Minimum Number of Coins for Fruits

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Dynamic ProgrammingGreedy Algorithms

You are given an 0-indexed integer array prices where prices[i] denotes the number of coins needed to purchase the (i + 1)^th fruit.

The fruit market has the following reward for each fruit:

If you purchase the (i + 1)^th fruit at prices[i] coins, you can get any number of the next i fruits for free.

Note that even if you can take fruit j for free, you can still purchase it for prices[j - 1] coins to receive its reward.

Return the minimum number of coins needed to acquire all the fruits.

Example 1:

Input: prices = [3,1,2]

Output: 4

Explanation:

Purchase the 1^st fruit with prices[0] = 3 coins, you are allowed to take the 2^nd fruit for free.
Purchase the 2^nd fruit with prices[1] = 1 coin, you are allowed to take the 3^rd fruit for free.
Take the 3^rd fruit for free.

Note that even though you could take the 2^nd fruit for free as a reward of buying 1^st fruit, you purchase it to receive its reward, which is more optimal.

Example 2:

Input: prices = [1,10,1,1]

Output: 2

Explanation:

Purchase the 1^st fruit with prices[0] = 1 coin, you are allowed to take the 2^nd fruit for free.
Take the 2^nd fruit for free.
Purchase the 3^rd fruit for prices[2] = 1 coin, you are allowed to take the 4^th fruit for free.
Take the 4^t^h fruit for free.

Example 3:

Input: prices = [26,18,6,12,49,7,45,45]

Output: 39

Explanation:

Purchase the 1^st fruit with prices[0] = 26 coin, you are allowed to take the 2^nd fruit for free.
Take the 2^nd fruit for free.
Purchase the 3^rd fruit for prices[2] = 6 coin, you are allowed to take the 4^th, 5^th and 6^th (the next three) fruits for free.
Take the 4^t^h fruit for free.
Take the 5^t^h fruit for free.
Purchase the 6^th fruit with prices[5] = 7 coin, you are allowed to take the 8^th and 9^th fruit for free.
Take the 7^t^h fruit for free.
Take the 8^t^h fruit for free.

Note that even though you could take the 6^th fruit for free as a reward of buying 3^rd fruit, you purchase it to receive its reward, which is more optimal.

Constraints:

1 <= prices.length <= 1000
1 <= prices[i] <= 10⁵

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the possible values for each fruit price?
Can the price of a fruit be zero?
Is there a maximum number of fruits that I would have to purchase?
If it's not possible to purchase all the fruits, what value should I return?
Can the input array be empty or null?

Brute Force Solution

Approach

We need to figure out the fewest coins to buy all the fruits, given a rule that buying one fruit also gets you the next one for free. The brute force way is to try every possible combination of buying or skipping fruits, calculating the total cost for each.

Here's how the algorithm would work step-by-step:

Start by considering the first fruit. We have two choices: either buy it or skip it.
If we buy the first fruit, we automatically get the second fruit for free. Now we need to decide what to do about the third fruit: buy it or skip it.
If we skip the first fruit, we move on to the second fruit and decide whether to buy it or skip it.
We continue making these choices for each fruit, always considering whether we buy it or skip it. Each choice creates a different path.
For each path (or combination of buy/skip decisions), calculate the total number of coins we would spend.
Once we've gone through every possible path, compare the total cost of all the paths.
The path with the lowest total cost is the answer: the minimum number of coins needed.

Code Implementation

def minimum_coins_for_fruits_brute_force(fruit_prices):
    minimum_cost = float('inf')

    def calculate_cost(index, current_cost):
        nonlocal minimum_cost

        # If we've reached the end of the fruits, update the minimum cost
        if index >= len(fruit_prices):
            minimum_cost = min(minimum_cost, current_cost)
            return

        # Option 1: Buy the current fruit
        # Buying a fruit gives the next one for free
        calculate_cost(index + 2, current_cost + fruit_prices[index])

        # Option 2: Skip the current fruit
        # If we skip, move to the next fruit
        calculate_cost(index + 1, current_cost)

    calculate_cost(0, 0)
    return minimum_cost

Big(O) Analysis

Time Complexity

O(2^n) – The algorithm explores all possible combinations of buying or skipping each fruit. For each fruit, there are two choices: buy or skip. Since there are n fruits, this leads to 2 * 2 * ... * 2 (n times) possible paths or combinations. Therefore, the total number of operations grows exponentially with the input size n, resulting in a time complexity of O(2^n).

Space Complexity

O(N) – The described brute force approach explores all possible combinations of buying or skipping fruits using a recursive strategy. Each level of recursion corresponds to a fruit, and the maximum depth of the recursion tree is proportional to the number of fruits, N. Therefore, the maximum space used by the call stack would be proportional to N, resulting in O(N) space complexity.

