# Maximum length Subsequence with alternating sign and maximum Sum

• Difficulty Level : Easy
• Last Updated : 28 Feb, 2022

Given an array arr[] of size n having both positive and negative integer excluding zero. The task is to find the subsequence with an alternating sign having maximum size and maximum sum 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[] = {2, 3, 7, -6, -4}
Output: 7 -4
Explanation:
Possible subsequences are [2, -6] [2, -4] [3, -6] [3, -4] [7, -6] [7, -4].
Out of these [7, -4] has the maximum sum.

Input: arr[] = {-4, 9, 4, 11, -5, -17, 9, -3, -5, 2}
Output: -4 11 -5 9 -3 2  ```

Approach:
The main idea to solve the above problem is to find the maximum element from the segments of the array which consists of the same sign which means we have to pick the maximum element among continuous positive and continuous negative elements. As we want maximum size we will only take one element from each segment and also to maximize the sum, we need to take the maximum element of each segment.
Below is the implementation of the above approach:

## CPP

 `// C++ implementation to find the``// subsequence with alternating sign``// having maximum size and maximum sum.` `#include ``using` `namespace` `std;` `// Function to find the subsequence``// with alternating sign having``// maximum size and maximum sum.``void` `findSubsequence(``int` `arr[], ``int` `n)``{``    ``int` `sign[n] = { 0 };` `    ``// Find whether each element``    ``// is positive or negative``    ``for` `(``int` `i = 0; i < n; i++) {``        ``if` `(arr[i] > 0)``            ``sign[i] = 1;``        ``else``            ``sign[i] = -1;``    ``}` `    ``int` `k = 0;``    ``int` `result[n] = { 0 };` `    ``// Find the required subsequence``    ``for` `(``int` `i = 0; i < n; i++) {` `        ``int` `cur = arr[i];``        ``int` `j = i;` `        ``while` `(j < n && sign[i] == sign[j]) {` `            ``// Find the maximum element``            ``// in the specified range``            ``cur = max(cur, arr[j]);``            ``++j;``        ``}` `        ``result[k++] = cur;` `        ``i = j - 1;``    ``}` `    ``// print the result``    ``for` `(``int` `i = 0; i < k; i++)``        ``cout << result[i] << ``" "``;``    ``cout << ``"\n"``;``}` `// Driver code``int` `main()``{``    ``// array declaration``    ``int` `arr[] = { -4, 9, 4, 11, -5, -17, 9, -3, -5, 2 };` `    ``// size of array``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr);` `    ``findSubsequence(arr, n);` `    ``return` `0;``}`

## Java

 `// Java implementation to find the``// subsequence with alternating sign``// having maximum size and maximum sum.``class` `GFG{`` ` `// Function to find the subsequence``// with alternating sign having``// maximum size and maximum sum.``static` `void` `findSubsequence(``int` `arr[], ``int` `n)``{``    ``int` `sign[] = ``new` `int``[n];`` ` `    ``// Find whether each element``    ``// is positive or negative``    ``for` `(``int` `i = ``0``; i < n; i++) {``        ``if` `(arr[i] > ``0``)``            ``sign[i] = ``1``;``        ``else``            ``sign[i] = -``1``;``    ``}`` ` `    ``int` `k = ``0``;``    ``int` `result[] = ``new` `int``[n];`` ` `    ``// Find the required subsequence``    ``for` `(``int` `i = ``0``; i < n; i++) {`` ` `        ``int` `cur = arr[i];``        ``int` `j = i;`` ` `        ``while` `(j < n && sign[i] == sign[j]) {`` ` `            ``// Find the maximum element``            ``// in the specified range``            ``cur = Math.max(cur, arr[j]);``            ``++j;``        ``}`` ` `        ``result[k++] = cur;`` ` `        ``i = j - ``1``;``    ``}`` ` `    ``// print the result``    ``for` `(``int` `i = ``0``; i < k; i++)``        ``System.out.print(result[i]+ ``" "``);``    ``System.out.print(``"\n"``);``}`` ` `// Driver code``public` `static` `void` `main(String[] args)``{``    ``// array declaration``    ``int` `arr[] = { -``4``, ``9``, ``4``, ``11``, -``5``, -``17``, ``9``, -``3``, -``5``, ``2` `};`` ` `    ``// size of array``    ``int` `n = arr.length;`` ` `    ``findSubsequence(arr, n);``}``}` `// This code is contributed by Princi Singh`

