Maximum length Subsequence with alternating sign and maximum Sum
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 <bits/stdc++.h> 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[0]); 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 <script> // Javascript 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. function findSubsequence(arr, n) { let sign = Array.from({length: n}, (_, i) => 0); // Find whether each element // is positive or negative for (let i = 0; i < n; i++) { if (arr[i] > 0) sign[i] = 1; else sign[i] = -1; } let k = 0; let result = Array.from({length: n}, (_, i) => 0); // Find the required subsequence for (let i = 0; i < n; i++) { let cur = arr[i]; let 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 (let i = 0; i < k; i++) document.write(result[i]+ " "); document.write("<br/>"); } // Driver Code // array declaration let arr = [ -4, 9, 4, 11, -5, -17, 9, -3, -5, 2 ]; // size of array let n = arr.length; findSubsequence(arr, n); // This code is contributed by sanjoy_62. </script>
Time Complexity :O(N)
Auxiliary Space: O(N)
Similar Reads
Longest alternating subsequence with maximum sum | Set 2 Given an array arr[] of size N, consisting of positive and negative integers, the task is to find the longest alternating subsequence(i.e. the sign of every element is opposite to that of its previous element) from the given array which has the maximum sum.Examples: Input: arr[] = {-2, 10, 3, -8, -4
7 min read
Longest alternating subsequence which has maximum sum of elements 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 maxi
10 min read
Maximum sum alternating subsequence Given an array, the task is to find sum of maximum sum alternating subsequence starting with first element. Here alternating sequence means first decreasing, then increasing, then decreasing, ... For example 10, 5, 14, 3 is an alternating sequence. Note that the reverse type of sequence (increasing
13 min read
Maximum sum subsequence of any size which is decreasing-increasing alternatively Given an array of integers arr[], find the subsequence with maximum sum whose elements are first decreasing, then increasing, or vice versa, The subsequence can start anywhere in the main sequence, not necessarily at the first element of the main sequence. A sequence {x1, x2, .. xn} is an alternatin
15+ min read
Maximum sum subsequence with at-least k distant elements Given an array and a number k, find a subsequence such that Sum of elements in subsequence is maximumIndices of elements of subsequence differ atleast by k Examples Input : arr[] = {4, 5, 8, 7, 5, 4, 3, 4, 6, 5} k = 2 Output: 19 Explanation: The highest value is obtained if you pick indices 1, 4, 7,
6 min read