 7)
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, s, res = len(A), set(A), 0
for i in range(n):
for j in range(i+1, n):
x, y, l = A[j], A[i] + A[j], 2
while y in s:
x, y = y, x + y
l += 1
res = max(res, l)
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:
exists = set(A)
max_length = 2
for i in range(len(A)):
for j in range(i):
first = A[i]
second = A[j]
start = second
curr_length = 2
while (first + second) in exists:
curr_length += 1
curr = first + second
second = first
first = curr
max_length = max(max_length, curr_length)
if max_length < 3: return 0
return max_length
def old_lenLongestFibSubseq(self, A: List[int]) -> int:
exists = set(A)
max_length = 2
for i in range(len(A)):
for j in range(i):
b = A[i]
a = A[j]
curr_length = 2
while (a + b) in exists:
curr_length += 1
curr = a + b
a = b
b = curr
max_length = max(max_length, curr_length)
if max_length <= 2: return 0
return max_length
# [1,2,3,4,5,6,7,8]
#
# **** START == 1 ****
#
# second = 1
# first = 2
# curr_length = 3
# curr = 3
# cache[1] = max(cache[1]+1, 3)
# cache[1] = max(0+1, 3)
# cache[1] = 3
#
# second = 2
# first = 3
# curr_length = 4
# curr = 5
#
# cache[1] = max(cache[1], curr_length)
# cache[1] = max(3, 3)
# cache[1] = 4
#
# second = 3
# first = 5
# curr_length = 5
# curr = 8
#
# cache[1] = max(cache[1], 5)
# cache[1] = 5
#
# second = 5
# first = 8
# curr_length = 6
# curr = 13
#
# 13 not in A, next pair
#
#
# **** START == 1 ****
#
# second = 1
# first = 3
# curr_length = 3
# curr = 4
#
# cache[1] = max(5, 3)
# cache[1] = 5
#
# second = 3
# first = 4
# curr = 7
# curr_length = 4
#
# cache[1] = max(5, 4)
# cache[1] = 5
#
# second = 4
# first = 7
# curr = 11
#
# 11 not in A, exit
#
|
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:
d1={x:i for i,x in enumerate(A)}
d2={}
ans=0
for i in range(1,len(A)):
for j in range(i):
d2[(j,i)]=2
val= A[i]-A[j]
if(val in d1 and d1[val]<j):
d2[(j,i)]=d2[(d1[val],j)]+1
ans=max(ans,d2[(j,i)])
if(ans>=3):
return ans
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:
n = len(A)
if n < 3:
return 0
# dp = [[2 for j in range(n)] for i in range(n)]
# ans = 2
# hmap = {A[0]:0, A[1]:1}
# for i in range(2, n):
# for j in range(1, i):
# pos = hmap.get(A[i]-A[j], -1)
# if pos >= 0 and pos < j:
# dp[i][j] = max(dp[i][j], dp[j][pos]+1)
# ans = max(ans, dp[i][j])
# hmap[A[i]] = i
# return ans if ans > 2 else 0
dp = [[2 for j in range(n)] for i in range(n)]
ans = 2
for i in range(2, n):
l, r = 0, i-1
while l < r:
if A[l] + A[r] == A[i]:
dp[i][r] = max(dp[i][r], dp[r][l] + 1)
ans = max(ans, dp[i][r])
l += 1
r -= 1
elif A[l] + A[r] < A[i]:
l += 1
else:
r -= 1
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:
def _helper(l1, l2, cache):
if A[l1] + A[l2] in table:
l1, l2 = l2, table[A[l1] + A[l2]]
return _helper(l1, l2, cache + 1)
else:
return cache if cache >= 3 else 0
ans = 0
table = {v:i for i, v in enumerate(A)}
for i in range(len(A)):
for j in range(i+1, len(A)):
if A[i] + A[j] > A[-1]: break
ans = max(ans, _helper(i, 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:
# dp = [[0 for i in range(len(A))] for j in range(len(len(A)))]
ans = 0
d={}
for i in range(len(A)):
d[A[i]] = 1
for i in range(len(A)):
for j in range(i+1,len(A)):
a = A[i]
b = A[j]
l=2
while(d.get(a+b)!=None):
l+=1
temp = b
b = a+b
a = temp
ans = max(l,ans)
if(ans>2):
return ans
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, nums: List[int]) -> int:
n=len(nums)
dp=[[0 for x in range(n)] for y in range(n)]
max_=0
for i in range(2,n):
l=0
r=i-1
while l<=r:
s=nums[l]+nums[r]
if s>nums[i]:
r-=1
elif s<nums[i]:
l+=1
else:
dp[r][i]=dp[l][r]+1
max_=max(max_,dp[r][i])
l+=1
r-=1
return max_ if max_==0 else max_+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:
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
|
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:
