Question

873. Length of Longest Fibonacci Subsequence

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Description

A sequence x1, x2, ..., xn is _Fibonacci-like_ if:

• n >= 3
• xi + 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 <= 1000
• 1 <= arr[i] < arr[i + 1] <= 109

Solutions

Solution 1

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

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;
  }
}

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;
  }
};

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
}

/**
 * @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;
};