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# Maximum sum possible by assigning alternate positive and negative sign to elements in a subsequence

Given an array arr[] consisting of N positive integers, the task is to find the maximum sum of subsequences from the given array such that elements in the subsequence are assigned positive and negative signs alternately.

Subsequence = {a, b, c, d, e, … },
Sum of the above subsequence = (a – b + c – d + e – …)

Examples:

Input: arr[] = {1, 2, 3, 4}
Output: 4
Explanation:
The subsequence having maximum sum is {4}.
The sum is 4.

Input: arr[]= {1, 2, 3, 4, 1, 2 }
Output: 5
Explanation:
The subsequence having maximum sum is {4, 1, 2}.
The sum = 4 -1 + 2 = 5.

Naive Approach: The simplest approach is to generate all the subsequences of the given array and then find the sum for every subsequence and print the maximum among all the sum of the subsequences.

Time Complexity: O(N*2N)
Auxiliary Space: O(1)

Efficient Approach: To optimize the above approach, the idea is to use Dynamic Programming. Initialize an auxiliary space dp[][] of size N*2 to store the Overlapping Subproblems. In each recursive call, add arr[i] or (-1)*arr[i] to the sum with the respective flag variable that denotes whether the current element is positive or negative. Below are the steps:

• Create a 2D dp[][] array of size N*2 and initialize the array with the  -1.
• Pass the variable flag that denotes the sign of the element have to pick in the next term. For Example, in the subsequence, {a, b, c}, then the maximum subsequence can be (a – b + c) or (b – c) or c. Instead of recurring for all the Overlapping Subproblems, again and again, store once in dp[][] array and use the recurring state.
• If the flag is 0 then the current element is to be considered as a positive element and if the flag is 1 then the current element is to be considered as a negative element.
• Store every result into the dp[][] array.
• Print the value of dp[N][flag] as the maximum sum after the above steps.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach``#include ``using` `namespace` `std;` `// Function to find the``// maximum sum subsequence``int` `findMax(vector<``int``>& a, ``int` `dp[],``            ``int` `i, ``int` `flag)``{``    ``// Base Case``    ``if` `(i == (``int``)a.size()) {``        ``return` `0;``    ``}` `    ``// If current state is already``    ``// calculated then use it``    ``if` `(dp[i][flag] != -1) {``        ``return` `dp[i][flag];``    ``}` `    ``int` `ans;` `    ``// If current element is positive``    ``if` `(flag == 0) {` `        ``// Update ans and recursively``        ``// call with update value of flag``        ``ans = max(findMax(a, dp, i + 1, 0),``                  ``a[i]``                      ``+ findMax(a, dp,``                                ``i + 1, 1));``    ``}` `    ``// Else current element is negative``    ``else` `{` `        ``// Update ans and recursively``        ``// call with update value of flag``        ``ans = max(findMax(a, dp, i + 1, 1),``                  ``-1 * a[i]``                      ``+ findMax(a, dp,``                                ``i + 1, 0));``    ``}` `    ``// Return maximum sum subsequence``    ``return` `dp[i][flag] = ans;``}` `// Function that finds the maximum``// sum of element of the subsequence``// with alternate +ve and -ve signs``void` `findMaxSumUtil(vector<``int``>& arr,``                    ``int` `N)``{``    ``// Create auxiliary array dp[][]``    ``int` `dp[N];` `    ``// Initialize dp[][]``    ``memset``(dp, -1, ``sizeof` `dp);` `    ``// Function Call``    ``cout << findMax(arr, dp, 0, 0);``}` `// Driver Code``int` `main()``{``    ``// Given array arr[]``    ``vector<``int``> arr = { 1, 2, 3, 4, 1, 2 };` `    ``int` `N = arr.size();` `    ``// Function Call``    ``findMaxSumUtil(arr, N);` `    ``return` `0;``}`

