java.lang.Object
g2901_3000.s2931_maximum_spending_after_buying_items.Solution

public class Solution extends java.lang.Object
2931 - Maximum Spending After Buying Items.

Hard

You are given a 0-indexed m * n integer matrix values, representing the values of m * n different items in m different shops. Each shop has n items where the jth item in the ith shop has a value of values[i][j]. Additionally, the items in the ith shop are sorted in non-increasing order of value. That is, values[i][j] >= values[i][j + 1] for all 0 <= j < n - 1.

On each day, you would like to buy a single item from one of the shops. Specifically, On the dth day you can:

  • Pick any shop i.
  • Buy the rightmost available item j for the price of values[i][j] * d. That is, find the greatest index j such that item j was never bought before, and buy it for the price of values[i][j] * d.

Note that all items are pairwise different. For example, if you have bought item 0 from shop 1, you can still buy item 0 from any other shop.

Return the maximum amount of money that can be spent on buying all m * n products.

Example 1:

Input: values = [[8,5,2],[6,4,1],[9,7,3]]

Output: 285

Explanation: On the first day, we buy product 2 from shop 1 for a price of values[1][2] * 1 = 1.

On the second day, we buy product 2 from shop 0 for a price of values[0][2] * 2 = 4.

On the third day, we buy product 2 from shop 2 for a price of values[2][2] * 3 = 9.

On the fourth day, we buy product 1 from shop 1 for a price of values[1][1] * 4 = 16.

On the fifth day, we buy product 1 from shop 0 for a price of values[0][1] * 5 = 25.

On the sixth day, we buy product 0 from shop 1 for a price of values[1][0] * 6 = 36.

On the seventh day, we buy product 1 from shop 2 for a price of values[2][1] * 7 = 49.

On the eighth day, we buy product 0 from shop 0 for a price of values[0][0] * 8 = 64.

On the ninth day, we buy product 0 from shop 2 for a price of values[2][0] * 9 = 81.

Hence, our total spending is equal to 285.

It can be shown that 285 is the maximum amount of money that can be spent buying all m * n products.

Example 2:

Input: values = [[10,8,6,4,2],[9,7,5,3,2]]

Output: 386

Explanation: On the first day, we buy product 4 from shop 0 for a price of values[0][4] * 1 = 2.

On the second day, we buy product 4 from shop 1 for a price of values[1][4] * 2 = 4.

On the third day, we buy product 3 from shop 1 for a price of values[1][3] * 3 = 9.

On the fourth day, we buy product 3 from shop 0 for a price of values[0][3] * 4 = 16.

On the fifth day, we buy product 2 from shop 1 for a price of values[1][2] * 5 = 25.

On the sixth day, we buy product 2 from shop 0 for a price of values[0][2] * 6 = 36.

On the seventh day, we buy product 1 from shop 1 for a price of values[1][1] * 7 = 49.

On the eighth day, we buy product 1 from shop 0 for a price of values[0][1] * 8 = 64

On the ninth day, we buy product 0 from shop 1 for a price of values[1][0] * 9 = 81.

On the tenth day, we buy product 0 from shop 0 for a price of values[0][0] * 10 = 100.

Hence, our total spending is equal to 386.

It can be shown that 386 is the maximum amount of money that can be spent buying all m * n products.

Constraints:

  • 1 <= m == values.length <= 10
  • 1 <= n == values[i].length <= 104
  • 1 <= values[i][j] <= 106
  • values[i] are sorted in non-increasing order.
  • Constructor Summary

    Constructors
    Constructor
    Description
     
  • Method Summary

    Modifier and Type
    Method
    Description
    long
    maxSpending(int[][] values)
     

    Methods inherited from class java.lang.Object

    clone, equals, finalize, getClass, hashCode, notify, notifyAll, toString, wait, wait, wait
  • Constructor Details

    • Solution

      public Solution()
  • Method Details

    • maxSpending

      public long maxSpending(int[][] values)