Maximum Sum Subsequence

Last Updated : 25 Jan, 2022

Given an array arr[] of size N, the task is to find the maximum sum non-empty subsequence present in the given array.

Examples:

Input: arr[] = { 2, 3, 7, 1, 9 }
Output: 22
Explanation:
Sum of the subsequence { arr[0], arr[1], arr[2], arr[3], arr[4] } is equal to 22, which is the maximum possible sum of any subsequence of the array.
Therefore, the required output is 22.

Input: arr[] = { -2, 11, -4, 2, -3, -10 }
Output: 13
Explanation:
Sum of the subsequence { arr[1], arr[3] } is equal to 13, which is the maximum possible sum of any subsequence of the array.
Therefore, the required output is 13.

Naive Approach: The simplest approach to solve this problem is to generate all possible non-empty subsequences of the array and calculate the sum of each subsequence of the array. Finally, print the maximum sum obtained from the subsequence.

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

Efficient Approach: The idea is to traverse the array and calculate the sum of positive elements of the array and print the sum obtained. Follow the steps below to solve the problem:

Check if the largest element of the array is greater than 0 or not. If found to be true, then traverse the array and print the sum of all positive elements of the array.
Otherwise, print the largest element present in the array.

Below is the implementation of the above approach:

C++

// C++ program to implement // the above approach #include <bits/stdc++.h> using namespace std;  // Function to print the maximum // non-empty subsequence sum int MaxNonEmpSubSeq(int a[], int n) {     // Stores the maximum non-empty     // subsequence sum in an array     int sum = 0;      // Stores the largest element     // in the array     int max = *max_element(a, a + n);      if (max <= 0) {          return max;     }      // Traverse the array     for (int i = 0; i < n; i++) {          // If a[i] is greater than 0         if (a[i] > 0) {              // Update sum             sum += a[i];         }     }     return sum; }  // Driver Code int main() {     int arr[] = { -2, 11, -4, 2, -3, -10 };     int N = sizeof(arr) / sizeof(arr[0]);      cout << MaxNonEmpSubSeq(arr, N);      return 0; }

Java

// Java program to implement // the above approach import java.util.*; class GFG {    // Function to print the maximum   // non-empty subsequence sum   static int MaxNonEmpSubSeq(int a[], int n)   {      // Stores the maximum non-empty     // subsequence sum in an array     int sum = 0;      // Stores the largest element     // in the array     int max = a[0];     for(int i = 1; i < n; i++)     {       if(max < a[i])       {         max = a[i];       }     }      if (max <= 0)      {            return max;     }      // Traverse the array     for (int i = 0; i < n; i++)     {        // If a[i] is greater than 0       if (a[i] > 0)        {          // Update sum         sum += a[i];       }     }     return sum;   }    // Driver code   public static void main(String[] args)   {     int arr[] = { -2, 11, -4, 2, -3, -10 };     int N = arr.length;      System.out.println(MaxNonEmpSubSeq(arr, N));   } }  // This code is contributed by divyesh072019

Python3

# Python3 program to implement # the above approach  # Function to print the maximum # non-empty subsequence sum def MaxNonEmpSubSeq(a, n):          # Stores the maximum non-empty     # subsequence sum in an array     sum = 0      # Stores the largest element     # in the array     maxm = max(a)      if (maxm <= 0):         return maxm      # Traverse the array     for i in range(n):                  # If a[i] is greater than 0         if (a[i] > 0):                          # Update sum             sum += a[i]                  return sum  # Driver Code if __name__ == '__main__':          arr = [ -2, 11, -4, 2, -3, -10 ]     N = len(arr)      print(MaxNonEmpSubSeq(arr, N))  # This code is contributed by mohit kumar 29

// C# program to implement // the above approach using System;  class GFG{      // Function to print the maximum // non-empty subsequence sum static int MaxNonEmpSubSeq(int[] a, int n) {          // Stores the maximum non-empty     // subsequence sum in an array     int sum = 0;          // Stores the largest element     // in the array     int max = a[0];     for(int i = 1; i < n; i++)     {         if (max < a[i])         {             max = a[i];         }     }          if (max <= 0)      {         return max;     }          // Traverse the array     for(int i = 0; i < n; i++)      {                  // If a[i] is greater than 0         if (a[i] > 0)         {                          // Update sum             sum += a[i];         }     }     return sum; }  // Driver Code static void Main()  {     int[] arr = { -2, 11, -4, 2, -3, -10 };     int N = arr.Length;          Console.WriteLine(MaxNonEmpSubSeq(arr, N)); } }  // This code is contributed by divyeshrabadiya07

JavaScript

<script>      // Javascript program to implement     // the above approach          // Function to print the maximum     // non-empty subsequence sum     function MaxNonEmpSubSeq(a, n)     {          // Stores the maximum non-empty         // subsequence sum in an array         let sum = 0;          // Stores the largest element         // in the array         let max = a[0];         for(let i = 1; i < n; i++)         {             if (max < a[i])             {                 max = a[i];             }         }          if (max <= 0)         {             return max;         }          // Traverse the array         for(let i = 0; i < n; i++)         {              // If a[i] is greater than 0             if (a[i] > 0)             {                  // Update sum                 sum += a[i];             }         }         return sum;     }          let arr = [ -2, 11, -4, 2, -3, -10 ];     let N = arr.length;           document.write(MaxNonEmpSubSeq(arr, N));    </script>

Output:

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

Maximum Sum Subsequence

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