g2901_3000.s2931_maximum_spending_after_buying_items.readme.md Maven / Gradle / Ivy
Go to download
Show more of this group Show more artifacts with this name
Show all versions of leetcode-in-kotlin Show documentation
Show all versions of leetcode-in-kotlin Show documentation
Kotlin-based LeetCode algorithm problem solutions, regularly updated
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.