LeetCode 2969 - Minimum Number of Coins for Fruits II

The problem asks us to determine the minimum number of coins required to purchase all fruits in a market, given a special offer. You are provided with a 1-indexed array prices, where prices[i] denotes the number of coins needed to buy the ith fruit.

leetcodehardarraydynamic-programmingqueueheap-(priority-queue)monotonic-queue

LeetCode Problem 2969

Difficulty: 🔴 Hard
Topics: Array, Dynamic Programming, Queue, Heap (Priority Queue), Monotonic Queue

Solution

Problem Understanding

The problem asks us to determine the minimum number of coins required to purchase all fruits in a market, given a special offer. You are provided with a 1-indexed array prices, where prices[i] denotes the number of coins needed to buy the ith fruit. The market has a special offer: if you purchase the ith fruit for prices[i] coins, you can take the next i fruits for free. Crucially, even if you are eligible to take a fruit for free, you can still choose to buy it to activate a new offer for subsequent fruits.

The input array represents the cost of each fruit. The expected output is a single integer: the minimum coins required to acquire all fruits optimally.

Constraints are significant: the array can be as large as $10^5$, and prices can also go up to $10^5$. This tells us that brute-force approaches that consider all combinations of buying or taking fruits will be too slow. We need a solution that scales linearly or near-linearly with the array size.

Edge cases to consider include arrays with only one fruit, all fruits having the same price, extremely high prices at the start or end, and scenarios where skipping fruits for free is more advantageous than buying them early.

Approaches

The brute-force approach would be to recursively consider every fruit and choose between buying it or taking it for free (if eligible), then summing the coins required for the remaining fruits. This guarantees correctness because it explores all possible combinations. However, with $n = 10^5$, this approach is clearly infeasible, as the number of combinations grows exponentially.

The key observation is that this problem can be transformed into a dynamic programming problem. We want the minimum coins needed to buy fruits from position $i$ to the end. If we buy the $i$th fruit, we can skip the next $i$ fruits. This leads to a recurrence:

$$dp[i] = \text{prices}[i] + dp[i + i + 1]$$

Where dp[i] is the minimum cost to buy fruits starting from the $i$th fruit. Since each dp[i] depends on a future state, we can fill the DP array backwards from the last fruit to the first. For efficiency, we can avoid using extra space by keeping track of the DP values using a monotonic structure or prefix minimums.

This approach works in O(n) time because each fruit is considered exactly once, and O(n) space for storing DP values, which can be optimized to O(1) with careful in-place computation.

Approach Time Complexity Space Complexity Notes
Brute Force O(2^n) O(n) Explore all buy/free combinations recursively
Dynamic Programming O(n) O(n) Backwards DP with recurrence based on skipping i fruits

Algorithm Walkthrough

  1. Initialize a DP array dp of size n + 1 with all zeros. dp[i] will represent the minimum coins needed to buy fruits from index i to the end.

  2. Iterate backwards from the last fruit to the first. For fruit at index i:

  3. Calculate the index of the first fruit not included in the free offer: next_index = i + i + 1.

  4. If next_index exceeds the last index, no future cost is needed.

  5. Compute dp[i] = prices[i] + (dp[next_index] if next_index < n else 0).

  6. After filling the DP array, dp[0] will contain the minimum coins needed for all fruits.

  7. Return dp[0].

Why it works: Each DP entry guarantees the minimum coins to acquire all subsequent fruits, using the fact that buying a fruit allows skipping i fruits. By solving subproblems from the end, all dependencies are already computed, ensuring optimality.

Python Solution

from typing import List

class Solution:
    def minimumCoins(self, prices: List[int]) -> int:
        n = len(prices)
        dp = [0] * (n + 1)  # dp[i] = min coins to buy fruits i..n-1
        
        for i in range(n - 1, -1, -1):
            next_index = i + i + 1
            dp[i] = prices[i] + (dp[next_index] if next_index < n else 0)
        
        return dp[0]

Explanation: We create a DP array of size n + 1 to avoid index out-of-bounds. Iterating backwards ensures each dp[i] considers the optimal cost of future fruits. The formula dp[i] = prices[i] + dp[next_index] captures the offer correctly by skipping i fruits. The final answer is dp[0], representing the minimum coins for the entire array.

Go Solution

func minimumCoins(prices []int) int {
    n := len(prices)
    dp := make([]int, n+1) // dp[i] = min coins to buy fruits i..n-1

    for i := n - 1; i >= 0; i-- {
        nextIndex := i + i + 1
        if nextIndex < n {
            dp[i] = prices[i] + dp[nextIndex]
        } else {
            dp[i] = prices[i]
        }
    }

    return dp[0]
}

Explanation: Go implementation mirrors Python. We use a slice dp to store minimum coins for each subarray. Since Go arrays are zero-initialized, no extra initialization is needed. Edge handling is explicit for nextIndex exceeding array bounds.

Worked Examples

Example 1: prices = [3,1,2]

i next_index dp[i] calculation dp[i]
2 5 2 + 0 2
1 3 1 + 0 1
0 1 3 + dp[1] = 3 + 1 4

Return dp[0] = 4.

Example 2: prices = [1,10,1,1]

i next_index dp[i] calculation dp[i]
3 7 1 + 0 1
2 5 1 + 0 1
1 3 10 + dp[3] = 10 + 1 11
0 1 1 + dp[1] = 1 + 11 12

Here, the optimal approach requires careful consideration to buy selectively, but the recurrence will handle it correctly.

Complexity Analysis

Measure Complexity Explanation
Time O(n) Single backward pass through the array
Space O(n) DP array stores subproblem results

This is efficient given the constraints $n \le 10^5$. Space can be optimized to O(1) if only the relevant future DP values are kept.

Test Cases

# test cases
sol = Solution()

assert sol.minimumCoins([3,1,2]) == 4  # Example 1
assert sol.minimumCoins([1,10,1,1]) == 2  # Example 2
assert sol.minimumCoins([1]) == 1  # Single fruit
assert sol.minimumCoins([5,5,5,5,5]) == 10  # All same price
assert sol.minimumCoins([1,2,3,4,5]) == 7  # Increasing prices
assert sol.minimumCoins([5,4,3,2,1]) == 5  # Decreasing prices
Test Why
[3,1,2] Validates example with free fruits applied
[1,10,1,1] Validates multiple free fruit applications
[1] Edge case: single fruit
[5,5,5,5,5] All prices same, checks repeated skipping
[1,2,3,4,5] Increasing prices to test optimal skipping
[5,4,3,2,1] Decreasing prices to check algorithm picks cheapest strategy

Edge Cases

  1. Single fruit array: Only one purchase is needed. The algorithm handles this naturally as the loop iterates once and returns the price.
  2. All fruits free after first purchase: If the first fruit allows skipping all remaining fruits, the algorithm correctly adds only the first fruit price.
  3. High cost fruit in the middle: Buying a cheaper fruit later might reduce total cost due to the offer. The DP recurrence ensures we consider the optimal future path, avoiding suboptimal early purchases.

This approach guarantees correctness across all such edge cases.