2921. Maximum Profitable Triplets With Increasing Prices II

Description

Given the 0-indexed 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] where i < 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 <= 50000
1 <= prices[i] <= 5000
1 <= profits[i] <= 10⁶

Solutions

class BinaryIndexedTree {
    private int n;
    private int[] c;

    public BinaryIndexedTree(int n) {
        this.n = n;
        c = new int[n + 1];
    }

    public void update(int x, int v) {
        while (x <= n) {
            c[x] = Math.max(c[x], v);
            x += x & -x;
        }
    }

    public int query(int x) {
        int mx = 0;
        while (x > 0) {
            mx = Math.max(mx, c[x]);
            x -= x & -x;
        }
        return mx;
    }
}

class Solution {
    public int maxProfit(int[] prices, int[] profits) {
        int n = prices.length;
        int[] left = new int[n];
        int[] right = new int[n];
        int m = 0;
        for (int x : prices) {
            m = Math.max(m, x);
        }
        BinaryIndexedTree tree1 = new BinaryIndexedTree(m + 1);
        BinaryIndexedTree tree2 = new BinaryIndexedTree(m + 1);
        for (int i = 0; i < n; ++i) {
            int x = prices[i];
            left[i] = tree1.query(x - 1);
            tree1.update(x, profits[i]);
        }
        for (int i = n - 1; i >= 0; --i) {
            int x = m + 1 - prices[i];
            right[i] = tree2.query(x - 1);
            tree2.update(x, profits[i]);
        }
        int ans = -1;
        for (int i = 0; i < n; ++i) {
            if (left[i] > 0 && right[i] > 0) {
                ans = Math.max(ans, left[i] + profits[i] + right[i]);
            }
        }
        return ans;
    }
}

class BinaryIndexedTree {
private:
    int n;
    vector<int> c;

public:
    BinaryIndexedTree(int n) {
        this->n = n;
        c.resize(n + 1, 0);
    }

    void update(int x, int v) {
        while (x <= n) {
            c[x] = max(c[x], v);
            x += x & -x;
        }
    }

    int query(int x) {
        int mx = 0;
        while (x > 0) {
            mx = max(mx, c[x]);
            x -= x & -x;
        }
        return mx;
    }
};

class Solution {
public:
    int maxProfit(vector<int>& prices, vector<int>& profits) {
        int n = prices.size();
        vector<int> left(n, 0);
        vector<int> right(n, 0);
        int m = *max_element(prices.begin(), prices.end());
        BinaryIndexedTree tree1(m + 1);
        BinaryIndexedTree tree2(m + 1);
        for (int i = 0; i < n; ++i) {
            int x = prices[i];
            left[i] = tree1.query(x - 1);
            tree1.update(x, profits[i]);
        }
        for (int i = n - 1; i >= 0; --i) {
            int x = m + 1 - prices[i];
            right[i] = tree2.query(x - 1);
            tree2.update(x, profits[i]);
        }
        int ans = -1;
        for (int i = 0; i < n; ++i) {
            if (left[i] > 0 && right[i] > 0) {
                ans = max(ans, left[i] + profits[i] + right[i]);
            }
        }
        return ans;
    }
};

class BinaryIndexedTree:
    def __init__(self, n: int):
        self.n = n
        self.c = [0] * (n + 1)

    def update(self, x: int, v: int):
        while x <= self.n:
            self.c[x] = max(self.c[x], v)
            x += x & -x

    def query(self, x: int) -> int:
        mx = 0
        while x:
            mx = max(mx, self.c[x])
            x -= x & -x
        return mx


class Solution:
    def maxProfit(self, prices: List[int], profits: List[int]) -> int:
        n = len(prices)
        left = [0] * n
        right = [0] * n

        m = max(prices)
        tree1 = BinaryIndexedTree(m + 1)
        tree2 = BinaryIndexedTree(m + 1)

        for i, x in enumerate(prices):
            left[i] = tree1.query(x - 1)
            tree1.update(x, profits[i])
        for i in range(n - 1, -1, -1):
            x = m + 1 - prices[i]
            right[i] = tree2.query(x - 1)
            tree2.update(x, profits[i])

        return max(
            (l + x + r for l, x, r in zip(left, profits, right) if l and r), default=-1
        )

type BinaryIndexedTree struct {
	n int
	c []int
}

func NewBinaryIndexedTree(n int) BinaryIndexedTree {
	c := make([]int, n+1)
	return BinaryIndexedTree{n: n, c: c}
}

func (bit *BinaryIndexedTree) update(x, v int) {
	for x <= bit.n {
		bit.c[x] = max(bit.c[x], v)
		x += x & -x
	}
}

