Create length-of-longest-fibonacci-subsequence.py

Author: kamyu104

Python/length-of-longest-fibonacci-subsequence.py

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+# Time:  O(n^2)
+# Space: O(n)
+
+# 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):
+        """
+        :type A: List[int]
+        :rtype: int
+        """
+        lookup = set(A)
+        result = 2
+        for i in xrange(len(A)):
+            for j in xrange(i+1, len(A)):
+                x, y, l = A[i], A[j], 2
+                while x+y in lookup:
+                    x, y, l = y, x+y, l+1
+                result = max(result, l)
+        return result if result > 2 else 0