## Java

 `// Java program for the above approach``import` `java.io.*;``import` `java.util.Arrays;` `class` `GFG{`` ` `// Function to find the``// maximum sum subsequence``static` `int` `findMax(``int``[] a, ``int` `dp[][],``                   ``int` `i, ``int` `flag)``{``    ` `    ``// Base Case``    ``if` `(i == (``int``)a.length)``    ``{``        ``return` `0``;``    ``}`` ` `    ``// If current state is already``    ``// calculated then use it``    ``if` `(dp[i][flag] != -``1``)``    ``{``        ``return` `dp[i][flag];``    ``}`` ` `    ``int` `ans;`` ` `    ``// If current element is positive``    ``if` `(flag == ``0``)``    ``{``        ` `        ``// Update ans and recursively``        ``// call with update value of flag``        ``ans = Math.max(findMax(a, dp, i + ``1``, ``0``),``                ``a[i] + findMax(a, dp, i + ``1``, ``1``));``    ``}`` ` `    ``// Else current element is negative``    ``else``    ``{``        ` `        ``// Update ans and recursively``        ``// call with update value of flag``        ``ans = Math.max(findMax(a, dp, i + ``1``, ``1``),``           ``-``1` `* a[i] + findMax(a, dp, i + ``1``, ``0``));``    ``}`` ` `    ``// Return maximum sum subsequence``    ``return` `dp[i][flag] = ans;``}`` ` `// Function that finds the maximum``// sum of element of the subsequence``// with alternate +ve and -ve signs``static` `void` `findMaxSumUtil(``int``[] arr,``                           ``int` `N)``{``    ` `    ``// Create auxiliary array dp[][]``    ``int` `dp[][] = ``new` `int``[N][``2``];`` ` `    ``// Initialize dp[][]``    ``for``(``int` `i = ``0``; i < N; i++)``    ``{``        ``for``(``int` `j = ``0``; j < ``2``; j++)``        ``{``            ``dp[i][j] = -``1``;``        ``}``    ``}``    ` `    ``// Function Call``    ``System.out.println(findMax(arr, dp, ``0``, ``0``));``}`` ` `// Driver Code``public` `static` `void` `main (String[] args)``{``    ` `    ``// Given array arr[]``    ``int``[] arr = { ``1``, ``2``, ``3``, ``4``, ``1``, ``2` `};`` ` `    ``int` `N = arr.length;`` ` `    ``// Function call``    ``findMaxSumUtil(arr, N);``}``}` `// This code is contributed by sanjoy_62`

## Python3

 `# Python3 program for the above approach` `# Function to find the``# maximum sum subsequence``def` `findMax(a, dp, i, flag):``    ` `    ``# Base Case``    ``if` `(i ``=``=` `len``(a)):``        ``return` `0` `    ``# If current state is already``    ``# calculated then use it``    ``if` `(dp[i][flag] !``=` `-``1``):``        ``return` `dp[i][flag]` `    ``ans ``=` `0` `    ``# If current element is positive``    ``if` `(flag ``=``=` `0``):` `        ``# Update ans and recursively``        ``# call with update value of flag``        ``ans ``=` `max``(findMax(a, dp, i ``+` `1``, ``0``),``           ``a[i] ``+` `findMax(a, dp, i ``+` `1``, ``1``))` `    ``# Else current element is negative``    ``else``:` `        ``# Update ans and recursively``        ``# call with update value of flag``        ``ans ``=` `max``(findMax(a, dp, i ``+` `1``, ``1``),``      ``-``1` `*` `a[i] ``+` `findMax(a, dp, i ``+` `1``, ``0``))` `    ``# Return maximum sum subsequence``    ``dp[i][flag] ``=` `ans``    ` `    ``return` `ans` `# Function that finds the maximum``# sum of element of the subsequence``# with alternate +ve and -ve signs``def` `findMaxSumUtil(arr, N):``    ` `    ``# Create auxiliary array dp[][]``    ``dp ``=` `[[``-``1` `for` `i ``in` `range``(``2``)]``              ``for` `i ``in` `range``(N)]` `    ``# Function call``    ``print``(findMax(arr, dp, ``0``, ``0``))` `# Driver Code``if` `__name__ ``=``=` `'__main__'``:``    ` `    ``# Given array arr[]``    ``arr ``=` `[ ``1``, ``2``, ``3``, ``4``, ``1``, ``2` `]` `    ``N ``=` `len``(arr)` `    ``# Function call``    ``findMaxSumUtil(arr, N)` `# This code is contributed by mohit kumar 29`

