Length Of Longest Fibonacci Subsequence Problem
Description
LeetCode Problem 873.
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 arr of positive integers forming a sequence, return the length of the longest Fibonacci-like subsequence of arr. If one does not exist, return 0.
A subsequence is derived from another sequence arr by deleting any number of elements (including none) from arr, 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:
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3
Input: arr = [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:
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Input: arr = [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].
Constraints:
- 3 <= arr.length <= 1000
- 1 <= arr[i] < arr[i + 1] <= 10^9
Sample C++ Code
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class Solution {
public:
int lenLongestFibSubseq(vector<int>& a) {
int n = a.size();
// dp[i][j] stores the length of the fibonnaci sequence ending with i, j.
vector<vector<int>> dp(n, vector<int>(n, 0));
int maxi = 0;
for (int i = 2; i < n; i++) {
int l = 0, r = i - 1;
// check if we can find a[l] + a[r] = a[i]
while (l < r) {
int sum = a[l] + a[r];
if (sum > a[i])
r--;
else if (sum < a[i])
l++;
else {
// we found a fibonacci sequence ending with indices r and i.
// we update dp[r][i] to dp[l][r] + 1.
dp[r][i] = dp[l][r] + 1;
maxi = max(maxi, dp[r][i]);
l++;
r--;
}
}
}
return maxi == 0 ? 0 : maxi + 2;
}
};