当一个项目总成本的其他项目免费时，获得最大项目

给定“N”件商品的价格列表。一个人只能以任何价格购买一件物品，他可以免费获得其他物品，以便其余物品的总价不超过所购买物品的价格。任务是找到这个人可以拥有的最大数量的物品。
例子：

Input: n = 5, arr = {5, 3, 1, 5, 6} 
Output: 3 
The person can buy any item of price 5 or 6 and download items of prices 1 and 3 for free. So, he can get at most 3 items.
Input: n = 2, arr = {7, 7} 
Output: 2 
 

方法：

The person should buy the most expensive item and then start taking the items starting from the least pricing (until the total price is lest than or equal to the bought item) in order to maximize the total number of items. 
Thus, we sort the list of prices and choose the last element, then we will iterate from 1st index to n-2 index and check if the total sum is less than or equal to the last element. 
 

下面是上述方法的实现：

C++

// C++ implementation of
// the above approach
#include 
using namespace std;
 
// Function to count the
// total number of items
int  items(int n, int a[]){
 
    // Sort the prices
    sort(a,a+n);
 
    // Choose the last element
    int z = a[n-1];
 
    // Initial count of item
    int x = 1;
 
    // Sum to keep track of
    // the total price
    // of free items
    int s = 0;
    for (int i=0;i

---

Java

// Java implementation of
// the above approach
import java.util.Arrays;
import java.io.*;
 
class GFG {
 
// Function to count the
// total number of items
static int items(int n, int a[]){
 
    // Sort the prices
    Arrays.sort(a);
 
    // Choose the last element
    int z = a[n-1];
 
    // Initial count of item
    int x = 1;
 
    // Sum to keep track of
    // the total price
    // of free items
    int s = 0;
    for (int i=0;i

---

Python3

# Python3 implementation of
# the above approach
 
# Function to count the
# total number of items
def items(n, a):
 
    # Sort the prices
    a.sort()
 
    # Choose the last element
    z = a[n-1]
 
    # Initial count of item
    x = 1
 
    # Sum to keep track of
    # the total price
    # of free items
    s = 0
    for i in range(0, n-1):
 
        s += a[i]
 
        # If total is less than
        # or equal to z then
        # we will add 1 to the answer
        if (s <= z):
            x+= 1
        else:
            break
    return x
 
n = 5
a = [5, 3, 1, 5, 6]
print(items(n, a))

C#

// C# implementation of the
// above approach
using System;
 
class GFG
{
// Function to count the
// total number of items
static int items(int n, int []a)
{
 
    // Sort the prices
    Array.Sort(a);
 
    // Choose the last element
    int z = a[n - 1];
 
    // Initial count of item
    int x = 1;
 
    // Sum to keep track of
    // the total price
    // of free items
    int s = 0;
    for (int i = 0; i < n - 1; i++)
    {
        s += a[i];
 
        // If total is less than or equal to z
        // then we will add 1 to the answer
        if (s <= z)
            x += 1;
        else
            break;
    }
    return x;
}
 
// Driver code
static public void Main ()
{
    int n = 5;
    int []a = {5, 3, 1, 5, 6};
    Console.WriteLine(items(n, a));
}
}
 
// This code is contributed
// by akt_mit

Javascript

    // Javascript implementation of the above approach
     
    // Function to count the
    // total number of items
    function items(n, a)
    {
 
        // Sort the prices
        a.sort(function(a, b){return a - b});
 
        // Choose the last element
        let z = a[n - 1];
 
        // Initial count of item
        let x = 1;
 
        // Sum to keep track of
        // the total price
        // of free items
        let s = 0;
        for (let i = 0; i < n - 1; i++)
        {
            s += a[i];
 
            // If total is less than or equal to z
            // then we will add 1 to the answer
            if (s <= z)
                x += 1;
            else
                break;
        }
        return x;
    }
     
    let n = 5;
    let a = [5, 3, 1, 5, 6];
    document.write(items(n, a));
     

输出：

3

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