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Longest subsequence having maximum sum
  • Difficulty Level : Medium
  • Last Updated : 21 Apr, 2021

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

Examples:

Input: arr[] = { 1, 2, -4, -2, 3, 0 } 
Output: 1 2 3 0 
Explanation: 
Sum of elements of the subsequence {1, 2, 3, 0} is 6 which is the maximum possible sum. 
Therefore, the required output is 1 2 3 0

Input: arr[] = { -10, -6, -2, -3, -4 } 
Output: -2

Naive Approach: The simplest approach to solve this problem is to traverse the array and generate all possible subsequence of the given array and calculate their sums. Print the longest of all subsequences with maximum sum. 



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

Efficient Approach: The problem can be solved using Greedy technique. 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 find the longest subsequence
// from the given array with maximum sum
void longestSubWithMaxSum(int arr[], int N)
{
    // Stores the largest element
    // of the array
    int Max = *max_element(arr,
                           arr + N);
 
    // If Max is less than 0
    if (Max < 0) {
 
        // Print the largest element
        // of the array
        cout << Max;
        return;
    }
 
    // Traverse the array
    for (int i = 0; i < N; i++) {
 
        // If arr[i] is greater
        // than or equal to 0
        if (arr[i] >= 0) {
 
            // Print elements of
            // the subsequence
            cout << arr[i] << " ";
        }
    }
}
 
// Driver Code
int main()
{
    int arr[] = { 1, 2, -4, -2, 3, 0 };
 
    int N = sizeof(arr) / sizeof(arr[0]);
 
    longestSubWithMaxSum(arr, N);
    return 0;
}

Java




// Java program to implement
// the above approach
import java.util.*;
  
class GFG{
  
// Function to find the longest subsequence
// from the given array with maximum sum
static void longestSubWithMaxSum(int arr[], int N)
{
     
    // Stores the largest element
    // of the array
    int Max = Arrays.stream(arr).max().getAsInt();
  
    // If Max is less than 0
    if (Max < 0)
    {
         
        // Print the largest element
        // of the array
        System.out.print(Max);
        return;
    }
  
    // Traverse the array
    for(int i = 0; i < N; i++)
    {
         
        // If arr[i] is greater
        // than or equal to 0
        if (arr[i] >= 0)
        {
             
            // Print elements of
            // the subsequence
            System.out.print(arr[i] + " ");
        }
    }
}
  
// Driver Code
public static void main(String[] args)
{
    int arr[] = { 1, 2, -4, -2, 3, 0 };
    int N = arr.length;
  
    longestSubWithMaxSum(arr, N);
}
}
 
// This code is contributed by code_hunt

Python3




# Python3 program to implement
# the above approach
 
# Function to find the longest subsequence
# from the given array with maximum sum
def longestSubWithMaxSum(arr, N):
 
    # Stores the largest element
    # of the array
    Max = max(arr)
 
    # If Max is less than 0
    if (Max < 0) :
 
        # Print the largest element
        # of the array
        print(Max)
        return
 
    # Traverse the array
    for i in range(N):
 
        # If arr[i] is greater
        # than or equal to 0
        if (arr[i] >= 0) :
 
            # Print elements of
            # the subsequence
            print(arr[i], end = " ")
 
# Driver code
arr = [ 1, 2, -4, -2, 3, 0 ]
 
N = len(arr)
 
longestSubWithMaxSum(arr, N)
 
# This code is contributed divyeshrabadiya07

C#




// C# program to implement
// the above approach
using System;
 
class GFG{
  
// Function to find the longest subsequence
// from the given array with maximum sum
static void longestSubWithMaxSum(int []arr,
                                 int N)
{
     
    // Stores the largest element
    // of the array
    int Max = arr[0];
     
    for(int i = 1; i < N; i++)
    {
        if (Max < arr[i])
            Max = arr[i];
    }
     
    // If Max is less than 0
    if (Max < 0)
    {
         
        // Print the largest element
        // of the array
        Console.Write(Max);
        return;
    }
     
    // Traverse the array
    for(int i = 0; i < N; i++)
    {
         
        // If arr[i] is greater
        // than or equal to 0
        if (arr[i] >= 0)
        {
             
            // Print elements of
            // the subsequence
            Console.Write(arr[i] + " ");
        }
    }
}
  
// Driver Code
public static void Main(String[] args)
{
    int []arr = { 1, 2, -4, -2, 3, 0 };
    int N = arr.Length;
  
    longestSubWithMaxSum(arr, N);
}
}
 
// This code is contributed by aashish1995

Javascript




<script>
 
// JavaScript program to implement
// the above approach
 
// Function to find the longest subsequence
// from the given array with maximum sum
function longestSubWithMaxSum(arr, N)
{    
     
    // Stores the largest element
    // of the array
    let Max = Math.max(...arr);
   
    // If Max is less than 0
    if (Max < 0)
    {
          
        // Print the largest element
        // of the array
        document.write(Max);
        return;
    }
   
    // Traverse the array
    for(let i = 0; i < N; i++)
    {
          
        // If arr[i] is greater
        // than or equal to 0
        if (arr[i] >= 0)
        {
              
            // Print the elements of
            // the subsequence
            document.write(arr[i] + " ");
        }
    }
}
  
// Driver code
let arr = [ 1, 2, -4, -2, 3, 0 ];
let N = arr.length;
 
longestSubWithMaxSum(arr, N);
 
// This code is contributed by avijitmondal1998
 
</script>
Output: 
1 2 3 0

 

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

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