from collections import defaultdict
used_tuples = set()
max_len = 0
hist = set(A)
for i in range(len(A) - 1):
for j in range(i + 1, len(A)):
curr_len = 0
s = A[i]
l = A[j]
if (s,l) in used_tuples:
continue
while s+l in hist:
curr_len += 1
used_tuples.add((s,l))
old_s = s
s = l
l = s + old_s
max_len = max(max_len, curr_len)
return max_len + 2 if max_len > 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:
n = len(A)
dp = collections.defaultdict(int)
ret = 0
a_set = set(A)
for i in range(n):
for j in range(i + 1, n):
dp[(A[i], A[j])] = 2
left = A[j] - A[i]
if left > A[i] or left not in a_set:
continue
dp[(A[i], A[j])] = max(dp[(A[i], A[j])], dp[(left, A[i])] + 1)
ret = max(ret, dp[(A[i], A[j])])
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:
s=set(a)
c=0
for i in range(len(a)):
for j in range(i+1,len(a)):
x,y,l=a[i],a[j],2
while x+y in s:
x,y=y,x+y
l+=1
c=max(c,l)
return c if c>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)
max_l = 0
cache = [[0 for j in range(n)] for i in range(n)]
a_dict = {number: index for (index, number) in enumerate(A)}
for i in range(n - 1, -1, -1):
for j in range(n - 1, i, -1):
if j == n - 1:
cache[i][j] = 2
else:
target = A[i] + A[j]
if target in a_dict:
cache[i][j] = cache[j][a_dict[target]] + 1
else:
cache[i][j] = 2
if cache[i][j] > 2:
max_l = max(cache[i][j], max_l)
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++.)
|
class Solution:
def lenLongestFibSubseq(self, A: List[int]) -> int:
pos = {A[i]: i for i in range(len(A))}
dp = [[2 for j in range(len(A))] for i in range(len(A))]
maxlen = 0
for z in range(2, len(A)):
for x in range(z):
y = pos.get(A[z] - A[x], -1)
if A[x] < (A[z] - A[x]) and y != -1:
dp[z][y] = dp[y][x] + 1
maxlen = max(maxlen, dp[z][y])
return maxlen if maxlen > 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)
index={n:i for i,n in enumerate(A)}
dp=collections.defaultdict(lambda: 2)
res=0
for i2 in range(N):
for i1 in range(i2-1,-1,-1):
n0=A[i2]-A[i1]
if n0>=A[i1]:
break
if n0 in index:
i0=index[n0]
dp[(i1,i2)]=dp[(i0,i1)]+1
res=max(res,dp[(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:
n = len(A)
dp = collections.defaultdict(int)
ret = 0
a_set = set(A)
for i in range(n):
for j in range(i + 1, n):
dp[(A[i], A[j])] = 2
left = A[j] - A[i]
if left >= A[i] or left not in a_set:
continue
#print(left, A[i], A[j], dp[(left, A[i])])
dp[(A[i], A[j])] = dp[(left, A[i])] + 1
ret = max(ret, dp[(A[i], A[j])])
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:
#from collections import deafal
def lenLongestFibSubseq(self, A: List[int]) -> int:
n = len(A)
s = set(A)
def helper(i,j):
ans = 2
while(True):
k = i+j
if(k in s and k<=10**9):
ans +=1
i = j
j = k
else:
break
return ans
c = 0
for i in reversed(list(range(n-2))):
for j in reversed(list(range(i+1,n-1))):
c = max(c,helper(A[i],A[j]))
return c if c>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:
longest = 0
coll = set(A)
n = len(A)
lookup = [[0 for _ in range(n)] for _ in range(n)]
j = 1
for i in range(n - 2):
if n - i + 2 < longest - lookup[i][j]:
break
for j in range(i + 1, n - 1):
if n - j + 1 < longest - lookup[i][j]:
break
target = A[i] + A[j]
prev = A[j]
while target in coll:
if lookup[i][j]:
lookup[i][j] += 1
else:
lookup[i][j] = 3
temp = target
target += prev
prev = temp
longest = max(longest, lookup[i][j])
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, A: List[int]) -> int:
d = {}
for i, v in enumerate(A):
d[v] = i
l = len(A)
res = -1
for i in range(0, l):
for j in range(i+1, l):
one_behind = A[j]
two_behind = A[i]
total = one_behind + two_behind
c = 0
while total in d:
print((total, one_behind, two_behind))
if c == 0:
c += 3
else:
c += 1
tmp = one_behind
one_behind = total
two_behind = tmp
total = one_behind + two_behind
res = max(res, c)
#break
#print(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++.)
