{ "data": { "question": { "questionId": "3107", "questionFrontendId": "100043", "categoryTitle": "Algorithms", "boundTopicId": 2520069, "title": "Maximum Spending After Buying Items", "titleSlug": "maximum-spending-after-buying-items", "content": "
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:
i
.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.
\n
Example 1:
\n\n\nInput: values = [[8,5,2],[6,4,1],[9,7,3]]\nOutput: 285\nExplanation: On the first day, we buy product 2 from shop 1 for a price of values[1][2] * 1 = 1.\nOn the second day, we buy product 2 from shop 0 for a price of values[0][2] * 2 = 4.\nOn the third day, we buy product 2 from shop 2 for a price of values[2][2] * 3 = 9.\nOn the fourth day, we buy product 1 from shop 1 for a price of values[1][1] * 4 = 16.\nOn the fifth day, we buy product 1 from shop 0 for a price of values[0][1] * 5 = 25.\nOn the sixth day, we buy product 0 from shop 1 for a price of values[1][0] * 6 = 36.\nOn the seventh day, we buy product 1 from shop 2 for a price of values[2][1] * 7 = 49.\nOn the eighth day, we buy product 0 from shop 0 for a price of values[0][0] * 8 = 64.\nOn the ninth day, we buy product 0 from shop 2 for a price of values[2][0] * 9 = 81.\nHence, our total spending is equal to 285.\nIt can be shown that 285 is the maximum amount of money that can be spent buying all m * n products. \n\n\n
Example 2:
\n\n\nInput: values = [[10,8,6,4,2],[9,7,5,3,2]]\nOutput: 386\nExplanation: On the first day, we buy product 4 from shop 0 for a price of values[0][4] * 1 = 2.\nOn the second day, we buy product 4 from shop 1 for a price of values[1][4] * 2 = 4.\nOn the third day, we buy product 3 from shop 1 for a price of values[1][3] * 3 = 9.\nOn the fourth day, we buy product 3 from shop 0 for a price of values[0][3] * 4 = 16.\nOn the fifth day, we buy product 2 from shop 1 for a price of values[1][2] * 5 = 25.\nOn the sixth day, we buy product 2 from shop 0 for a price of values[0][2] * 6 = 36.\nOn the seventh day, we buy product 1 from shop 1 for a price of values[1][1] * 7 = 49.\nOn the eighth day, we buy product 1 from shop 0 for a price of values[0][1] * 8 = 64\nOn the ninth day, we buy product 0 from shop 1 for a price of values[1][0] * 9 = 81.\nOn the tenth day, we buy product 0 from shop 0 for a price of values[0][0] * 10 = 100.\nHence, our total spending is equal to 386.\nIt can be shown that 386 is the maximum amount of money that can be spent buying all m * n products.\n\n\n
\n
Constraints:
\n\n1 <= m == values.length <= 10
1 <= n == values[i].length <= 104
1 <= values[i][j] <= 106
values[i]
are sorted in non-increasing order.给你一个下标从 0 开始大小为 m * n
的整数矩阵 values
,表示 m
个不同商店里 m * n
件不同的物品。每个商店有 n
件物品,第 i
个商店的第 j
件物品的价值为 values[i][j]
。除此以外,第 i
个商店的物品已经按照价值非递增排好序了,也就是说对于所有 0 <= j < n - 1
都有 values[i][j] >= values[i][j + 1]
。
每一天,你可以在一个商店里购买一件物品。具体来说,在第 d
天,你可以:
i
。j
,开销为 values[i][j] * d
。换句话说,选择该商店中还没购买过的物品中最大的下标 j
,并且花费 values[i][j] * d
去购买。注意,所有物品都视为不同的物品。比方说如果你已经从商店 1
购买了物品 0
,你还可以在别的商店里购买其他商店的物品 0
。
请你返回购买所有 m * n
件物品需要的 最大开销 。
\n\n
示例 1:
\n\n\n输入:values = [[8,5,2],[6,4,1],[9,7,3]]\n输出:285\n解释:第一天,从商店 1 购买物品 2 ,开销为 values[1][2] * 1 = 1 。\n第二天,从商店 0 购买物品 2 ,开销为 values[0][2] * 2 = 4 。\n第三天,从商店 2 购买物品 2 ,开销为 values[2][2] * 3 = 9 。\n第四天,从商店 1 购买物品 1 ,开销为 values[1][1] * 4 = 16 。\n第五天,从商店 0 购买物品 1 ,开销为 values[0][1] * 5 = 25 。\n第六天,从商店 1 购买物品 0 ,开销为 values[1][0] * 6 = 36 。\n第七天,从商店 2 购买物品 1 ,开销为 values[2][1] * 7 = 49 。\n第八天,从商店 0 购买物品 0 ,开销为 values[0][0] * 8 = 64 。\n第九天,从商店 2 购买物品 0 ,开销为 values[2][0] * 9 = 81 。\n所以总开销为 285 。\n285 是购买所有 m * n 件物品的最大总开销。\n\n\n
示例 2:
\n\n\n输入:values = [[10,8,6,4,2],[9,7,5,3,2]]\n输出:386\n解释:第一天,从商店 0 购买物品 4 ,开销为 values[0][4] * 1 = 2 。\n第二天,从商店 1 购买物品 4 ,开销为 values[1][4] * 2 = 4 。\n第三天,从商店 1 购买物品 3 ,开销为 values[1][3] * 3 = 9 。\n第四天,从商店 0 购买物品 3 ,开销为 values[0][3] * 4 = 16 。\n第五天,从商店 1 购买物品 2 ,开销为 values[1][2] * 5 = 25 。\n第六天,从商店 0 购买物品 2 ,开销为 values[0][2] * 6 = 36 。\n第七天,从商店 1 购买物品 1 ,开销为 values[1][1] * 7 = 49 。\n第八天,从商店 0 购买物品 1 ,开销为 values[0][1] * 8 = 64 。\n第九天,从商店 1 购买物品 0 ,开销为 values[1][0] * 9 = 81 。\n第十天,从商店 0 购买物品 0 ,开销为 values[0][0] * 10 = 100 。\n所以总开销为 386 。\n386 是购买所有 m * n 件物品的最大总开销。\n\n\n
\n\n
提示:
\n\n1 <= m == values.length <= 10
1 <= n == values[i].length <= 104
1 <= values[i][j] <= 106
values[i]
按照非递增顺序排序。1
to m * n
.",
"On each day, buy the product that minimizes values[i][values[i].length - 1]
, and pop it from values[i]
."
],
"solution": null,
"status": null,
"sampleTestCase": "[[8,5,2],[6,4,1],[9,7,3]]",
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"judgeType": "large",
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