func (bit *BinaryIndexedTree) query(x int) int {
	mx := 0
	for x > 0 {
		mx = max(mx, bit.c[x])
		x -= x & -x
	}
	return mx
}

func maxProfit(prices []int, profits []int) int {
	n := len(prices)
	left := make([]int, n)
	right := make([]int, n)
	m := 0

	for _, x := range prices {
		m = max(m, x)
	}

	tree1 := NewBinaryIndexedTree(m + 1)
	tree2 := NewBinaryIndexedTree(m + 1)

	for i, x := range prices {
		left[i] = tree1.query(x - 1)
		tree1.update(x, profits[i])
	}

	for i := n - 1; i >= 0; i-- {
		x := m + 1 - prices[i]
		right[i] = tree2.query(x - 1)
		tree2.update(x, profits[i])
	}

	ans := -1

	for i := 0; i < n; i++ {
		if left[i] > 0 && right[i] > 0 {
			ans = max(ans, left[i]+profits[i]+right[i])
		}
	}

	return ans
}

class BinaryIndexedTree {
    private n: number;
    private c: number[];

    constructor(n: number) {
        this.n = n;
        this.c = Array(n + 1).fill(0);
    }

    update(x: number, v: number): void {
        while (x <= this.n) {
            this.c[x] = Math.max(this.c[x], v);
            x += x & -x;
        }
    }

    query(x: number): number {
        let mx = 0;
        while (x > 0) {
            mx = Math.max(mx, this.c[x]);
            x -= x & -x;
        }
        return mx;
    }
}

function maxProfit(prices: number[], profits: number[]): number {
    const n: number = prices.length;
    const left: number[] = Array(n).fill(0);
    const right: number[] = Array(n).fill(0);
    const m = Math.max(...prices);

    const tree1: BinaryIndexedTree = new BinaryIndexedTree(m + 1);
    const tree2: BinaryIndexedTree = new BinaryIndexedTree(m + 1);

    for (let i = 0; i < n; i++) {
        const x: number = prices[i];
        left[i] = tree1.query(x - 1);
        tree1.update(x, profits[i]);
    }

    for (let i = n - 1; i >= 0; i--) {
        const x: number = m + 1 - prices[i];
        right[i] = tree2.query(x - 1);
        tree2.update(x, profits[i]);
    }

    let ans: number = -1;

    for (let i = 0; i < n; i++) {
        if (left[i] > 0 && right[i] > 0) {
            ans = Math.max(ans, left[i] + profits[i] + right[i]);
        }
    }

    return ans;
}

struct BinaryIndexedTree {
    n: usize,
    c: Vec<i32>,
}

impl BinaryIndexedTree {
    fn new(n: usize) -> BinaryIndexedTree {
        BinaryIndexedTree {
            n,
            c: vec![0; n + 1],
        }
    }

    fn update(&mut self, x: usize, v: i32) {
        let mut x = x;
        while x <= self.n {
            self.c[x] = self.c[x].max(v);
            x += x & x.wrapping_neg();
        }
    }

    fn query(&self, x: usize) -> i32 {
        let mut x = x;
        let mut mx = 0;
        while x > 0 {
            mx = mx.max(self.c[x]);
            x -= x & x.wrapping_neg();
        }
        mx
    }
}

impl Solution {
    pub fn max_profit(prices: Vec<i32>, profits: Vec<i32>) -> i32 {
        let n = prices.len();
        let mut left = vec![0; n];
        let mut right = vec![0; n];
        let m = prices.iter().cloned().max().unwrap_or(0);

        let mut tree1 = BinaryIndexedTree::new(m as usize + 1);
        let mut tree2 = BinaryIndexedTree::new(m as usize + 1);

        for i in 0..n {
            let x = prices[i] as usize;
            left[i] = tree1.query(x - 1);
            tree1.update(x, profits[i]);
        }

        for i in (0..n).rev() {
            let x = (m + 1 - prices[i]) as usize;
            right[i] = tree2.query(x - 1);
            tree2.update(x, profits[i]);
        }

        let mut ans = -1;
        for i in 0..n {
            if left[i] > 0 && right[i] > 0 {
                ans = ans.max(left[i] + profits[i] + right[i]);
            }
        }

        ans
    }
}

2921 - Maximum Profitable Triplets With Increasing Prices II

2921. Maximum Profitable Triplets With Increasing Prices II

Description

Solutions

All Problems

All Solutions

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2921. Maximum Profitable Triplets With Increasing Prices II

Description

Solutions

All Problems

All Solutions