Given a strictly increasing array A of positive integers where,
![1 \leq A[i] \leq 10^{18}](https://geeksforgeeks.org/wp-content/uploads/2023/10/neveropen_quicklatex.com-789c532a9bb5753436283ecb233cd00f_l3.png "Rendered by QuickLaTeX.com")
The task is to find the length of the longest Fibonacci-like subsequence of A. If such subsequence does not exist, return 0.
Examples:
Input: A = [1, 3, 7, 11, 12, 14, 18]
Output: 3
Explanation:
The longest subsequence that is Fibonacci-like: [1, 11, 12]. Other possible subsequences are [3, 11, 14] or [7, 11, 18].Input: A = [1, 2, 3, 4, 5, 6, 7, 8]
Output: 5
Explanation:
The longest subsequence that is Fibonacci-like: [1, 2, 3, 5, 8].
Naive Approach: A Fibonacci-like sequence is such that it has each two adjacent terms that determine the next expected term.
For example, with 1, 1, we expect that the sequence must continue 2, 3, 5, 8, 13, … and so on.
- Use Set or Map to determine quickly whether the next term of Fibonacci sequence is present in the array A or not. Because of the exponential growth of these terms, there will be not more than log(M) searches to get next element on each iteration.
- For each starting pair A[i], A[j], we maintain the next expected value y = A[i] + A[j] and the previously seen largest value x = A[j]. If y is in the array, then we can then update these values (x, y) -> (y, x+y) otherwise we stop immediately.
Below is the implementation of above approach:
C++
// CPP implementation of above approach#include <bits/stdc++.h>using namespace std;// Function to return the max Length of// Fibonacci subsequenceint LongestFibSubseq(int A[], int n){ // Store all array elements in a hash // table unordered_set<int> S(A, A + n); int maxLen = 0, x, y; for (int i = 0; i < n; ++i) { for (int j = i + 1; j < n; ++j) { x = A[j]; y = A[i] + A[j]; int length = 2; // check until next fib element is found while (S.find(y) != S.end()) { // next element of fib subseq int z = x + y; x = y; y = z; maxLen = max(maxLen, ++length); } } } return maxLen >= 3 ? maxLen : 0;}// Driver programint main(){ int A[] = { 1, 2, 3, 4, 5, 6, 7, 8 }; int n = sizeof(A) / sizeof(A[0]); cout << LongestFibSubseq(A, n); return 0;}// This code is written by Sanjit_Prasad |
Java
// Java implementation of above approach import java.util.*;public class GFG {// Function to return the max Length of // Fibonacci subsequence static int LongestFibSubseq(int A[], int n) { // Store all array elements in a hash // table TreeSet<Integer> S = new TreeSet<>(); for (int t : A) { // Add each element into the set S.add(t); } int maxLen = 0, x, y; for (int i = 0; i < n; ++i) { for (int j = i + 1; j < n; ++j) { x = A[j]; y = A[i] + A[j]; int length = 3; // check until next fib element is found while (S.contains(y) && (y != S.last())) { // next element of fib subseq int z = x + y; x = y; y = z; maxLen = Math.max(maxLen, ++length); } } } return maxLen >= 3 ? maxLen : 0; }// Driver program public static void main(String[] args) { int A[] = {1, 2, 3, 4, 5, 6, 7, 8}; int n = A.length; System.out.print(LongestFibSubseq(A, n)); }}// This code is contributed by 29AjayKumar |
Python3
# Python3 implementation of the # above approach # Function to return the max Length # of Fibonacci subsequence def LongestFibSubseq(A, n): # Store all array elements in # a hash table S = set(A) maxLen = 0 for i in range(0, n): for j in range(i + 1, n): x = A[j] y = A[i] + A[j] length = 2 # check until next fib # element is found while y in S: # next element of fib subseq z = x + y x = y y = z length += 1 maxLen = max(maxLen, length) return maxLen if maxLen >= 3 else 0# Driver Codeif __name__ == "__main__": A = [1, 2, 3, 4, 5, 6, 7, 8] n = len(A) print(LongestFibSubseq(A, n)) # This code is contributed # by Rituraj Jain |
C#
