How do you solve Minimum Number of Coins for Fruits on LeetCode?

Minimum Number of Coins for Fruits (LeetCode #2944) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n²) time complexity and O(n) space complexity. Key insight: Dynamic programming helps in breaking down the problem into smaller subproblems.

What is the brute force solution for Minimum Number of Coins for Fruits?

The brute force approach for Minimum Number of Coins for Fruits is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves checking every possible way to purchase fruits and calculating the total cost. This means trying to buy each fruit and considering the free fruits you can get from that purchase. The optimal Optimal Solution approach improves this to O(n²).

What are the interview tips for Minimum Number of Coins for Fruits?

Always explain your thought process while coding. Consider edge cases, such as empty arrays or single fruit. Practice explaining your solution clearly and concisely.

What are common mistakes when solving Minimum Number of Coins for Fruits?

Not considering all possible free fruits after a purchase. Forgetting to initialize the dp array correctly.

#2944

Minimum Number of Coins for Fruits

Medium

ArrayDynamic ProgrammingQueueHeap (Priority Queue)Monotonic QueueDynamic ProgrammingGreedy Algorithms

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n²)
Space	O(1)	O(n)

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Intuition

Time O(n²)Space O(n)

Using dynamic programming, we can build a solution that calculates the minimum coins needed to buy all fruits by considering the best fruit to buy at each step. This way, we avoid redundant calculations.

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Algorithm

3 steps

1Step 1: Initialize a dp array where dp[i] represents the minimum coins needed to buy fruits from index i to the end.
2Step 2: Iterate from the last fruit to the first fruit, calculating the minimum cost for each fruit based on previous results.
3Step 3: For each fruit, consider buying it and adding the cost of the next fruits that can be taken for free.

solution.py9 lines

1def minCoins(prices):
2    n = len(prices)
3    dp = [0] * n
4    dp[n - 1] = prices[n - 1]
5    for i in range(n - 2, -1, -1):
6        dp[i] = prices[i] + dp[i + 1]
7        for j in range(i + 1, n):
8            dp[i] = min(dp[i], prices[i] + dp[j])
9    return dp[0]

ℹ

Complexity note: The time complexity remains O(n²) due to the nested loop structure, but we are storing results in the dp array to avoid redundant calculations. The space complexity is O(n) because we use an array to store the minimum costs.

1Dynamic programming helps in breaking down the problem into smaller subproblems.
2Choosing the right fruit to buy can lead to significant savings.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.