花花酱 LeetCode 873. Length of Longest Fibonacci Subsequence

zxi — Sun, 22 Jul 2018 10:19:58 +0000

Problem

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 A of positive integers forming a sequence, find the length of the longest fibonacci-like subsequence of A. If one does not exist, return 0.

(Recall that a subsequence is derived from another sequence A by deleting any number of elements (including none) from A, 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: [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: [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].

Note:

3 <= A.length <= 1000
1 <= A[0] < A[1] < ... < A[A.length - 1] <= 10^9
(The time limit has been reduced by 50% for submissions in Java, C, and C++.)

Solution 1: DP

C++

// Author: Huahua
// Running time: 68 ms
//
// w/ Hashtable
// Time complexity: O(n^2)
// Space complexity: O(n^2)
class Solution {
public:
  int lenLongestFibSubseq(vector& A) {
    const int n = A.size();        
    unordered_map m;
    for (int i = 0; i < n; ++i)
      m[A[i]] = i;
    vector> dp(n, vector(n, 2));
    int ans = 0;
    for (int j = 0; j < n; ++j)
      for (int k = j + 1; k < n; ++k) {
        int a_i = A[k] - A[j];
        if (a_i >= A[j]) break; // pruning 168 ms -> 68 ms
        auto it = m.find(a_i);
        if (it == end(m)) continue;
        int i = it->second;
        dp[j][k] = dp[i][j] + 1;        
        ans = max(ans, dp[j][k]);
      }
    return ans;
  }
};

C++ V2

// Author: Huahua
// Running time: 96 ms
// w / Binary Search
// Time complexity: O(n^2logn)
// Space complexity: O(n)
class Solution {
public:
  int lenLongestFibSubseq(vector& A) {
    const int n = A.size();    
    vector> dp(n, vector(n, 2));
    int ans = 0;
    for (int j = 0; j < n; ++j)
      for (int k = j + 1; k < n; ++k) {
        int a_i = A[k] - A[j];
        if (a_i >= A[j]) break;
        auto it = lower_bound(begin(A), begin(A) + j, a_i);
        int i = it - begin(A);
        if (A[i] != a_i) continue;
        dp[j][k] = dp[i][j] + 1;  
        ans = max(ans, dp[j][k]);
      }
    return ans;          
  }
};

Solution 2: HashTable

Time complexity: O(n^3)

Space complexity: O(n)

C++

// Author: Huahua
// Running time: 172 ms
class Solution {
public:
  int lenLongestFibSubseq(vector& A) {
    const int n = A.size();    
    unordered_set m(begin(A), end(A));    
    int ans = 0;
    for (int i = 0; i < n; ++i)
      for (int j = i + 1; j < n; ++j) {
        int a = A[i];
        int b = A[j];
        int c = a + b;
        int l = 2;
        while (m.count(c)) {
          a = b;
          b = c;
          c = a + b;
          ans = max(ans, ++l);
        }
      }
    return ans;
  }
};

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Fibonacci Archives - Huahua's Tech Road

花花酱 LeetCode 873. Length of Longest Fibonacci Subsequence

Problem

Solution 1: DP

C++

C++ V2

Solution 2: HashTable

C++