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leetcode-problemset/leetcode-cn/originData/[no content]maximum-profitable-triplets-with-increasing-prices-ii.json

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"question": {
"questionId": "3204",
"questionFrontendId": "2921",
"categoryTitle": "Algorithms",
"boundTopicId": 2508831,
"title": "Maximum Profitable Triplets With Increasing Prices II",
"titleSlug": "maximum-profitable-triplets-with-increasing-prices-ii",
"content": null,
"translatedTitle": "具有递增价格的最大利润三元组 II",
"translatedContent": null,
"isPaidOnly": true,
"difficulty": "Hard",
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"topicTags": [
{
"name": "Binary Indexed Tree",
"slug": "binary-indexed-tree",
"translatedName": "树状数组",
"__typename": "TopicTagNode"
},
{
"name": "Array",
"slug": "array",
"translatedName": "数组",
"__typename": "TopicTagNode"
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"hints": [
"Let's fix the middle chosen item for instance index <code>j</code>.",
"Lets define an array <code>max_right</code>, where <code>max_right[j]</code> represents the maximum <code>profit[k]</code> for every index <code>k > j</code> such that <code>prices[k] > prices[j]</code>.",
"Consider using a Fenwick tree to fill the <code>max_right</code>.",
"Do the same for items with an index <code>i < j</code> such that <code>prices[i] < prices[j]</code> and find the maximum <code>profit[i]</code> among them. Let's call this array <code>max_left</code>.",
"Now the profit when an item with the index <code>j</code> is the middle one would be <code>profit[j] + max_right[j] + max_left[j]</code>.",
"Finally, do the above procedure for all <code>j</code>'s and find the maximum profit among them. That would be the final answer."
],
"solution": null,
"status": null,
"sampleTestCase": "[10,2,3,4]\n[100,2,7,10]",
"metaData": "{\n \"name\": \"maxProfit\",\n \"params\": [\n {\n \"name\": \"prices\",\n \"type\": \"integer[]\"\n },\n {\n \"type\": \"integer[]\",\n \"name\": \"profits\"\n }\n ],\n \"return\": {\n \"type\": \"integer\"\n }\n}",
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