|
class Solution:
def lenLongestFibSubseq(self, nums: List[int]) -> int:
n, ans = len(nums), 0
idx = {n: i for i, n in enumerate(nums)}
dp = [[0]*n for _ in range(n)]
for j in range(n):
dp[0][j] = 2
dp[j][j] = 1
for i in range(1, n):
for j in range(i+1, n):
dp[i][j] = 2
if nums[i] > nums[j]//2 and nums[j]-nums[i] in idx:
ii = idx[nums[j]-nums[i]]
dp[i][j] = dp[ii][i] + 1
ans = max(ans, dp[i][j])
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:
s = 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]
l = 2
while y in s:
x,y = y,x+y
l += 1
ans = max(ans,l)
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:
d={}
for i in range(len(A)):
d[A[i]]=i
dp=[[2 for i in range(len(A))] for j in range(len(A))]
ans=0
for j in range(0,len(A)):
for i in range(0,j):
if A[j]-A[i] in d and d[A[j]-A[i]]<i:
k= d[A[j]-A[i]]
dp[i][j]=dp[k][i]+1
ans=max(dp[i][j],ans)
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, nums: List[int]) -> int:
n = len(nums)
idx = {n: i for i, n in enumerate(nums)}
dp = [[0]*n for _ in range(n)]
for j in range(n):
dp[0][j] = 2
dp[j][j] = 1
ans = 0
for i in range(1, n):
for j in range(i+1, n):
if nums[i] <= nums[j]//2:
dp[i][j] = 2
else:
prev = nums[j]-nums[i]
if prev in idx:
dp[i][j] = dp[idx[prev]][i] + 1
else:
dp[i][j] = 2
ans = max(ans, dp[i][j])
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++.)
|
from collections import defaultdict
class Solution:
def lenLongestFibSubseq(self, A: List[int]) -> int:
index_map = { x:i for i, x in enumerate(A) }
longest = defaultdict(lambda: 2)
ans = 0
for k in range(len(A)):
for j in range(len(A)):
target = A[k] - A[j]
i = index_map.get(target, -1)
if i >= 0 and i < j:
cand = longest[j,k] = longest[i,j] + 1
ans = max(ans, cand)
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(object):
def lenLongestFibSubseq(self, A):
S = 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 S:
x, y = y, x + y
length += 1
ans = max(ans, length)
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:
res = 2
S = set(A)
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(cnt, 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:
maxlen = 0
dp = [[0 for _ in range(len(A))] for _ in range(len(A))]
for i in range(2, len(A)):
l, r = 0, i - 1
while l < r:
s = A[l] + A[r]
if s > A[i]:
r -= 1
elif s < A[i]:
l += 1
else:
dp[r][i] = max(dp[r][i], dp[l][r] + 1)
maxlen = max(maxlen, dp[r][i])
l += 1
r -= 1
return maxlen if maxlen == 0 else maxlen + 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:
length = len(A)
s = set(A)
result = 0
for i in range(length-1):
for j in range(i+1, length):
a=A[i]
b=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)
n = len(A)
count = 0
for i in range(n):
for j in range(i+1, n):
x, y = A[j], A[i]+A[j]
length = 2
while y in s:
x, y = y, x+y
length +=1
count = max(count, length)
return count if count >= 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:
n = len(A)
dp = [[2]*n for i in range(n)]
map_ = collections.defaultdict(int)
for i,j in enumerate(A):
map_[j] = i
# i:end j:start
for i in range(2, n):
for j in range(1, i):
diff = A[i] - A[j]
if diff in list(map_.keys()) and map_[diff] < j:
k = map_[diff]
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++.)
|
class Solution:
def lenLongestFibSubseq(self, A: List[int]) -> int:
start = 0
longest = 0
cur_total = 0
root = A[0]
coll = set(A)
n = len(A)
cur = 2
target = [[None for _ in range(n)] for _ in range(n)]
lookup = [[None for _ in range(n)] for _ in range(n)]
for i in range(n - 2):
for j in range(i + 1, n - 1):
target[i][j] = A[i] + A[j]
prev = A[j]
while target[i][j] in coll:
if lookup[i][j]:
lookup[i][j] += 1
else:
lookup[i][j] = 3
temp = target[i][j]
target[i][j] += prev
prev = temp
longest = max(longest, lookup[i][j])
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, A: List[int]) -> int:
s=set(A)
maxi=2
for i in range(len(A)-2):
for j in range(i+1, len(A)-1):
c=2
a=A[j]
b=A[i]+A[j]
while(b in s):
c+=1
a,b= b, a+b
maxi=max(maxi, c)
return maxi if maxi>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:
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:
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:
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++.)
