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2907. Maximum Profitable Triplets With Increasing Prices I
Description
Given the 0indexed arrays prices
and profits
of length n
. There are n
items in an store where the i^{th}
item has a price of prices[i]
and a profit of profits[i]
.
We have to pick three items with the following condition:
prices[i] < prices[j] < prices[k]
wherei < j < k
.
If we pick items with indices i
, j
and k
satisfying the above condition, the profit would be profits[i] + profits[j] + profits[k]
.
Return the maximum profit we can get, and 1
if it's not possible to pick three items with the given condition.
Example 1:
Input: prices = [10,2,3,4], profits = [100,2,7,10] Output: 19 Explanation: We can't pick the item with index i=0 since there are no indices j and k such that the condition holds. So the only triplet we can pick, are the items with indices 1, 2 and 3 and it's a valid pick since prices[1] < prices[2] < prices[3]. The answer would be sum of their profits which is 2 + 7 + 10 = 19.
Example 2:
Input: prices = [1,2,3,4,5], profits = [1,5,3,4,6] Output: 15 Explanation: We can select any triplet of items since for each triplet of indices i, j and k such that i < j < k, the condition holds. Therefore the maximum profit we can get would be the 3 most profitable items which are indices 1, 3 and 4. The answer would be sum of their profits which is 5 + 4 + 6 = 15.
Example 3:
Input: prices = [4,3,2,1], profits = [33,20,19,87] Output: 1 Explanation: We can't select any triplet of indices such that the condition holds, so we return 1.
Constraints:
3 <= prices.length == profits.length <= 2000
1 <= prices[i] <= 10^{6}
1 <= profits[i] <= 10^{6}
Solutions
Solution 1: Enumerate the Middle Element
We can enumerate the middle element $profits[j]$, and then enumerate the left element $profits[i]$ and the right element $profits[k]$. For each $profits[j]$, we need to find the maximum $profits[i]$ and the maximum $profits[k]$ such that $prices[i] < prices[j] < prices[k]$. We define $left$ as the maximum value on the left of $profits[j]$, and $right$ as the maximum value on the right of $profits[j]$. If they exist, we update the answer as $ans = \max(ans, left + profits[j] + right)$.
The time complexity is $O(n^2)$, where $n$ is the length of the array. The space complexity is $O(1)$.
Solution 2: Binary Indexed Tree
We can use two Binary Indexed Trees (BITs) to maintain the maximum profit on the left and right of each price, respectively. Then, we enumerate the middle price, query the maximum profit on both sides through the BIT, and finally take the maximum value.
The time complexity is $O(n \times \log M)$, and the space complexity is $O(M)$. Here, $n$ is the length of the array $prices$, and $M$ is the maximum value in the array $prices$. In this problem, $M \le 10^6$.
Since the length of $prices$ does not exceed $2000$, and the value of $prices[i]$ reaches $10^6$, we can discretize $prices$, which can reduce the space complexity to $O(n)$.

class Solution { public int maxProfit(int[] prices, int[] profits) { int n = prices.length; int ans = 1; for (int j = 0; j < n; ++j) { int left = 0, right = 0; for (int i = 0; i < j; ++i) { if (prices[i] < prices[j]) { left = Math.max(left, profits[i]); } } for (int k = j + 1; k < n; ++k) { if (prices[j] < prices[k]) { right = Math.max(right, profits[k]); } } if (left > 0 && right > 0) { ans = Math.max(ans, left + profits[j] + right); } } return ans; } }

class Solution { public: int maxProfit(vector<int>& prices, vector<int>& profits) { int n = prices.size(); int ans = 1; for (int j = 0; j < n; ++j) { int left = 0, right = 0; for (int i = 0; i < j; ++i) { if (prices[i] < prices[j]) { left = max(left, profits[i]); } } for (int k = j + 1; k < n; ++k) { if (prices[j] < prices[k]) { right = max(right, profits[k]); } } if (left && right) { ans = max(ans, left + profits[j] + right); } } return ans; } };

class Solution: def maxProfit(self, prices: List[int], profits: List[int]) > int: n = len(prices) ans = 1 for j, x in enumerate(profits): left = right = 0 for i in range(j): if prices[i] < prices[j] and left < profits[i]: left = profits[i] for k in range(j + 1, n): if prices[j] < prices[k] and right < profits[k]: right = profits[k] if left and right: ans = max(ans, left + x + right) return ans

func maxProfit(prices []int, profits []int) int { n := len(prices) ans := 1 for j, x := range profits { left, right := 0, 0 for i := 0; i < j; i++ { if prices[i] < prices[j] { left = max(left, profits[i]) } } for k := j + 1; k < n; k++ { if prices[j] < prices[k] { right = max(right, profits[k]) } } if left > 0 && right > 0 { ans = max(ans, left+x+right) } } return ans }

function maxProfit(prices: number[], profits: number[]): number { const n = prices.length; let ans = 1; for (let j = 0; j < n; ++j) { let [left, right] = [0, 0]; for (let i = 0; i < j; ++i) { if (prices[i] < prices[j]) { left = Math.max(left, profits[i]); } } for (let k = j + 1; k < n; ++k) { if (prices[j] < prices[k]) { right = Math.max(right, profits[k]); } } if (left > 0 && right > 0) { ans = Math.max(ans, left + profits[j] + right); } } return ans; }

impl Solution { pub fn max_profit(prices: Vec<i32>, profits: Vec<i32>) > i32 { let n = prices.len(); let mut ans = 1; for j in 0..n { let mut left = 0; let mut right = 0; for i in 0..j { if prices[i] < prices[j] { left = left.max(profits[i]); } } for k in j + 1..n { if prices[j] < prices[k] { right = right.max(profits[k]); } } if left > 0 && right > 0 { ans = ans.max(left + profits[j] + right); } } ans } }