Longest Prefix Subsequence matching Fibonacci Sequence

Given an array arr of size N. Also, there is another array B of the infinite size where B[0] = B[1] = 1, and for all (i >= 2), B[i] = B[i-1]+B[i-2], the task is to find the length of the longest subsequence which is the prefix of array B. If there is no subsequence which is the prefix of array B then return 0.

Examples:

Input: N = 6, arr = {1, 2, 3, 1, 2, 3}
Output: 4
Explanation: Subsequence {1,1,2,3} which is a prefix array of B of length 4.

Input: N = 5, arr = {2, 3, 1, 2, 5}
Output: 1
Explanation: Subsequence {1} which is a prefix array of B of length 1.

Approach: To solve the problem follow the below idea:

Using Dynamic Programming, we can iterate over both the arrays A and B, and see the total common length until A length is finished.

Below are the steps involved:

• Create an array dp as B of length A where each element is as dp[i] = dp[i - 1] + dp[i-2].
• Iterate over both the arrays:
  • If(A[i] == B[j])
    • count++, increase the lcs.
  • Otherwise, increase i++, As the prefix of B should be matched.

Below is the implementation of the code:

#include <bits/stdc++.h>
#include <iostream>
using namespace std;

// Maximum Length of longest Common
// subsequence
int solve(int N, int A[])
{
    // Create a array dp as B
    vector<int> dp(N + 1);
    dp[0] = 1;
    dp[1] = 1;
    for (int i = 2; i <= N; i++) {
        dp[i] = dp[i - 1] + dp[i - 2];
    }
    int count = 0, j = 0;
    for (int i = 0; i < N; i++) {

        // If elements are matched
        if (A[i] == dp[j]) {
            j++;
            count++;
        }
        if (j == dp.size()) {
            dp[i] = dp[i - 1] + dp[i - 2];
        }
    }
    // Return the lcs length
    return count;
}

// Driver code
int main()
{

    int N = 6;
    int A[] = { 1, 2, 3, 1, 2, 3 };

    // Function call
    cout << solve(N, A);
    return 0;
}

import java.util.Arrays;

public class Main {

    // Function to find the maximum length of the longest common subsequence
    static int solve(int N, int[] A) {
        // Create an array dp as B
        int[] dp = new int[N + 1];
        dp[0] = 1;
        dp[1] = 1;

        for (int i = 2; i <= N; i++) {
            dp[i] = dp[i - 1] + dp[i - 2];
        }

        int count = 0, j = 0;
        for (int i = 0; i < N; i++) {
            // If elements are matched
            if (A[i] == dp[j]) {
                j++;
                count++;
            }
            if (j == dp.length) {
                dp[i] = dp[i - 1] + dp[i - 2];
            }
        }
        // Return the LCS length
        return count;
    }

    // Driver code
    public static void main(String[] args) {
        int N = 6;
        int[] A = { 1, 2, 3, 1, 2, 3 };

        // Function call
        System.out.println(solve(N, A));
    }
}

# Maximum Length of longest Common subsequence
def solve(N, A):
    # Create an array dp as B
    dp = [0] * (N + 1)
    dp[0] = 1
    dp[1] = 1
    for i in range(2, N + 1):
        dp[i] = dp[i - 1] + dp[i - 2]

    count = 0
    j = 0
    for i in range(N):
        # If elements are matched
        if A[i] == dp[j]:
            j += 1
            count += 1
        if j == len(dp):
            dp.append(dp[-1] + dp[-2])

    # Return the LCS length
    return count

# Driver code
if __name__ == "__main__":
    N = 6
    A = [1, 2, 3, 1, 2, 3]

    # Function call
    print(solve(N, A))

// C# code for the above approach

using System;

public class GFG {

    // Function to find the maximum length of the longest
    // common subsequence
    static int solve(int N, int[] A)
    {
        // Create an array dp as B
        int[] dp = new int[N + 1];
        dp[0] = 1;
        dp[1] = 1;

        for (int i = 2; i <= N; i++) {
            dp[i] = dp[i - 1] + dp[i - 2];
        }

        int count = 0, j = 0;
        for (int i = 0; i < N; i++) {
            // If elements are matched
            if (A[i] == dp[j]) {
                j++;
                count++;
            }
            if (j == dp.Length) {
                dp[i] = dp[i - 1] + dp[i - 2];
            }
        }
        // Return the LCS length
        return count;
    }

    // Driver code
    public static void Main()
    {
        int N = 6;
        int[] A = { 1, 2, 3, 1, 2, 3 };

        // Function call
        Console.WriteLine(solve(N, A));
    }
}

// This code is contributed by ragul21

// Javascript code for the above approach
// Maximum Length of longest Common
// subsequence

function solve(N, A) {
    // Create a array dp
    let dp = new Array(N + 1).fill(0);
    dp[0] = 1;
    dp[1] = 1;
    
    // finding the fibonacci series
    // by iteration method
    for (let i = 2; i <= N; i++) {
        dp[i] = dp[i - 1] + dp[i - 2];
    }
    
    let count = 0, j = 0;
    
    for (let i = 0; i < N; i++) {

        // If elements are matched
        if (A[i] == dp[j]) {
            j++;
            count++;
        }
        if (j == dp.length) {
            dp[i] = dp[i - 1] + dp[i - 2];
        }
    }
    // Return the lcs length
    return count;
}

// Driver code
let N = 6;
let A = [1, 2, 3, 1, 2, 3];

// Function call
console.log(solve(N, A));

// This code is contributed by ragul21

4

Time Complexity: O (N + M)
Auxiliary Space: O(N)