## Python3

 `# Python3 implementation to find the``# subsequence with alternating sign``# having maximum size and maximum sum.` `# Function to find the subsequence``# with alternating sign having``# maximum size and maximum sum.``def` `findSubsequence(arr, n):``    ``sign ``=` `[``0``]``*``n` `    ``# Find whether each element``    ``# is positive or negative``    ``for` `i ``in` `range``(n):``        ``if` `(arr[i] > ``0``):``            ``sign[i] ``=` `1``        ``else``:``            ``sign[i] ``=` `-``1` `    ``k ``=` `0``    ``result ``=` `[``0``]``*``n` `    ``# Find the required subsequence``    ``i ``=` `0``    ``while` `i < n:` `        ``cur ``=` `arr[i]``        ``j ``=` `i` `        ``while` `(j < n ``and` `sign[i] ``=``=` `sign[j]):` `            ``# Find the maximum element``            ``# in the specified range``            ``cur ``=` `max``(cur, arr[j])``            ``j ``+``=` `1` `        ``result[k] ``=` `cur``        ``k ``+``=` `1` `        ``i ``=` `j ``-` `1``        ``i ``+``=` `1` `    ``# print the result``    ``for` `i ``in` `range``(k):``        ``print``(result[i],end``=``" "``)` `# Driver code``if` `__name__ ``=``=` `'__main__'``:``    ``# array declaration``    ``arr``=``[``-``4``, ``9``, ``4``, ``11``, ``-``5``, ``-``17``, ``9``, ``-``3``, ``-``5``, ``2``]` `    ``# size of array``    ``n ``=` `len``(arr)` `    ``findSubsequence(arr, n)` `# This code is contributed by mohit kumar 29`

## C#

 `// C# implementation to find the``// subsequence with alternating sign``// having maximum size and maximum sum.``using` `System;` `public` `class` `GFG{``  ` `// Function to find the subsequence``// with alternating sign having``// maximum size and maximum sum.``static` `void` `findSubsequence(``int` `[]arr, ``int` `n)``{``    ``int` `[]sign = ``new` `int``[n];``  ` `    ``// Find whether each element``    ``// is positive or negative``    ``for` `(``int` `i = 0; i < n; i++) {``        ``if` `(arr[i] > 0)``            ``sign[i] = 1;``        ``else``            ``sign[i] = -1;``    ``}``  ` `    ``int` `k = 0;``    ``int` `[]result = ``new` `int``[n];``  ` `    ``// Find the required subsequence``    ``for` `(``int` `i = 0; i < n; i++) {``  ` `        ``int` `cur = arr[i];``        ``int` `j = i;``  ` `        ``while` `(j < n && sign[i] == sign[j]) {``  ` `            ``// Find the maximum element``            ``// in the specified range``            ``cur = Math.Max(cur, arr[j]);``            ``++j;``        ``}``  ` `        ``result[k++] = cur;``  ` `        ``i = j - 1;``    ``}``  ` `    ``// print the result``    ``for` `(``int` `i = 0; i < k; i++)``        ``Console.Write(result[i]+ ``" "``);``    ``Console.Write(``"\n"``);``}``  ` `// Driver code``public` `static` `void` `Main(String[] args)``{``    ``// array declaration``    ``int` `[]arr = { -4, 9, 4, 11, -5, -17, 9, -3, -5, 2 };``  ` `    ``// size of array``    ``int` `n = arr.Length;``  ` `    ``findSubsequence(arr, n);``}``}``// This code contributed by Rajput-Ji`

## Javascript

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Output:

`-4 11 -5 9 -3 2`

Time Complexity :O(N)

Auxiliary Space: O(N)

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