|
class Solution:
def lenLongestFibSubseq(self, A: List[int]) -> int:
d = {}
for i, v in enumerate(A):
d[v] = i
l = len(A)
res = -1
for i in range(0, l):
for j in range(i+1, l):
one_behind = A[j]
two_behind = A[i]
total = one_behind + two_behind
c = 0
while total in d:
#print(total, one_behind, two_behind)
if c == 0:
c += 3
else:
c += 1
one_behind, two_behind = total, one_behind
total = one_behind + two_behind
res = max(res, c)
#break
#print(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++.)
|
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):
if (m := n - A[j]) not in indices:
continue
if (i := indices[m]) >= j:
break
d[j, k] = d[i, j] + 1
ans = max(ans, d[j, k])
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:
s = set(A)
ans = 0
for i in range(len(A)):
for j in range(i+1,len(A)):
x,y = A[i],A[j]
l = 2
while x+y in s:
x,y = y,x+y
l += 1
ans = max(ans,l)
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:
d = {}
for i, v in enumerate(A):
d[v] = i
l = len(A)
res = -1
for i in range(0, l):
for j in range(i+1, l):
one_behind = A[j]
two_behind = A[i]
total = one_behind + two_behind
c = 0
while total in d:
#print(total, one_behind, two_behind)
if c == 0:
c += 3
else:
c += 1
tmp = one_behind
one_behind = total
two_behind = tmp
total = one_behind + two_behind
res = max(res, c)
#break
#print(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++.)
|
class Solution:
def lenLongestFibSubseq(self, A: List[int]) -> int:
N = len(A)
st = set(A)
res = 2
for i in range(N):
for j in range(i+1, N):
a, b, l = A[i], A[j], 2
while a + b in st:
a, b, l = b, a+b, l+1
res = max(res, l)
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)
two_sum = {}
result = 0
for i in range(n - 1):
for j in range(i + 1, n):
curr_pair = (A[i], A[j])
prev_pair = (A[j] - A[i], A[i])
two_sum[curr_pair] = 2
if prev_pair in two_sum:
two_sum[curr_pair] = max(two_sum[curr_pair], two_sum[prev_pair] + 1)
result = max(result, two_sum[curr_pair])
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 collections import Counter
class Solution:
def lenLongestFibSubseq(self, A: List[int]) -> int:
res=0
d={x:i for i,x in enumerate(A)}
n=len(A)
dp=[[2 for j in range(n)] for i in range(n)]
for i in range(2,n):
for j in range(i):
z=A[i]
x=A[j]
y=z-x
k=d.get(y,-1)
#x<y<z
if x<y and k!=-1:
dp[i][k]=1+dp[k][j]
res=max(res,dp[i][k])
if res>2:
return res
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 bisect import bisect_left
def BinarySearch(a, x):
i = bisect_left(a, x)
if i != len(a) and a[i] == x:
return i
else:
return 0
class Solution:
def lenLongestFibSubseq(self, A: List[int]) -> int:
def get_length(x,y):
s=0
while x+y in A:
s+=1
x,y = y, x+y
return s
if len(A) < 3:
return 0
ans = 0
sum_lim = A[-1]
for fi in range(len(A)-2):
curr_ans = 0
temp_fi = fi
si = temp_fi+1
# print(\"======={}====================\".format(fi))
while si< len(A):
reqd_sum = A[temp_fi]+A[si]
if reqd_sum > sum_lim:
break
pres = BinarySearch(A, reqd_sum)
if pres:
# temp_fi = si
# si = pres
curr_ans = 1+ get_length(A[si], A[pres])
ans = max(ans, curr_ans)
si += 1
return ans+2 if ans > 0 else 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:
s, memo = set(A), {}
for i in range(len(A)):
for j in range(i):
if (A[i] - A[j]) in s and (A[i] - A[j]) < A[j]:
memo[A[j], A[i]] = memo.get( (A[i] - A[j], A[j]), 2) + 1
return max(memo.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:
# only need the first 2 elements to define a fib seq
setA = set(A)
rint = 0
def getLen(i, j):
a, b = A[i], A[j]
if a + b not in setA:
return 0
length = 2
while a + b in setA:
length += 1
a, b = b, a + b
return length
for i in range(len(A) - 2):
for j in range(i + 1, len(A) - 1):
if A[j] - A[i] in setA and A[j] - A[i] < A[i]:
continue
