873. Length of Longest Fibonacci Subsequence
Difficulty: Medium
Topics: Array
, Hash Table
, Dynamic Programming
A sequence x1, x2, ..., xn
is Fibonacci-like if:
n >= 3
-
xi + xi+1 == xi+2
for alli + 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:
- 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:
- 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] <= 109
Solution:
We need to find the length of the longest Fibonacci-like subsequence in a strictly increasing array. A Fibonacci-like sequence is defined such that each element after the first two is the sum of the two preceding ones.
Approach
-
Dynamic Programming (DP) with Hash Map:
- Hash Map: Use a hash map to store the indices of elements for quick lookup. This allows us to check if the difference between two elements (which would be the preceding element in a Fibonacci sequence) exists in the array.
-
DP Table: Use a 2D DP array where
dp[j][k]
represents the length of the longest Fibonacci-like subsequence ending with elements at indicesj
andk
. Initialize all elements to 2 since any two elements can start a potential sequence.
-
Iterate Over Pairs:
- For each pair of indices
(j, k)
wherej < k
, check if there exists an elementi
such thatarr[i] + arr[j] = arr[k]
. If such an element exists andi < j
, update the DP value fordp[j][k]
based on the DP value ofdp[i][j]
.
- For each pair of indices
-
Track Maximum Length:
- Keep track of the maximum length of the subsequence found during the iteration. If no valid subsequence of length 3 or more is found, return 0.
Let's implement this solution in PHP: 873. Length of Longest Fibonacci Subsequence
<?php
/**
* @param Integer[] $arr
* @return Integer
*/
function lenLongestFibSubseq($arr) {
...
...
...
/**
* go to ./solution.php
*/
}
// Example test cases
$arr1 = [1,2,3,4,5,6,7,8];
echo lenLongestFibSubseq($arr1) . "\n"; // Output: 5
$arr2 = [1,3,7,11,12,14,18];
echo lenLongestFibSubseq($arr2) . "\n"; // Output: 3
?>
Explanation:
- Hash Map Initialization: We first create a hash map to store the indices of the elements in the array. This allows O(1) time complexity for checking the existence of elements.
- DP Array Initialization: The DP array is initialized to 2 for all pairs since any two elements can form the start of a potential sequence.
-
Iterate Over Pairs: For each pair
(j, k)
, we compute the difference to check if it exists in the array. If it does and the index of the difference is less thanj
, we update the DP value for(j, k)
based on the DP value of(i, j)
. - Track Maximum Length: During the iteration, we keep track of the maximum length of any valid subsequence found. If the maximum length is at least 3, we return it; otherwise, we return 0.
This approach efficiently checks all possible pairs and uses dynamic programming to build up the solution, resulting in an O(n^2) time complexity and O(n^2) space complexity, which is feasible for the given problem constraints.
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