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 of length 5. There can be many more subsequence 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 subsequence of this length.
Approach:
This problem can be solved using Dynamic Programming. It is a variation Longest Increasing Subsequence(LIS). Following are the steps:
- 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 current element needs to be positive and if pos = false then current element needs to be negative.
- If we include the current element, change the value of pos and recurr for the next element because we need the next element of opposite sign than the previous included element.
- 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];
- 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.
- 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; // Recurr 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; // Recurr for the next // iteration inc = 1 + solve(arr, n, i + 1, pos); } // If current element is // excluded, reccur 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)); } |
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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; // Recurr 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; // Recurr for the next // iteration inc = 1 + solve(arr, n, i + 1, pos); } // If current element is // excluded, reccur 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 |
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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; # Recurr 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; # Recurr for the next # iteration inc = 1 + solve(arr, n, i + 1, pos); # If current element is # excluded, reccur 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 |
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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; // Recurr 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; // Recurr for the next // iteration inc = 1 + solve(arr, n, i + 1, pos); } // If current element is // excluded, reccur 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 |
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Output:
5
Time Complexity: O(N) where N is the length of array.
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Improved By : AnkitRai01
