Buying Cars in Python

Suppose we have a list of prices of cars for sale, and we also have a budget k, we have to find the maximum number of cars we can buy.

So, if the input is like [80, 20, 10, 30, 80], k = 85, then the output will be 3 as we can buy three cars with prices 20, 10, 30.

Algorithm

To solve this, we will follow these steps −

• Initialize count := 0

• Sort the list of prices in ascending order

• For i in range 0 to size of prices, do

  • If prices[i] <= k, then

    • k := k - prices[i]

    • count := count + 1

  • Otherwise,

    • Break from the loop

• Return count

The key insight is to sort the prices first and buy the cheapest cars to maximize the count.

Example

Let us see the following implementation to get better understanding −

class Solution:
    def solve(self, prices, k):
        count = 0
        prices.sort()
        for i in range(len(prices)):
            if prices[i] <= k:
                k = k - prices[i]
                count += 1
            else:
                break
        return count

ob = Solution()
prices = [80, 20, 10, 30, 80]
budget = 85
print(ob.solve(prices, budget))

3

How It Works

The algorithm works as follows ?

• Sort prices: [10, 20, 30, 80, 80]

• Buy car 1: Price = 10, Budget left = 85 - 10 = 75, Count = 1

• Buy car 2: Price = 20, Budget left = 75 - 20 = 55, Count = 2

• Buy car 3: Price = 30, Budget left = 55 - 30 = 25, Count = 3

• Cannot buy: Price = 80 > 25, so we stop

Alternative Implementation

Here's a more Pythonic version using a simple loop ?

def max_cars(prices, budget):
    prices.sort()
    count = 0
    
    for price in prices:
        if price <= budget:
            budget -= price
            count += 1
        else:
            break
    
    return count

# Test the function
prices = [80, 20, 10, 30, 80]
budget = 85
result = max_cars(prices, budget)
print(f"Maximum cars you can buy: {result}")

Maximum cars you can buy: 3

Conclusion

This greedy algorithm maximizes the number of cars by always buying the cheapest available car first. The time complexity is O(n log n) due to sorting, and space complexity is O(1).