Suppose we have one sequence like 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
Now suppose a strictly increasing array A forming a sequence, we have to find the length of the longest fibonacci-like subsequence of A. If there is no such sequence, then return 0.
So, if the input is like A = [1,2,3,4,5,6,7,8], then the output will be 5 because there is a sequence [1,2,3,5,8] of length 5.
To solve this, we will follow these steps −
sA := a new set from elements of A
last := last element of A
B := a map containing each element present in A and their frequencies
best := 0
for i in size of A down to 0, do
a := A[i]
for each b in subarray of A[from index i+1 to end], do
c := a+b
if c is present in sA, then
B[a,b] := 1 + B[b,c]
best := maximum of best and B[a,b]+2
otherwise when c > last, then
come out from loop
return best
Let us see the following implementation to get better understanding −
from collections import Counter def solve(A): sA = set(A) last = A[-1] B = Counter() best = 0 for i in reversed(range(len(A))): a = A[i] for b in A[i+1:]: c = a+b if c in sA: B[a,b] = 1 + B[b,c] best = max(best , B[a,b]+2) elif c>last: break return best A = [1,2,3,4,5,6,7,8] print(solve(A))
[1,2,3,4,5,6,7,8]
5