Given a list of length N with positive and negative integers. The task is to choose the longest alternating subsequence of the given sequence (i.e. the sign of each next element is the opposite of the sign of the current element). Among all such subsequences, we have to choose one which has the maximum sum of elements and display that sum.
Examples:
Input: list = [-2 10 3 -8 -4 -1 5 -2 -3 1]
Output: 11
Explanation:
The largest subsequence with the greatest sum is [-2 10 -1 5 -2 1] with length 6.Input: list=[12 4 -5 7 -9]
Output: 5
Explanation:
The largest subsequence with greatest sum is [12 -5 7 -9] with length 4.
Approach: The solution can be reached by the following approach:-
- To get alternating subsequences with maximum length and the largest sum, we will be traversing the whole list (length of list)-1 times for comparing signs of consecutive elements.
- During traversal, if we are getting more than 1 consecutive element of the same sign(exp. 1 2 4), then we will append the maximum element out of them to another list named large. so from 1, 2 and 4 we will append 4 to another list.
- If we have consecutive elements of opposite sign, we will simply add those elements to that list named large.
- Finally, the list named large will have the longest alternating subsequence with the largest elements.
- Now, we will have to calculate the sum of all elements from that list named large.
Lets take an example, we have a list [1, 2, 3, -2, -5, 1, -7, -1].
- In traversing this list length-1 times, we are getting 1, 2, 3 with the same sign so we will append greatest of these (i.e 3) to another list named large here.
Hence large=[3]- Now -2 and -5 have the same sign so we will append -2 to another List.
large=[3, -2]- Now, the sign of 1 and -7 are opposite, so we will append 1 to large.
large=[3, -2, 1]- For -7, -1 signs are same, Hence append -1 to large.
large=[3, -2, 1, -1]- Calculate the sum = 3 – 2 + 1 – 1 = 1
Below is the implementation of the above approach:
C++
// C++ implementation to find the // longest alternating subsequence // which has the maximum sum#include<bits/stdc++.h>using namespace std;int calculateMaxSum(int n, int li[]){ // Creating a temporary list ar to // every time store same sign element // to calculate maximum element from // that list ar vector<int> ar; // Appending 1st element of list li // to the ar ar.push_back(li[0]); // Creating list to store maximum // values vector<int> large; for(int j = 0; j < n - 1; j++) { // If both number are positive // then append (j + 1)th element // to temporary list ar if(li[j] > 0 and li[j + 1] > 0) { ar.push_back(li[j + 1]); } else if(li[j] > 0 and li[j + 1] < 0) { // If opposite elements found // then append maximum element // to large list large.push_back(*max_element(ar.begin(), ar.end())); // Empty ar list to re-append // next elements ar.clear(); ar.push_back(li[j + 1]); } else if(li[j] < 0 and li[j + 1] > 0) { // If opposite elements found // then append maximum element // to large list large.push_back(*max_element(ar.begin(), ar.end())); // Empty ar list to re-append // next elements ar.clear(); ar.push_back(li[j + 1]); } else { // If both number are negative // then append (j + 1)th element // to temporary list ar ar.push_back(li[j + 1]); } } // The final Maximum element in ar list // also needs to be appended to large list large.push_back(*max_element(ar.begin(), ar.end())); // Returning the sum of all elements // from largest elements list with // largest alternating subsequence size int sum = 0; for(int i = 0; i < large.size(); i++) sum += large[i]; return sum;} // Driver codeint main(){ int list[] = { -2, 8, 3, 8, -4, -15, 5, -2, -3, 1 }; int N = sizeof(list) / sizeof(list[0]); cout << (calculateMaxSum(N, list));}// This code is contributed by Bhupendra_Singh |
Java
// Java implementation to find the // longest alternating subsequence // which has the maximum sum import java.util.*;class GFG{static int calculateMaxSum(int n, int li[]) { // Creating a temporary list ar to // every time store same sign element // to calculate maximum element from // that list ar Vector<Integer> ar = new Vector<>(); // Appending 1st element of list li // to the ar ar.add(li[0]); // Creating list to store maximum // values Vector<Integer> large = new Vector<>(); for(int j = 0; j < n - 1; j++) { // If both number are positive // then append (j + 1)th element // to temporary list ar if(li[j] > 0 && li[j + 1] > 0) { ar.add(li[j + 1]); } else if(li[j] > 0 && li[j + 1] < 0) { // If opposite elements found // then append maximum element // to large list large.add(Collections.max(ar)); // Empty ar list to re-append // next elements ar.clear(); ar.add(li[j + 1]); } else if(li[j] < 0 && li[j + 1] > 0) { // If opposite elements found // then append maximum element // to large list large.add(Collections.max(ar)); // Empty ar list to re-append // next elements ar.clear(); ar.add(li[j + 1]); } else { // If both number are negative // then append (j + 1)th element // to temporary list ar ar.add(li[j + 1]); } } // The final Maximum element in ar list // also needs to be appended to large list large.add(Collections.max(ar)); // Returning the sum of all elements // from largest elements list with // largest alternating subsequence size int sum = 0; for(int i = 0; i < large.size(); i++) sum += (int)large.get(i); return sum; } // Driver codepublic static void main(String args[]) { int list[] = { -2, 8, 3, 8, -4, -15, 5, -2, -3, 1 }; int N = (list.length); System.out.print(calculateMaxSum(N, list));}}// This code is contributed by Stream_Cipher |
