Minimum Number of Coins for Fruits

This problem exhibits optimal substructure and overlapping subproblems, making it a good candidate for dynamic programming. We can define a DP state dp[i] as the minimum cost to acquire the first i fruits. Our goal is to find dp[n], the minimum cost for all n fruits.

The base case is dp[0] = 0, as acquiring no fruits costs nothing.

To compute dp[i], we must ensure fruit i is acquired. This can happen if we buy fruit i, or if we get it for free from a prior purchase. We can generalize this by considering that to acquire fruit i, we must have bought some fruit k (where 1 <= k <= i) that 'covers' fruit i. Buying fruit k (at index k-1) gives fruits k+1 through 2k for free. Thus, fruit k covers fruit i if k=i or k+1 <= i <= 2k. The latter condition simplifies to ceil(i/2) <= k <= i-1.

So, to find dp[i], we can take the minimum over all valid choices of k from ceil(i/2) to i. For each such k, the cost is the price of fruit k (prices[k-1]) plus the minimum cost to acquire fruits 1 to k-1 (dp[k-1]). This gives the recurrence relation:

dp[i] = min_{k=ceil(i/2)}^{i} (dp[k-1] + prices[k-1])

We can implement this by iterating i from 1 to n and, for each i, iterating k through its valid range to find the minimum cost.

• Create a DP array dp of size n+1, where n is the number of fruits.
• Initialize dp[0] = 0, and other elements to a large value.
• Iterate i from 1 to n to compute dp[i], the minimum cost to acquire fruits 1 through i.
• For each i, calculate the cost by considering all possible fruits k (from ceil(i/2) to i) that could be purchased to cover fruit i.
• The cost of purchasing fruit k is prices[k-1], and we must have already acquired fruits 1 to k-1 with a minimum cost of dp[k-1]. The total cost for this choice is dp[k-1] + prices[k-1].
• Update dp[i] with the minimum cost found among all valid choices of k.
• The final answer is dp[n].

We use a bottom-up dynamic programming approach. Let dp[i] be the minimum cost to obtain the first i fruits (using 1-based indexing for fruits, so i ranges from 1 to n). The size of our dp array will be n+1.

• Initialization: dp[0] = 0. All other dp values can be initialized to infinity.

• Recurrence: To calculate dp[i], we consider all possible fruits k we could have purchased that would cover fruit i. A purchase of fruit k covers fruit i if i is k itself, or if i is one of the free fruits obtained from buying k (i.e., k+1 <= i <= 2k). This gives a range for k from ceil(i/2) to i. For any such k, the total cost would be the cost to acquire fruits up to k-1 (dp[k-1]) plus the cost of fruit k (prices[k-1]). We take the minimum over all these possibilities.

dp[i] = min(dp[k-1] + prices[k-1]) for k in [ceil(i/2), i]

• Final Answer: The minimum cost to acquire all n fruits is dp[n].

Here is the Java implementation:

class Solution {
    public int minimumCoins(int[] prices) {
        int n = prices.length;
        int[] dp = new int[n + 1];
        // dp[i] = min cost to acquire first i fruits (1-based index)
        
        dp[0] = 0;
        
        for (int i = 1; i <= n; i++) {
            dp[i] = Integer.MAX_VALUE;
            // To acquire fruit i, we must buy some fruit k (1-based)
            // such that buying k covers i.
            // This means k=i, or k+1 <= i <= 2k.
            // The range for k is [ceil(i/2), i].
            // For positive integers, ceil(i/2) is (i+1)/2 using integer division.
            int start_k = (i + 1) / 2;
            for (int k = start_k; k <= i; k++) {
                // Cost of buying fruit k is prices[k-1].
                // We must have acquired fruits 1...k-1, which costs dp[k-1].
                // Total cost for this choice is dp[k-1] + prices[k-1].
                dp[i] = Math.min(dp[i], dp[k - 1] + prices[k - 1]);
            }
        }
        
        return dp[n];
    }
}