// C# implementation of above approachusing System;using System.Collections.Generic;class GFG { // Function to return the max Length of // Fibonacci subsequence static int LongestFibSubseq(int []A, int n) { // Store all array elements in a hash // table SortedSet<int> S = new SortedSet<int>(); foreach (int t in A) { // Add each element into the set S.Add(t); } int maxLen = 0, x, y; for (int i = 0; i < n; ++i) { for (int j = i + 1; j < n; ++j) { x = A[j]; y = A[i] + A[j]; int length = 3; // check until next fib element is found while (S.Contains(y) && y != last(S)) { // next element of fib subseq int z = x + y; x = y; y = z; maxLen = Math.Max(maxLen, ++length); } } } return maxLen >= 3 ? maxLen : 0; } static int last(SortedSet<int> S) { int ans = 0; foreach(int a in S) ans = a; return ans; } // Driver Code public static void Main(String[] args) { int []A = {1, 2, 3, 4, 5, 6, 7, 8}; int n = A.Length; Console.Write(LongestFibSubseq(A, n)); } } // This code is contributed by 29AjayKumar |
Javascript
<script>// Javascript implementation of above approach// Function to return the max Length of// Fibonacci subsequencefunction LongestFibSubseq(A, n){ // Store all array elements in a hash // table var S = new Set(A); var maxLen = 0, x, y; for (var i = 0; i < n; ++i) { for (var j = i + 1; j < n; ++j) { x = A[j]; y = A[i] + A[j]; var length = 2; // check until next fib element is found while (S.has(y)) { // next element of fib subseq var z = x + y; x = y; y = z; maxLen = Math.max(maxLen, ++length); } } } return maxLen >= 3 ? maxLen : 0;}// Driver programvar A = [1, 2, 3, 4, 5, 6, 7, 8];var n = A.length;document.write( LongestFibSubseq(A, n));// This code is contributed by famously.</script> |
Output
5
Complexity Analysis:
- Time Complexity: O(N2 * log(M)), where N is the length of array and M is max(A).
- Auxiliary Space: O(N)
Efficient Approach: To optimize the above approach the idea is to implement Dynamic Programming. Initialize a dp table, dp[a, b] that represents the length of Fibonacci sequence ends up with (a, b). Then update the table as dp[a, b] = (dp[b – a, a] + 1 ) or 2
Below is the implementation of the above approach:
C++
// CPP program for the above approach#include <bits/stdc++.h>using namespace std;// Function to return the max Length of// Fibonacci subsequenceint LongestFibSubseq(int A[], int n){ // Initialize the unordered map unordered_map<int, int> m; int N = n, res = 0; // Initialize dp table int dp[N][N]; // Iterate till N for (int j = 0; j < N; ++j) { m[A[j]] = j; for (int i = 0; i < j; ++i) { // Check if the current integer // forms a fibonacci sequence int k = m.find(A[j] - A[i]) == m.end() ? -1 : m[A[j] - A[i]]; // Update the dp table dp[i][j] = (A[j] - A[i] < A[i] && k >= 0) ? dp[k][i] + 1 : 2; res = max(res, dp[i][j]); } } // Return the answer return res > 2 ? res : 0;}// Driver programint main(){ int A[] = { 1, 3, 7, 11, 12, 14, 18 }; int n = sizeof(A) / sizeof(A[0]); cout << LongestFibSubseq(A, n); return 0;} |
Java
