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 subsequence 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 <bits/stdc++.h>using namespace std;// Function to find the// maximum sum subsequenceint findMax(vector<int>& a, int dp[][2], 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 signsvoid findMaxSumUtil(vector<int>& arr, int N){ // Create auxiliary array dp[][] int dp[N][2]; // Initialize dp[][] memset(dp, -1, sizeof dp); // Function Call cout << findMax(arr, dp, 0, 0);}// Driver Codeint 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 approachimport java.io.*;import java.util.Arrays;class GFG{ // Function to find the// maximum sum subsequencestatic 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 signsstatic 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 Codepublic 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 subsequencedef 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 signsdef 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 Codeif __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 approachusing System;class GFG { // Function to find the// maximum sum subsequencestatic 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 signsstatic 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 Codepublic 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
<script>// Javascript program for the above approach// Function to find the// maximum sum subsequencefunction findMax(a, dp, i, flag){ // Base Case if (i == a.length) { return 0; } // If current state is already // calculated then use it if (dp[i][flag] != -1) { return dp[i][flag]; } let 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 signsfunction findMaxSumUtil(arr, N){ // Create auxiliary array dp[][] let dp = new Array(N); // Loop to create 2D array using 1D array for(var i = 0; i < dp.length; i++) { dp[i] = new Array(2); } // Initialize dp[][] for(let i = 0; i < N; i++) { for(let j = 0; j < 2; j++) { dp[i][j] = -1; } } // Function Call document.write(findMax(arr, dp, 0, 0));}// Driver code// Given array arr[]let arr = [ 1, 2, 3, 4, 1, 2 ];let N = arr.length;// Function callfindMaxSumUtil(arr, N);// This code is contributed by splevel62</script> |
Output:
5
Time Complexity: O(N)
Auxiliary Space: O(N)
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