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Length of Longest Fibonacci Subsequence in C++
Suppose we have 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
Now suppose a strictly increasing array A of positive integers forming a sequence, we have to find the length of the longest fibonacci-like subsequence of A. If one does not exist, then return 0. So if the number is like [1,2,3,4,5,6,7,8], then the output will be 5. The longest subsequence that is Fibonacci is like [1,2,3,5,8].
To solve this, we will follow these steps −
ret := 0
create a map m, n := size of array A
create a matrix called dp of size n x n
for i in range 0 to n – 1
m[A[i]] := i
for j in range i – 1, down to 0
req := A[i] – A[j]
when A[i] – A[j] < A[j] and m has (A[i] – A[j]), then
dp[i, j] := max of dp[i, j], dp[j, m[A[i] – A[j]]] + 1
otherwise dp[i,j] := max of dp[i, j] and 2
ret := max of ret and dp[i,j]
return ret when ret >= 3 otherwise return 0.
Let us see the following implementation to get better understanding −
Example
#include <bits/stdc++.h>
using namespace std;
class Solution {
public:
int lenLongestFibSubseq(vector<int> & A) {
int ret = 0;
unordered_map <int, int> m;
int n = A.size();
vector < vector <int> > dp(n, vector <int>(n));
for(int i = 0; i < n; i++){
m[A[i]] = i;
for(int j = i - 1; j >= 0; j--){
int req = A[i] - A[j];
if(A[i] - A[j] < A[j] && m.count(A[i] - A[j])){
dp[i][j] = max(dp[i][j], dp[j][m[A[i] - A[j]]] + 1);
}else{
dp[i][j] = max(dp[i][j], 2);
}
ret = max(ret, dp[i][j]);
}
}
return ret >= 3 ? ret : 0;
}
};
main(){
vector<int> v = {1,2,3,4,5,6,7,8};
Solution ob;
cout << (ob.lenLongestFibSubseq(v));
}Input
[1,2,3,4,5,6,7,8]
Output
5
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