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873. Length of Longest Fibonacci Subsequence

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

Dynamic programming.

  • 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;
        }
    };
    
  • 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
    
    
  • 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;
    };
    
    
  • function lenLongestFibSubseq(arr: number[]): number {
        const n = arr.length;
        const f: number[][] = Array.from({ length: n }, () => Array(n).fill(0));
        const d: Map<number, number> = new Map();
        for (let i = 0; i < n; ++i) {
            d.set(arr[i], i);
            for (let j = 0; j < i; ++j) {
                f[i][j] = 2;
            }
        }
        let ans = 0;
        for (let i = 2; i < n; ++i) {
            for (let j = 1; j < i; ++j) {
                const t = arr[i] - arr[j];
                const k = d.get(t);
                if (k !== undefined && k < j) {
                    f[i][j] = Math.max(f[i][j], f[j][k] + 1);
                    ans = Math.max(ans, f[i][j]);
                }
            }
        }
        return ans;
    }
    
    
  • use std::collections::HashMap;
    impl Solution {
        pub fn len_longest_fib_subseq(arr: Vec<i32>) -> i32 {
            let n = arr.len();
            let mut f = vec![vec![0; n]; n];
            let mut d = HashMap::new();
            for i in 0..n {
                d.insert(arr[i], i);
                for j in 0..i {
                    f[i][j] = 2;
                }
            }
            let mut ans = 0;
            for i in 2..n {
                for j in 1..i {
                    let t = arr[i] - arr[j];
                    if let Some(&k) = d.get(&t) {
                        if k < j {
                            f[i][j] = f[i][j].max(f[j][k] + 1);
                            ans = ans.max(f[i][j]);
                        }
                    }
                }
            }
            ans
        }
    }
    
    

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