## C#

 `// C# program for the above approach``using` `System;` `class` `GFG {`` ` `// Function to find the``// maximum sum subsequence``static` `int` `findMax(``int``[] a, ``int``[,] dp,``                   ``int` `i, ``int` `flag)``{``    ` `    ``// Base Case``    ``if` `(i == (``int``)a.Length)``    ``{``        ``return` `0;``    ``}``  ` `    ``// If current state is already``    ``// calculated then use it``    ``if` `(dp[i, flag] != -1)``    ``{``        ``return` `dp[i, flag];``    ``}``  ` `    ``int` `ans;``  ` `    ``// If current element is positive``    ``if` `(flag == 0)``    ``{``        ` `        ``// Update ans and recursively``        ``// call with update value of flag``        ``ans = Math.Max(findMax(a, dp, i + 1, 0),``                ``a[i] + findMax(a, dp, i + 1, 1));``    ``}``  ` `    ``// Else current element is negative``    ``else``    ``{``        ` `        ``// Update ans and recursively``        ``// call with update value of flag``        ``ans = Math.Max(findMax(a, dp, i + 1, 1),``           ``-1 * a[i] + findMax(a, dp, i + 1, 0));``    ``}``  ` `    ``// Return maximum sum subsequence``    ``return` `dp[i, flag] = ans;``}``  ` `// Function that finds the maximum``// sum of element of the subsequence``// with alternate +ve and -ve signs``static` `void` `findMaxSumUtil(``int``[] arr,``                           ``int` `N)``{``     ` `    ``// Create auxiliary array dp[][]``    ``int``[,] dp = ``new` `int``[N, 2];``  ` `    ``// Initialize dp[][]``    ``for``(``int` `i = 0; i < N; i++)``    ``{``        ``for``(``int` `j = 0; j < 2; j++)``        ``{``            ``dp[i, j] = -1;``        ``}``    ``}``     ` `    ``// Function Call``    ``Console.WriteLine(findMax(arr, dp, 0, 0));``}`` ` `// Driver Code``public` `static` `void` `Main()``{``    ` `    ``// Given array arr[]``    ``int``[] arr = { 1, 2, 3, 4, 1, 2 };``  ` `    ``int` `N = arr.Length;``  ` `    ``// Function call``    ``findMaxSumUtil(arr, N);``}``}` `// This code is contributed by code_hunt`

## Javascript

 ``

Output:

`5`

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

Efficient Approach: Using the DP Tabulation method ( Iterative approach )

The approach to solving this problem is the same but the DP tabulation(bottom-up) method is better then Dp + memoization(top-down) because the memoization method needs extra stack space for recursion calls.

Steps to solve this problem :

• Create a table DP to store the solution of the subproblems.
• Initialize the table with base cases
• Now Iterate over subproblems to get the value of the current problem from the previous computation of subproblems stored in DP.
• Return the final solution stored in dp0].

Implementation :

## C++

 `// C++ program for above approach` `#include ``using` `namespace` `std;`  `// Function to find the``// maximum sum subsequence``int` `findMax(vector<``int``>& a, ``int` `N)``{``    ``int` `dp[N];` `    ``// Initializing the base case``    ``dp[N-1] = a[N-1];``    ``dp[N-1] = 0;``    ` `  ` `      ``// iterate over subproblems to get the current value``    ``// from previous computation stored in DP``    ``for` `(``int` `i = N-2; i >= 0; i--) {``      ` `          ``// Update current value in DP``        ``dp[i] = max(dp[i+1], a[i]+dp[i+1]);``        ``dp[i] = max(dp[i+1], -1*a[i]+dp[i+1]);``    ``}``  ` `      ``// return ans``    ``return` `dp;``}` `// Driver Code``int` `main()``{``    ``vector<``int``> arr = { 1, 2, 3, 4, 1, 2 };``    ``int` `N = arr.size();``  ` `      ``// function Call``    ``cout << findMax(arr, N);``    ``return` `0;``}` `// this code is contributed by bhardwajji`

