LeetCode in Kotlin
873. Length of Longest Fibonacci Subsequence
Medium
A sequence x1, x2, ..., xn is Fibonacci-like if:
n >= 3xi + xi+1 == xi+2for 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 <= 10001 <= arr[i] < arr[i + 1] <= 109
Solution
class Solution {
fun lenLongestFibSubseq(arr: IntArray?): Int {
if (arr == null || arr.size < 3) {
return 0
}
val len = arr.size
val dp = Array(len) { IntArray(len) }
var ans = 0
for (i in 2 until len) {
var left = 0
var right = i - 1
while (left < right) {
if (arr[left] + arr[right] < arr[i]) {
left++
} else if (arr[left] + arr[right] > arr[i]) {
right--
} else {
// dp[right][i] = Math.max(dp[right][i], dp[left][right] + 1);
dp[right][i] = dp[left][right] + 1
ans = Math.max(ans, dp[right][i])
left++
right--
}
}
}
return if (ans > 0) ans + 2 else 0
}
}