rint = max(rint, getLen(i, j))
# print(i, j, getLen(i, j))
return rint
|
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:
longest = 0
coll = set(A)
n = len(A)
lookup = [[0 for _ in range(n)] for _ in range(n)]
for i in range(n - 2):
for j in range(i + 1, n - 1):
if n - j + 1 < longest - lookup[i][j]:
break
target = A[i] + A[j]
prev = A[j]
while target in coll:
if lookup[i][j]:
lookup[i][j] += 1
else:
lookup[i][j] = 3
temp = target
target += prev
prev = temp
longest = max(longest, lookup[i][j])
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, nums: List[int]) -> int:
n=len(nums)
dp=[[0 for x in range(n)] for y in range(n)]
max_=0
for i in range(2,n):
l=0
r=i-1
min_possible=nums[0]+nums[1]
max_possible=nums[r]+nums[r-1]
if nums[i]<min_possible or nums[i]>max_possible:
continue
while l<=r:
s=nums[l]+nums[r]
if s>nums[i]:
r-=1
elif s<nums[i]:
l+=1
else:
dp[r][i]=dp[l][r]+1
max_=max(max_,dp[r][i])
l+=1
r-=1
return max_ if max_==0 else max_+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:
if len(A) <= 2: return 0
dp = {(A[1], A[0]): 2}
maxx = 0
for i in range(2, len(A)):
j = i-1
while j >= 0:
v = A[i] - A[j]
if v < A[j] and (A[j], v) in dp:
maxx = max(maxx, dp[(A[j], v)]+1)
dp[(A[i], A[j])] = dp[(A[j], v)]+1
else:
dp[(A[i], A[j])] = 2
j -= 1
#print(dp)
return maxx
|
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, ordered_nums: List[int]) -> int:
def dfs(prev, curr):
if (prev, curr) in explored:
return -1
elif prev + curr not in unordered_nums:
return 2
else:
explored.add((prev, curr))
return 1 + dfs(curr, prev + curr)
unordered_nums = {n for n in ordered_nums}
explored = set()
path_len = 0
for i in range(len(unordered_nums)):
for j in range(i + 1, len(unordered_nums)):
path_len = max(path_len, dfs(ordered_nums[i], ordered_nums[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++.)
|
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 = [[0 for _ in range(len(nums))] for _ in range(len(nums))]
path_len = 0
for i in range(len(mat)):
for j in range(i + 1, len(mat)):
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:
S = set(A)
res = 0
for i in range(len(A)):
for j in range(i+1,len(A)):
f1 ,f2= A[i],A[j]
length = 0
if(f1+f2 in S):
length = 2
while(f1+f2 in S):
length+=1
tmp = f1+f2
f1 = f2
f2 = tmp
res = max(length,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++.)
|
class Solution:
def lenLongestFibSubseq(self, A: List[int]) -> int:
S = set(A)
ans = 2
for i in range(len(A)):
for j in range(i + 1, len(A)):
a = A[i]
b = A[j]
l = 2
while((a+b) in S):
temp = a
a = b
b = temp + b
l += 1
ans = max(ans, l)
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:
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:
A.sort()
S = (set(A))
maxlen = 0
for i in range(len(A)):
for j in range(i+1,len(A)):
a,b = A[j],A[i]+A[j]
count = 2
while b in S:
count+=1
a,b = b,a+b
maxlen = max(maxlen,count)
if maxlen >= 3:
return maxlen
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 = dict()
for i in A:
dp[(i,i)] = 0
n = len(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]>=0 and (A[j]-A[i],A[i]) in dp:
val = dp[(A[j]-A[i],A[i])] + 1
else:
val = 2
dp[(A[i],A[j])] = val
a = max(list(dp.values()))
if a>2:
return a
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:
longest = 0
coll = set(A)
n = len(A)
target = [[None for _ in range(n)] for _ in range(n)]
lookup = [[None for _ in range(n)] for _ in range(n)]
for i in range(n - 2):
for j in range(i + 1, n - 1):
target[i][j] = A[i] + A[j]
prev = A[j]
while target[i][j] in coll:
if lookup[i][j]:
lookup[i][j] += 1
else:
lookup[i][j] = 3
temp = target[i][j]
target[i][j] += prev
prev = temp
longest = max(longest, lookup[i][j])
return longest
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.