Given an array arr[] of length N, the task is to find the length of the largest sub-sequence with non-negative sum.
Examples:
Input: arr[] = {1, 2, -3}
Output: 3
The complete array has a non-negative sum.
Input: arr[] = {1, 2, -4}
Output: 2
{1, 2} is the required subsequence.
Approach: The idea is that all the non-negative numbers must be included in the sub-sequence because such numbers will only increase the value of the total sum.
Now, it’s not hard to see among negative numbers, the larger ones must be chosen first. So, keep adding the negative numbers in non-increasing order of their values as long as they don’t decrease the value of the total sum below 0. This can be done after sorting the array.
Below is the implementation of the above approach:
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std;
// Function to return the length of // the largest subsequence // with non-negative sum int maxLen( int * arr, int n)
{ // To store the current sum
int c_sum = 0;
// Sort the input array in
// non-increasing order
sort(arr, arr + n, greater< int >());
// Traverse through the array
for ( int i = 0; i < n; i++) {
// Add the current element to the sum
c_sum += arr[i];
// Condition when c_sum falls
// below zero
if (c_sum < 0)
return i;
}
// Complete array has a non-negative sum
return n;
} // Driver code int main()
{ int arr[] = { 3, 5, -6 };
int n = sizeof (arr) / sizeof ( int );
cout << maxLen(arr, n);
return 0;
} |
// Java implementation of the approach import java.util.*;
class GFG
{ // Function to return the length of // the largest subsequence // with non-negative sum static int maxLen( int [] arr, int n)
{ // To store the current sum
int c_sum = 0 ;
// Sort the input array in
// non-increasing order
Arrays.sort(arr);
// Traverse through the array
for ( int i = n- 1 ; i >= 0 ; i--)
{
// Add the current element to the sum
c_sum += arr[i];
// Condition when c_sum falls
// below zero
if (c_sum < 0 )
return i;
}
// Complete array has a non-negative sum
return n;
} // Driver code public static void main(String []args)
{ int arr[] = { 3 , 5 , - 6 };
int n = arr.length;
System.out.println(maxLen(arr, n));
} } // This code is contributed by Rajput-Ji |
# Python3 implementation of the approach # Function to return the length of # the largest subsequence # with non-negative sum def maxLen(arr, n) :
# To store the current sum
c_sum = 0 ;
# Sort the input array in
# non-increasing order
arr.sort(reverse = True );
# Traverse through the array
for i in range (n) :
# Add the current element to the sum
c_sum + = arr[i];
# Condition when c_sum falls
# below zero
if (c_sum < 0 ) :
return i;
# Complete array has a non-negative sum
return n;
# Driver code if __name__ = = "__main__" :
arr = [ 3 , 5 , - 6 ];
n = len (arr);
print (maxLen(arr, n));
# This code is contributed by AnkitRai01 |
// C# implementation of the approach using System;
class GFG
{ // Function to return the length of // the largest subsequence // with non-negative sum static int maxLen( int [] arr, int n)
{ // To store the current sum
int c_sum = 0;
// Sort the input array in
// non-increasing order
Array.Sort(arr);
// Traverse through the array
for ( int i = n - 1; i >= 0; i--)
{
// Add the current element to the sum
c_sum += arr[i];
// Condition when c_sum falls
// below zero
if (c_sum < 0)
return i;
}
// Complete array has a non-negative sum
return n;
} // Driver code public static void Main(String []args)
{ int []arr = { 3, 5, -6 };
int n = arr.Length;
Console.WriteLine(maxLen(arr, n));
} } // This code is contributed by PrinciRaj1992 |
<script> // Javascript implementation of the approach // Function to return the length of // the largest subsequence // with non-negative sum function maxLen(arr, n)
{ // To store the current sum
var c_sum = 0;
// Sort the input array in
// non-increasing order
arr.sort((a,b)=> b-a)
// Traverse through the array
for ( var i = 0; i < n; i++) {
// Add the current element to the sum
c_sum += arr[i];
// Condition when c_sum falls
// below zero
if (c_sum < 0)
return i;
}
// Complete array has a non-negative sum
return n;
} // Driver code var arr = [3, 5, -6];
var n = arr.length;
document.write( maxLen(arr, n)); </script> |
3
Time Complexity: O(n*log(n))
Auxiliary Space: O(1)