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2473. Minimum Cost to Buy Apples

ArrayGraphHeap (Priority Queue)Shortest Path

Explanation:

This problem can be solved using dynamic programming. We can create a 2D DP array to keep track of the minimum cost to buy i kilograms of apples. We will iterate through the array and update the minimum cost based on the cost of buying single apples or bags of apples. At the end, we return the minimum cost to buy the required amount of apples.

Here are the steps:

  1. Initialize a 1D DP array dp of size n+1 where n is the required amount of apples.
  2. Initialize dp[0] = 0 since the cost to buy 0 kilograms of apples is 0.
  3. Iterate from 1 to n and for each i:
    • Calculate the cost of buying a single apple and a bag of apples.
    • Update dp[i] as the minimum of these two costs.
  4. Return dp[n] which represents the minimum cost to buy n kilograms of apples.

Time complexity: O(n) Space complexity: O(n)

:

View Leetcode Problem
public int minCostToBuyApples(int[] prices, int[] bags) {
    int n = bags.length;
    int[] dp = new int[n + 1];
    dp[0] = 0;
    for (int i = 1; i <= n; i++) {
        int singleAppleCost = i <= prices.length ? prices[i - 1] : Integer.MAX_VALUE;
        int bagCost = bags[i - 1];
        dp[i] = Math.min(singleAppleCost, bagCost + (i > 1 ? dp[i - 1] : 0));
    }
    return dp[n];
}

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