/*package whatever //do not write package name here */import java.io.*;import java.util.*;class GFG{ // Function to return the max Length of // Fibonacci subsequence static int LongestFibSubseq(int[] A, int n) { // Initialize the unordered map HashMap<Integer,Integer>m = new HashMap<>(); int N = n, res = 0; // Initialize dp table int[][] dp = new int[N][N]; // Iterate till N for (int j = 0; j < N; ++j) { m.put(A[j], j); for (int i = 0; i < j; ++i) { // Check if the current integer // forms a fibonacci sequence int k = m.containsKey(A[j] - A[i])? m.get(A[j] - A[i]):-1; // Update the dp table dp[i][j] = (A[j] - A[i] < A[i] && k >= 0) ? dp[k][i] + 1 : 2; res = Math.max(res, dp[i][j]); } } // Return the answer return res > 2 ? res : 0; } // Drivers code public static void main(String args[]){ int[] A = { 1, 3, 7, 11, 12, 14, 18 }; int n = A.length; System.out.println(LongestFibSubseq(A, n)); }}// This code is contributed by shinjanpatra |
Python3
# Python program for the above approach# Function to return the max Length of# Fibonacci subsequencedef LongestFibSubseq(A, n): # Initialize the unordered map m = {} N, res = n, 0 # Initialize dp table dp = [ [0 for i in range(N) ] for J in range(N) ] # Iterate till N for j in range(N): m[A[j]] = j for i in range(j): # Check if the current integer # forms a fibonacci sequence k = -1 if ((A[j] - A[i]) not in m) else m[A[j] - A[i]] # Update the dp table dp[i][j] = dp[k][i] + 1 if (A[j] - A[i] < A[i] and k >= 0) else 2 res = max(res, dp[i][j]) # Return the answer return res if res > 2 else 0# Driver programA = [ 1, 3, 7, 11, 12, 14, 18 ]n = len(A)print(LongestFibSubseq(A, n))# This code is contributed by shinjanpatra |
C#
// C# implementation of above approachusing System;using System.Collections.Generic;class HelloWorld { // Function to return the max Length of // Fibonacci subsequence static int LongestFibSubseq(int[] A, int n) { // Initialize the unordered map var m = new Dictionary<int, int>(); int N = n; int res = 0; // Initialize dp table int[, ] dp = new int[N, N]; // Iterate till N for (int j = 0; j < N; ++j) { m[A[j]] = j; for (int i = 0; i < j; ++i) { // Check if the current integer // forms a fibonacci sequence int k = m.ContainsKey(A[j] - A[i]) ? m[A[j] - A[i]] : -1; // Update the dp table dp[i, j] = (A[j] - A[i] < A[i] && k >= 0) ? dp[k, i] + 1 : 2; res = Math.Max(res, dp[i, j]); } } // Return the answer return res > 2 ? res : 0; } // Driver program static void Main() { int[] A = { 1, 3, 7, 11, 12, 14, 18 }; int n = A.Length; Console.WriteLine(LongestFibSubseq(A, n)); }}// The code is contributed by Gautam goel (gautamgoel962) |
Javascript
<script>// JavaScript program for the above approach// Function to return the max Length of// Fibonacci subsequencefunction LongestFibSubseq(A,n){ // Initialize the unordered map let m = new Map(); let N = n, res = 0; // Initialize dp table let dp = new Array(N); for(let i=0;i<N;i++){ dp[i] = new Array(N); } // Iterate till N for (let j = 0; j < N; ++j) { m.set(A[j],j); for (let i = 0; i < j; ++i) { // Check if the current integer // forms a fibonacci sequence let k = m.has(A[j] - A[i]) == false ? -1 : m.get(A[j] - A[i]); // Update the dp table dp[i][j] = (A[j] - A[i] < A[i] && k >= 0) ? dp[k][i] + 1 : 2; res = Math.max(res, dp[i][j]); } } // Return the answer return res > 2 ? res : 0;}// Driver programlet A = [ 1, 3, 7, 11, 12, 14, 18 ];let n = A.length;document.write(LongestFibSubseq(A, n));// code is contributed by shinjanpatra</script> |
Output
3
Complexity Analysis:
- Time Complexity: O(N2), where N is the length of the array.
- Auxiliary Space: O(N2)
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