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Description

Solutions

Count Maximum Profitable Groups 🍑

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A team of analysts at Amazon needs to analyze the stock prices of Amazon over a period of several months.

A group of consecutively chosen months is said to be maximum profitable if the price in its first or last month is the maximum for the group. More formally, a group of consecutive months [l, r] (1 ≤ l ≤ r ≤ n) is said to be maximum profitable if either:

stockPrice[l] = max(stockPrice[l], stockPrice[l + 1], ..., stockPrice[r])

or, stockPrice[r] = max(stockPrice[l], stockPrice[l + 1], ..., stockPrice[r])

Given prices over n consecutive months, find the number of maximum profitable groups which can be formed. Note that the months chosen must be consecutive, i.e., you must choose a subarray of the given array.

Function Description

Complete the function countMaximumProfitableGroups function in the editor below.

countMaximumProfitableGroups has the following parameter:

int stockPrice[n]: the stock prices

Returns

long integer: the number of maximum profitable groups

🍓 1000 thanks to spike for spike's incredible help! 🥑

Example 1:

Input:  stockPrice = [3, 1, 3, 5]
Output: 10 
Explanation:
The 10 possible groups are [3], [3, 1], [3, 1, 3], [3, 1, 3, 5], [1], [1, 3], [1, 3, 5], [3], [3, 5], [5]
In each group, the maximum price is in either the first or last position.

Example 2:

Input:  stockPrice = [1, 5, 2]
Output: 5 
Explanation:
There are 6 possible groups: [1], [1, 5], [1, 5, 2], [5], [5, 2], [2].
Only [1, 5, 2], is not maximum profitable because its maximum value 5 is not at either end of the group.

Example 3:

Input:  stockPrice = [2, 3, 2]
Output: 5 
Explanation:
All 5 groups other than prices [2, 3, 2] are maximum profitable. In [2, 3, 2], the 
maximum value 3 is neither the first nor the last element. Return 5.

Constraints:

1 ≤ n ≤ 5 * 10⁵
1 ≤ stockPrice[i] ≤ 10⁸