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Maximum Sum Subsequence
  • Difficulty Level : Easy
  • Last Updated : 23 Apr, 2021

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 * 2N) 
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:

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-emepty subsequence sum
int MaxNonEmpSubSeq(int a[], int n)
{
    // Stores the maximum non-emepty
    // 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-emepty subsequence sum
  static int MaxNonEmpSubSeq(int a[], int n)
  {
 
    // Stores the maximum non-emepty
    // 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 prthe maxmimum
# non-emepty subsequence sum
def MaxNonEmpSubSeq(a, n):
     
    # Stores the maxmimum non-emepty
    # 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#




// C# program to implement
// the above approach
using System;
 
class GFG{
     
// Function to print the maximum
// non-emepty subsequence sum
static int MaxNonEmpSubSeq(int[] a, int n)
{
     
    // Stores the maximum non-emepty
    // 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-emepty subsequence sum
    function MaxNonEmpSubSeq(a, n)
    {
 
        // Stores the maximum non-emepty
        // 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: 
13

 

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

 

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