Longest alternating subsequence in terms of positive and negative integers

Given an array arr[] of positive and negative numbers only. The task is to find the length of the longest alternating (means negative-positive-negative or positive-negative-positive) subsequence present in the array. 

Examples:

Input: arr[] = {-4, 3, -5, 9, 10, 12, 2, -1} 
Output: 5 
Explanation:
The longest sequence is {-4, 3, -5, 9, -1}, which is of length 5. There can be many more subsequences of this length.

Input: arr[] = {10, 12, 2, -1} 
Output: 2 
Explanation:
The longest sequence is {10, -1}, which is 2. There can be many more subsequences of this length. 

Approach:
This problem can be solved using Dynamic Programming. It is a variation Longest Increasing Subsequence(LIS). The following are the steps: 

1. For including and excluding an element in the given array arr[] for LAS(Longest Alternative Subsequence), a variable pos is used, when pos = true means the current element needs to be positive and if pos = false then current element needs to be negative.
2. If we include the current element, change the value of pos and recur for the next element because we need the next element of the opposite to the previously included element.
3. Now LAS[i][pos] can be recursively written as: 
   • Base case: If the index called recursively is greater than the last element, then return 0, as there is no such element left to form LAS and if LAS[i][pos] is calculated, then return the value. 

if(i == N) { 
return 0; 
} 
if(LAS[i][pos]) { 
return LAS[i][pos]; 
} 

• Recursive call: If the base case is not met, then recursively call when current element is included and excluded, then find the maximum of those to find LAS at that index. 
  

LAS[i][pos] = Longest Alternating Subsequence at index i by including or excluding that element for the value of pos, 
LAS[i][pos] = max(1 + recursive_function(i+1, pos), recursive_function(i+1, pos));  

• Return statement: At each recursive call(except the base case), return the value of LAS[i][pos]. 
  

return LAS[i][pos];  

1. The LAS for the given array arr[] is the maximum of LAS[0][0] and LAS[0][1].

Below is the implementation of the above approach: 

C++

// C++ program to find the
// length of longest alternate
// subsequence
#include <bits/stdc++.h>
using namespace std;

// LAS[i][pos] array to find
// the length of LAS till
// index i by including or
// excluding element arr[i]
// on the basis of value of pos
int LAS[1000][2] = { false };

int solve(int arr[], int n, int i, bool pos)
{
    // Base Case
    if (i == n)
        return 0;

    if (LAS[i][pos])
        return LAS[i][pos];

    int inc = 0, exc = 0;

    // If current element is
    // positive and pos is true
    // Include the current element
    // and change pos to false
    if (arr[i] > 0 && pos == true) {
        pos = false;

        // Recur for the next
        // iteration
        inc = 1 + solve(arr, n, i + 1, pos);
    }

    // If current element is
    // negative and pos is false
    // Include the current element
    // and change pos to true
    else if (arr[i] < 0 && pos == false) {
        pos = true;

        // Recur for the next
        // iteration
        inc = 1 + solve(arr, n, i + 1, pos);
    }

    // If current element is
    // excluded, recur for
    // next iteration
    exc = solve(arr, n, i + 1, pos);

    LAS[i][pos] = max(inc, exc);

    return LAS[i][pos];
}

// Driver's Code
int main()
{
    int arr[] = { -1, 2, 3, 4, 5,
                  -6, 8, -99 };
    int n = sizeof(arr) / sizeof(arr[0]);

    // Print LAS
    cout << max(solve(arr, n, 0, 0),
                solve(arr, n, 0, 1));
}

Java

// Java program to find the
// length of longest alternate
// subsequence
class GFG {

// LAS[i][pos] array to find
// the length of LAS till
// index i by including or
// excluding element arr[i]
// on the basis of value of pos
static int LAS[][] = new int[1000][2];

static int solve(int arr[], int n, int i,int pos)
{

    // Base Case
    if (i == n)
        return 0;

    if (LAS[i][pos]== 1)
        return LAS[i][pos];

    int inc = 0, exc = 0;

    // If current element is
    // positive and pos is 1
    // Include the current element
    // and change pos to 0
    if (arr[i] > 0 && pos == 1) {
        pos = 0;

        // Recur for the next
        // iteration
        inc = 1 + solve(arr, n, i + 1, pos);
    }

    // If current element is
    // negative and pos is o
    // Include the current element
    // and change pos to 1
    else if (arr[i] < 0 && pos == 0) {
        pos = 1;

        // Recur for the next
        // iteration
        inc = 1 + solve(arr, n, i + 1, pos);
    }

    // If current element is
    // excluded, recur for
    // next iteration
    exc = solve(arr, n, i + 1, pos);

    LAS[i][pos] = Math.max(inc, exc);

    return LAS[i][pos];
}

// Driver's Code
public static void main (String[] args)
{
    int arr[] = { -1, 2, 3, 4, 5, -6, 8, -99 };
    int n = arr.length;

    // Print LAS
    System.out.println(Math.max(solve(arr, n, 0, 0),solve(arr, n, 0, 1)));
}
}

