Description#
A sequence x1, x2, ..., xn is Fibonacci-like if:
n >= 3xi + xi+1 == xi+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:
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
Solutions#
Solution 1#
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| class Solution:
def lenLongestFibSubseq(self, arr: List[int]) -> int:
mp = {v: i for i, v in enumerate(arr)}
n = len(arr)
dp = [[0] * n for _ in range(n)]
for i in range(n):
for j in range(i):
dp[j][i] = 2
ans = 0
for i in range(n):
for j in range(i):
d = arr[i] - arr[j]
if d in mp and (k := mp[d]) < j:
dp[j][i] = max(dp[j][i], dp[k][j] + 1)
ans = max(ans, dp[j][i])
return ans
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| class Solution {
public int lenLongestFibSubseq(int[] arr) {
int n = arr.length;
Map<Integer, Integer> mp = new HashMap<>();
for (int i = 0; i < n; ++i) {
mp.put(arr[i], i);
}
int[][] dp = new int[n][n];
for (int i = 0; i < n; ++i) {
for (int j = 0; j < i; ++j) {
dp[j][i] = 2;
}
}
int ans = 0;
for (int i = 0; i < n; ++i) {
for (int j = 0; j < i; ++j) {
int d = arr[i] - arr[j];
if (mp.containsKey(d)) {
int k = mp.get(d);
if (k < j) {
dp[j][i] = Math.max(dp[j][i], dp[k][j] + 1);
ans = Math.max(ans, dp[j][i]);
}
}
}
}
return ans;
}
}
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| class Solution {
public:
int lenLongestFibSubseq(vector<int>& arr) {
unordered_map<int, int> mp;
int n = arr.size();
for (int i = 0; i < n; ++i) mp[arr[i]] = i;
vector<vector<int>> dp(n, vector<int>(n));
for (int i = 0; i < n; ++i)
for (int j = 0; j < i; ++j)
dp[j][i] = 2;
int ans = 0;
for (int i = 0; i < n; ++i) {
for (int j = 0; j < i; ++j) {
int d = arr[i] - arr[j];
if (mp.count(d)) {
int k = mp[d];
if (k < j) {
dp[j][i] = max(dp[j][i], dp[k][j] + 1);
ans = max(ans, dp[j][i]);
}
}
}
}
return ans;
}
};
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| func lenLongestFibSubseq(arr []int) int {
n := len(arr)
mp := make(map[int]int, n)
for i, v := range arr {
mp[v] = i + 1
}
dp := make([][]int, n)
for i := 0; i < n; i++ {
dp[i] = make([]int, n)
for j := 0; j < i; j++ {
dp[j][i] = 2
}
}
ans := 0
for i := 0; i < n; i++ {
for j := 0; j < i; j++ {
d := arr[i] - arr[j]
k := mp[d] - 1
if k >= 0 && k < j {
dp[j][i] = max(dp[j][i], dp[k][j]+1)
ans = max(ans, dp[j][i])
}
}
}
return ans
}
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| /**
* @param {number[]} arr
* @return {number}
*/
var lenLongestFibSubseq = function (arr) {
const mp = new Map();
const n = arr.length;
const dp = new Array(n).fill(0).map(() => new Array(n).fill(0));
for (let i = 0; i < n; ++i) {
mp.set(arr[i], i);
for (let j = 0; j < i; ++j) {
dp[j][i] = 2;
}
}
let ans = 0;
for (let i = 0; i < n; ++i) {
for (let j = 0; j < i; ++j) {
const d = arr[i] - arr[j];
if (mp.has(d)) {
const k = mp.get(d);
if (k < j) {
dp[j][i] = Math.max(dp[j][i], dp[k][j] + 1);
ans = Math.max(ans, dp[j][i]);
}
}
}
}
return ans;
};
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