# Find the Maximum Alternate Subsequence Sum from a given array

Given an array arr[] of size n having both positive and negative integer excluding zero. Find the maximum sum of maximum size alternate subsequence of a given sequence that is, in a subsequence sign of each adjacent element is opposite for example if the first one is positive then the second one has to be negative followed by another positive integer and so on.
Examples:

Input : arr[] = {1, 2, 3, 4, -1, -2}
Output :
Explanation:
Here, maximum size subsequences are [1, -1] [1, -2] [2, -1] [2, -2] [3, -1]
[3, -2] [4, -1] [4, -2] but the subsequence which have maximum sum is [4, -1]
and the sum of this subsequence is 3. Hence, the output is 3.

Input : arr[] = {-1, -6, 12, -4, -5}
Output :
Explanation:
Here, maximum size subsequences are [-1, 12, -4] [-1, 12, -5] [-6, -12, -4]
[-6, 12, -5] but the subsequence which have maximum sum is [-1, 12, -4]
and the sum of this subsequence is 7. Hence, the output is 7.

Naive approach:
To solve the problem mentioned above the simple approach is to find all the alternating subsequences and their sum and return the maximum sum among all of them.
Efficient Approach:
The efficient approach to solve the above problem is to pick the maximum element among continuous positive and continuous negative elements every time and add it to the maximum sum so far. The variable which stores the maximum sum will hold the final answer.

Below is the implementation of the above-mentioned approach:

## Java

 `// Java program to find the maximum ` `// alternating subsequence sum for  ` `// a given array  ` `import` `java.io.*; ` `import` `java.util.*;  ` ` `  `class` `GFG{ ` ` `  `// Function to find maximum  ` `// alternating subequence sum  ` `static` `int` `maxAlternatingSum(``int``[] arr, ``int` `n) ` `{ ` `     `  `    ``// Initialize sum to 0  ` `    ``int` `max_sum = ``0``; ` `     `  `    ``int` `i = ``0``; ` `     `  `    ``while` `(i < n) ` `    ``{  ` `        ``int` `current_max = arr[i];  ` `        ``int` `k = i; ` `         `  `         ``while` `(k < n &&  ` `              ``((arr[i] > ``0` `&& arr[k] > ``0``) ||  ` `               ``(arr[i] < ``0` `&& arr[k] < ``0``))) ` `         ``{  ` `            ``current_max = Math.max(current_max, arr[k]); ` `            ``k += ``1``; ` `        ``} ` `         `  `        ``// Calculate the sum  ` `        ``max_sum += current_max; ` `         `  `        ``i = k; ` `    ``}  ` `     `  `    ``// Return the final result  ` `    ``return` `max_sum; ` `}  ` ` `  `// Driver Code  ` `public` `static` `void` `main(String args[]) ` `{  ` `     `  `    ``// Array initialization  ` `    ``int``[] arr = { ``1``, ``2``, ``3``, ``4``, -``1``, -``2` `}; ` `     `  `    ``// Length of array  ` `    ``int` `n = arr.length;  ` `     `  `    ``System.out.println(maxAlternatingSum(arr, n));  ` `} ` `}  ` ` `  `// This code is contributed by offbeat `

## Python3

 `# Python3 program to find the maximum alternating  ` `# subsequence sum for a given array  ` ` `  `# Function to find maximum  ` `# alternating subequence sum  ` `def` `maxAlternatingSum(arr, n):  ` ` `  `    ``# initialize sum to 0  ` `    ``max_sum ``=` `0` `     `  `    ``i ``=` `0` `     `  `    ``while` `i``0` `and` `arr[k]>``0``)  ` `        ``or` `(arr[i]<``0` `and` `arr[k]<``0``)):  ` `             `  `            ``current_max ``=` `max``(current_max, arr[k])  ` `             `  `            ``k``+``=` `1` `         `  `        ``# calculate the sum  ` `        ``max_sum``+``=` `current_max  ` `         `  `        ``i ``=` `k  ` `         `  `    ``# return the final result  ` `    ``return` `max_sum  ` ` `  ` `  `# Driver code  ` ` `  `# array initiaisation  ` `arr ``=` `[``1``, ``2``, ``3``, ``4``, ``-``1``, ``-``2``]  ` ` `  `# length of array  ` `n ``=` `len``(arr)  ` ` `  `print``(maxAlternatingSum(arr, n))  `

Output:

```3
```

Time Complexity : O(n)

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Improved By : offbeat

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