// This code is contributed by AnkitRai01

Python3

# Python3 program to find the
# length of longest alternate
# subsequence
import numpy as np

# LAS[i][pos] array to find
# the length of LAS till
# index i by including or
# excluding element arr[i]
# on the basis of value of pos
LAS = np.zeros((1000, 2))

for i in range(1000) :
    for j in range(2) :
        LAS[i][j] = False

def solve(arr, n, i, pos) :

    # Base Case
    if (i == n) :
        return 0;

    if (LAS[i][pos]) :
        return LAS[i][pos];

    inc = 0; exc = 0;

    # If current element is
    # positive and pos is true
    # Include the current element
    # and change pos to false
    if (arr[i] > 0 and pos == True) :
        pos = False;

        # Recur for the next
        # iteration
        inc = 1 + solve(arr, n, i + 1, pos);

    # If current element is
    # negative and pos is false
    # Include the current element
    # and change pos to true
    elif (arr[i] < 0 and pos == False) :
        pos = True;

        # Recur for the next
        # iteration
        inc = 1 + solve(arr, n, i + 1, pos);

    # If current element is
    # excluded, recur for
    # next iteration
    exc = solve(arr, n, i + 1, pos);

    LAS[i][pos] = max(inc, exc);

    return LAS[i][pos];

# Driver's Code
if __name__ == "__main__" :

    arr = [ -1, 2, 3, 4, 5, -6, 8, -99 ];
    n = len(arr);

    # Print LAS
    print(max(solve(arr, n, 0, 0), solve(arr, n, 0, 1)));

# This code is contributed by AnkitRai01

C#

// C# program to find the
// length of longest alternate
// subsequence

using System;

public class GFG {

// LAS[i][pos] array to find
// the length of LAS till
// index i by including or
// excluding element arr[i]
// on the basis of value of pos
static int [,]LAS = new int[1000,2];

static int solve(int []arr, int n, int i,int pos)
{

    // Base Case
    if (i == n)
        return 0;

    if (LAS[i,pos]== 1)
        return LAS[i,pos];

    int inc = 0, exc = 0;

    // If current element is
    // positive and pos is 1
    // Include the current element
    // and change pos to 0
    if (arr[i] > 0 && pos == 1) {
        pos = 0;

        // Recur for the next
        // iteration
        inc = 1 + solve(arr, n, i + 1, pos);
    }

    // If current element is
    // negative and pos is o
    // Include the current element
    // and change pos to 1
    else if (arr[i] < 0 && pos == 0) {
        pos = 1;

        // Recur for the next
        // iteration
        inc = 1 + solve(arr, n, i + 1, pos);
    }

    // If current element is
    // excluded, recur for
    // next iteration
    exc = solve(arr, n, i + 1, pos);

    LAS[i,pos] = Math.Max(inc, exc);

    return LAS[i,pos];
}

// Driver's Code
public static void Main()
{
    int []arr = { -1, 2, 3, 4, 5, -6, 8, -99 };
    int n = arr.Length;

    // Print LAS
    Console.WriteLine(Math.Max(solve(arr, n, 0, 0),solve(arr, n, 0, 1)));
}
}

// This code is contributed by AnkitRai01

Javascript

<script>
// Javascript program to find the
// length of longest alternate
// subsequence


// LAS[i][pos] array to find
// the length of LAS till
// index i by including or
// excluding element arr[i]
// on the basis of value of pos
let LAS = new Array();

for(let i = 0; i < 1000; i++){
    let temp = new Array()
    for(let j = 0; j < 2; j++){
        temp.push([])
    }
    LAS.push(temp);
}

for(let i = 0; i < 1000; i++){
    for(let j = 0; j < 2; j++){
        LAS[i][j] = false
    }
}

function solve(arr, n, i, pos)
{
    // Base Case
    if (i == n)
        return 0;

    if (LAS[i][pos])
        return LAS[i][pos];

    let inc = 0, exc = 0;

    // If current element is
    // positive and pos is true
    // Include the current element
    // and change pos to false
    if (arr[i] > 0 && pos == true) {
        pos = false;

        // Recur for the next
        // iteration
        inc = 1 + solve(arr, n, i + 1, pos);
    }

    // If current element is
    // negative and pos is false
    // Include the current element
    // and change pos to true
    else if (arr[i] < 0 && pos == false) {
        pos = true;

        // Recur for the next
        // iteration
        inc = 1 + solve(arr, n, i + 1, pos);
    }

    // If current element is
    // excluded, recur for
    // next iteration
    exc = solve(arr, n, i + 1, pos);

    LAS[i][pos] = Math.max(inc, exc);

    return LAS[i][pos];
}

// Driver's Code

let arr = [ -1, 2, 3, 4, 5,
            -6, 8, -99 ];
let n = arr.length;

// Print LAS
document.write(Math.max(solve(arr, n, 0, 0), solve(arr, n, 0, 1)));

// This code is contributed by _saurabh_jaiswal
</script>

Output:

5

Time Complexity: O(N) where N is the length of the array.

Auxiliary Space: O(n) for call stack