Given a list of prices of ‘N’ items. A person can buy only one item of any price and he can get other items for free such that the total price of the rest of the items doesn’t exceed the price of the bought item. The task is to find the maximum number of items the person can have.
Examples:
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
Approach: The person should buy the most expensive item and then start taking the items starting from the least pricing (until the total price is less 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.
Below is the implementation of the above approach:
// C++ implementation of // the above approach #include <bits/stdc++.h> 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<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;
} int main()
{ int n = 5;
int a[]= {5, 3, 1, 5, 6};
cout<<items(n, a); } //contributed by Arnab Kundu |
// 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<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
public static void main (String[] args) {
int n = 5 ;
int a[]= { 5 , 3 , 1 , 5 , 6 };
System.out.println(items(n, a));
}
//This code is contributed by ajit } |
# 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# 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 |
<?php //PHP implementation of // the above approach // Function to count the // total number of items function items( $n , $a ){
// Sort the prices
sort( $a );
// 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 =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 ;
} //Code driven $n = 5;
$a = array (5, 3, 1, 5, 6);
echo items( $n , $a );
//This code is contributed by ajit ?> |
<script> // 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));
</script> |
3
Complexity Analysis:
- Time Complexity: O(N*logN), as we are using inbuilt sort function.
- Auxiliary Space: O(1), as we are not using any extra space.