Maximum Sum Subsequence (original) (raw)

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 * 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:

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#

// 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

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