Python3
# Python3 implementation to find the # longest alternating subsequence # which has the maximum sumdef calculateMaxSum(n, li): # Creating a temporary list ar to every # time store same sign element to # calculate maximum element from # that list ar ar =[] # Appending 1st element of list li # to the ar ar.append(li[0]) # Creating list to store maximum # values large =[] for j in range(0, n-1): # If both number are positive # then append (j + 1)th element # to temporary list ar if(li[j]>0 and li[j + 1]>0): ar.append(li[j + 1]) elif(li[j]>0 and li[j + 1]<0): # If opposite elements found # then append maximum element # to large list large.append(max(ar)) # Empty ar list to re-append # next elements ar =[] ar.append(li[j + 1]) elif(li[j]<0 and li[j + 1]>0): # If opposite elements found # then append maximum element # to large list large.append(max(ar)) # Empty ar list to re-append # next elements ar =[] ar.append(li[j + 1]) else: # If both number are negative # then append (j + 1)th element # to temporary list ar ar.append(li[j + 1]) # The final Maximum element in ar list # also needs to be appended to large list large.append(max(ar)) # returning the sum of all elements # from largest elements list with # largest alternating subsequence size return sum(large)# Driver codelist =[-2, 8, 3, 8, -4, -15, 5, -2, -3, 1]N = len(list)print(calculateMaxSum(N, list)) |
C#
// C# implementation to find the // longest alternating subsequence // which has the maximum sum using System; using System.Collections.Generic;class GFG{static int find_max(List<int> ar){ int mx = -1000000; foreach(var i in ar) { if(i > mx) mx = i; } return mx;}static int calculateMaxSum(int n, int []li) { // Creating a temporary list ar to // every time store same sign element // to calculate maximum element from // that list ar List<int> ar = new List<int>(); // Appending 1st element of list li // to the ar ar.Add(li[0]); // Creating list to store maximum // values List<int> large = new List<int>(); for(int j = 0; j < n - 1; j++) { // If both number are positive // then append (j + 1)th element // to temporary list ar if(li[j] > 0 && li[j + 1] > 0) { ar.Add(li[j + 1]); } else if(li[j] > 0 && li[j + 1] < 0) { // If opposite elements found // then append maximum element // to large list large.Add(find_max(ar)); // Empty ar list to re-append // next elements ar.Clear(); ar.Add(li[j + 1]); } else if(li[j] < 0 && li[j + 1] > 0) { // If opposite elements found // then append maximum element // to large list large.Add(find_max(ar)); // Empty ar list to re-append // next elements ar.Clear(); ar.Add(li[j + 1]); } else { // If both number are negative // then append (j + 1)th element // to temporary list ar ar.Add(li[j + 1]); } } // The final Maximum element in ar list // also needs to be appended to large list large.Add(find_max(ar)); // Returning the sum of all elements // from largest elements list with // largest alternating subsequence size int sum = 0; foreach(var i in large) { sum += i; } return sum; } // Driver codepublic static void Main() { int []list = { -2, 8, 3, 8, -4, -15, 5, -2, -3, 1 }; int N = (list.Length); Console.WriteLine(calculateMaxSum(N, list));}}// This code is contributed by Stream_Cipher |
Javascript
<script> // Javascript implementation to find the // longest alternating subsequence // which has the maximum sum function find_max(ar) { let mx = -1000000; for(let i = 0; i < ar.length; i++) { if(ar[i] > mx) mx = ar[i]; } return mx; } function calculateMaxSum(n, li) { // Creating a temporary list ar to // every time store same sign element // to calculate maximum element from // that list ar let ar = []; // Appending 1st element of list li // to the ar ar.push(li[0]); // Creating list to store maximum // values let large = []; for(let j = 0; j < n - 1; j++) { // If both number are positive // then append (j + 1)th element // to temporary list ar if(li[j] > 0 && li[j + 1] > 0) { ar.push(li[j + 1]); } else if(li[j] > 0 && li[j + 1] < 0) { // If opposite elements found // then append maximum element // to large list large.push(find_max(ar)); // Empty ar list to re-append // next elements ar = []; ar.push(li[j + 1]); } else if(li[j] < 0 && li[j + 1] > 0) { // If opposite elements found // then append maximum element // to large list large.push(find_max(ar)); // Empty ar list to re-append // next elements ar = []; ar.push(li[j + 1]); } else { // If both number are negative // then append (j + 1)th element // to temporary list ar ar.push(li[j + 1]); } } // The final Maximum element in ar list // also needs to be appended to large list large.push(find_max(ar)); // Returning the sum of all elements // from largest elements list with // largest alternating subsequence size let sum = 0; for(let i = 0; i < large.length; i++) { sum += large[i]; } return sum; } let list = [ -2, 8, 3, 8, -4, -15, 5, -2, -3, 1 ]; let N = (list.length); document.write(calculateMaxSum(N, list)); </script> |
Output:
6
Time Complexity:O(N2)
Auxiliary Space: O(N)
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