Optimal Solution

Approach

The key to solving this problem efficiently is to use a dynamic programming approach. We build up the solution for larger problems from the solutions of smaller subproblems. Specifically, we calculate the minimum cost to buy fruits up to a certain point by considering whether or not to buy a coin at each point.

Here's how the algorithm would work step-by-step:

Imagine you're going through the fruits one by one, from left to right.
For each fruit, you have two choices: either buy a coin that covers it and the next *k* fruits, or don't buy a coin for it.
Figure out the cost of buying a coin at the current fruit. This includes the coin's cost plus the minimum cost to buy coins for the remaining fruits after the coin's effect ends.
Figure out the cost of *not* buying a coin at the current fruit. This means you have to have bought a coin in the last *k* fruits in order to cover the current fruit. Thus, you'll need to look back up to *k* fruits to see the previous coin cost.
Choose the cheaper option (either buying the coin or not buying it) for each fruit. Remember the cheapest cost so far.
Continue this process for all the fruits. The final result is the minimum total cost to buy all the fruits.

Code Implementation

def minimum_coins_for_fruits(prices, discount_interval, coin_cost):
    number_of_fruits = len(prices)
    minimum_cost = [0] * (number_of_fruits + 1)

    for i in range(1, number_of_fruits + 1):
        minimum_cost[i] = float('inf')

    for i in range(1, number_of_fruits + 1):
        # Option 1: Buy a coin at this fruit.
        buy_coin_cost = coin_cost + minimum_cost[min(number_of_fruits, i + discount_interval)]

        # Option 2: Do not buy a coin at this fruit. We must have bought a coin in the last k fruits.
        no_coin_cost = float('inf')

        for j in range(max(0, i - discount_interval), i):
            no_coin_cost = min(no_coin_cost, minimum_cost[j])

        # If we didn't buy a coin in the last k fruits, then it's impossible.
        if no_coin_cost == float('inf') and i > discount_interval:
            continue

        if i <= discount_interval:
            no_coin_cost = 0

        minimum_cost[i] = min(buy_coin_cost, no_coin_cost)

    return minimum_cost[number_of_fruits]

Big(O) Analysis

Time Complexity

O(n*k) – The algorithm iterates through each of the n fruits. For each fruit, it potentially looks back up to k previous fruits to determine the minimum cost of not buying a coin at the current fruit. This look back involves a constant time operation for each of the k previous fruits. Because the outer loop iterates n times and the inner (look-back) loop iterates at most k times, the overall time complexity is O(n*k).

Space Complexity

O(N) – The dynamic programming approach builds a solution from smaller subproblems, implying the storage of intermediate results. The explanation mentions remembering the cheapest cost so far for each fruit, suggesting an auxiliary array (or similar data structure) is used to store these minimum costs. This array's size is directly proportional to the number of fruits, N. Therefore, the space complexity is O(N).

Edge Cases

Case	How to Handle
Null or empty prices array	Return 0 immediately as no fruits can be bought.
Coins array is null or empty	Return 0 immediately as we have no coins to use.
Price of a fruit is negative	Invalid input; either throw an exception or treat as zero.
A fruit price is zero	If zero price fruits exist, treat them as free and solve normally or return 0 if all fruits are zero priced.
No combination of coins can sum up to a fruit price	Return -1 indicating that the desired combination is impossible.
Large number of coins and fruit prices leading to integer overflow when calculating subproblems	Use larger data types (e.g., long) to store intermediate results to prevent potential integer overflows.
Very large prices array, leading to excessive memory usage	Optimize the DP table to be one dimensional (if possible) or use other space-saving techniques.
Coins with value zero exist	Handle this case by filtering out the zero value coins before the algorithm, or skipping them in the main logic since they don't contribute.

Null or empty prices array

How to Handle:

Return 0 immediately as no fruits can be bought.

Coins array is null or empty

How to Handle:

Return 0 immediately as we have no coins to use.

Price of a fruit is negative

How to Handle:

Invalid input; either throw an exception or treat as zero.

A fruit price is zero

How to Handle:

If zero price fruits exist, treat them as free and solve normally or return 0 if all fruits are zero priced.

No combination of coins can sum up to a fruit price

How to Handle:

Return -1 indicating that the desired combination is impossible.

Large number of coins and fruit prices leading to integer overflow when calculating subproblems

How to Handle:

Use larger data types (e.g., long) to store intermediate results to prevent potential integer overflows.

Very large prices array, leading to excessive memory usage

How to Handle:

Optimize the DP table to be one dimensional (if possible) or use other space-saving techniques.

Coins with value zero exist

How to Handle:

Handle this case by filtering out the zero value coins before the algorithm, or skipping them in the main logic since they don't contribute.