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Formatted question description: https://leetcode.ca/all/873.html

873. Length of Longest Fibonacci Subsequence

Level

Medium

Description

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

Use a set to store all the numbers in array A. For each indices pair (i, j) where i < j, try to generate a fibonacci-like subsequence of A, where all the numbers from the third number to the end are determined by the sum of the two previous numbers. A fibonacci-like subsequence ends until a number is not in the set. If the subsequence has length greater than or equal to 3, update the longest length. Finally, return the longest length.

  • class Solution {
        public int lenLongestFibSubseq(int[] A) {
            int length = A.length;
            Set<Integer> set = new HashSet<Integer>();
            for (int i = 0; i < length; i++)
                set.add(A[i]);
            int longestLength = 0;
            int end1 = length - 2, end2 = length - 1;
            for (int i = 0; i < end1; i++) {
                for (int j = i + 1; j < end2; j++) {
                    int num1 = A[i];
                    int num2 = A[j];
                    int curLength = 2;
                    while (set.contains(num1 + num2)) {
                        int num3 = num1 + num2;
                        curLength++;
                        num1 = num2;
                        num2 = num3;
                    }
                    if (curLength >= 3)
                        longestLength = Math.max(longestLength, curLength);
                }
            }
            return longestLength;
        }
    }
    
  • 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(object):
        def lenLongestFibSubseq(self, A):
            """
            :type A: List[int]
            :rtype: int
            """
            s = set(A)
            n = len(A)
            res = 0
            for i in range(n - 1):
                for j in range(i + 1, n):
                    a, b = A[i], A[j]
                    count = 2
                    while a + b in s:
                        a, b = b, a + b
                        count += 1
                        res = max(res, count)
            return res if res > 2 else 0
    
  • 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
    }
    
    func max(a, b int) int {
    	if a > b {
    		return a
    	}
    	return b
    }
    
  • /**
     * @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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