LeetCode 2994: Friday Purchases II Solution

Master LeetCode problem 2994 (Friday Purchases II ), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

2994. Friday Purchases II

Hard

Database

Problem Explanation

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 certain items on each day. At each day, we can either choose to buy the item at that day or skip it. We will compare the cost of buying the item with the total cost of buying it on the previous Friday and add it to the DP array. Finally, we return the minimum cost of buying all items on Friday.

Let's say dp[i][j] represents the minimum cost to buy j items by Friday (inclusive) considering the first i days.
At each day i, we have two choices:
1. Buy the item on that day: dp[i][j] = min(dp[i][j], dp[i-1][j-1] + cost[i])
2. Skip the item on that day: dp[i][j] = min(dp[i][j], dp[i-1][j])

Time Complexity: O(nk) where n is the number of days and k is the total number of items to buy. Space Complexity: O(nk) for the DP array.

Solution Code

public int minCost(int[] cost, int friday) {
    int n = cost.length;
    int[][] dp = new int[n + 1][friday + 1];
    
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= friday; j++) {
            dp[i][j] = Math.min(dp[i - 1][j - 1] + cost[i - 1], dp[i - 1][j]);
        }
    }
    
    return dp[n][friday];
}

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