## Java

 `import` `java.util.*;` `public` `class` `Main {``    ``// Function to find the maximum sum subsequence``    ``public` `static` `int` `findMax(List a, ``int` `N) {``        ``int``[][] dp = ``new` `int``[N][``2``];``        ` `        ``// Initializing the base case``        ``dp[N-``1``][``0``] = a.get(N-``1``);``        ``dp[N-``1``][``1``] = ``0``;``          ` `        ``// iterate over subproblems to get the current value``        ``// from previous computation stored in DP``        ``for` `(``int` `i = N-``2``; i >= ``0``; i--) {``            ``// Update current value in DP``            ``dp[i][``0``] = Math.max(dp[i+``1``][``0``], a.get(i)+dp[i+``1``][``1``]);``            ``dp[i][``1``] = Math.max(dp[i+``1``][``1``], -``1``*a.get(i)+dp[i+``1``][``0``]);``        ``}``      ` `        ``// return ans``        ``return` `dp[``0``][``0``];``    ``}``      ` `    ``// Driver Code``    ``public` `static` `void` `main(String[] args) {``        ``List arr = ``new` `ArrayList<>(Arrays.asList(``1``, ``2``, ``3``, ``4``, ``1``, ``2``));``        ``int` `N = arr.size();``          ` `        ``// function Call``        ``System.out.println(findMax(arr, N));``    ``}``}`

## Python3

 `# Python program for the above approach` `# Function to find the maximum element``def` `findMax(a, N):``    ``dp ``=` `[[``0` `for` `i ``in` `range``(``2``)] ``for` `j ``in` `range``(N)]` `    ``# Initializing the base case``    ``dp[N``-``1``][``0``] ``=` `a[N``-``1``]``    ``dp[N``-``1``][``1``] ``=` `0` `    ``# Iterate over subproblems to get the``    ``# current value from previous computation``    ``# stored in DP``    ``for` `i ``in` `range``(N``-``2``, ``-``1``, ``-``1``):` `        ``# Update current value in DP``        ``dp[i][``0``] ``=` `max``(dp[i``+``1``][``0``], a[i]``+``dp[i``+``1``][``1``])``        ``dp[i][``1``] ``=` `max``(dp[i``+``1``][``1``], ``-``1``*``a[i]``+``dp[i``+``1``][``0``])` `        ``# return ans``    ``return` `dp[``0``][``0``]`  `# Driver Code``arr ``=` `[``1``, ``2``, ``3``, ``4``, ``1``, ``2``]``N ``=` `len``(arr)` `print``(findMax(arr, N))`

## C#

 `using` `System;` `class` `MaxSumSubsequence {` `    ``// Function to find the maximum sum``    ``// of the subsequence``    ``static` `int` `findMax(``int``[] a, ``int` `N)``    ``{``        ``int``[, ] dp = ``new` `int``[N, 2];` `        ``// Initializing the base case``        ``dp[N - 1, 0] = a[N - 1];``        ``dp[N - 1, 1] = 0;` `        ``// iterate over subproblems to get``        ``// the current value from previous``        ``// computation stored in DP``        ``for` `(``int` `i = N - 2; i >= 0; i--) {` `            ``// Update current value in DP``            ``dp[i, 0] = Math.Max(dp[i + 1, 0],``                                ``a[i] + dp[i + 1, 1]);``            ``dp[i, 1] = Math.Max(dp[i + 1, 1],``                                ``-1 * a[i] + dp[i + 1, 0]);``        ``}` `        ``// Return the ans``        ``return` `dp[0, 0];``    ``}` `    ``// Driver Code``    ``static` `void` `Main()``    ``{` `        ``int``[] arr = { 1, 2, 3, 4, 1, 2 };``        ``int` `N = arr.Length;``        ``Console.WriteLine(findMax(arr, N));``    ``}``}`

## Javascript

 `function` `findMax(arr, N) {``  ``let dp = ``new` `Array(N);``  ``for` `(let i = 0; i < N; i++) {``    ``dp[i] = ``new` `Array(2);``  ``}` `  ``// Initializing the base case``  ``dp[N - 1] = arr[N - 1];``  ``dp[N - 1] = 0;` `  ``// iterate over subproblems to get``  ``// the current value from previous``  ``// computation stored in DP``  ``for` `(let i = N - 2; i >= 0; i--) {``    ``// Update current value in DP``    ``dp[i] = Math.max(dp[i + 1], arr[i] + dp[i + 1]);``    ``dp[i] = Math.max(dp[i + 1], -1 * arr[i] + dp[i + 1]);``  ``}` `  ``// Return the ans``  ``return` `dp;``}` `// Driver Code``let arr = [1, 2, 3, 4, 1, 2];``let N = arr.length;``console.log(findMax(arr, N));`

Output:

`5`

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

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