2280. Minimum Lines to Represent a Line Chart

Description

You are given a 2D integer array stockPrices where stockPrices[i] = [dayi, pricei] indicates the price of the stock on day dayi is pricei. A line chart is created from the array by plotting the points on an XY plane with the X-axis representing the day and the Y-axis representing the price and connecting adjacent points. One such example is shown below:

Return the minimum number of lines needed to represent the line chart.

 

Example 1:

Input: stockPrices = [[1,7],[2,6],[3,5],[4,4],[5,4],[6,3],[7,2],[8,1]]
Output: 3
Explanation:
The diagram above represents the input, with the X-axis representing the day and Y-axis representing the price.
The following 3 lines can be drawn to represent the line chart:
- Line 1 (in red) from (1,7) to (4,4) passing through (1,7), (2,6), (3,5), and (4,4).
- Line 2 (in blue) from (4,4) to (5,4).
- Line 3 (in green) from (5,4) to (8,1) passing through (5,4), (6,3), (7,2), and (8,1).
It can be shown that it is not possible to represent the line chart using less than 3 lines.

Example 2:

Input: stockPrices = [[3,4],[1,2],[7,8],[2,3]]
Output: 1
Explanation:
As shown in the diagram above, the line chart can be represented with a single line.

 

Constraints:

  • 1 <= stockPrices.length <= 105
  • stockPrices[i].length == 2
  • 1 <= dayi, pricei <= 109
  • All dayi are distinct.

Solutions

Solution 1

Python Code
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class Solution:
    def minimumLines(self, stockPrices: List[List[int]]) -> int:
        stockPrices.sort()
        dx, dy = 0, 1
        ans = 0
        for (x, y), (x1, y1) in pairwise(stockPrices):
            dx1, dy1 = x1 - x, y1 - y
            if dy * dx1 != dx * dy1:
                ans += 1
            dx, dy = dx1, dy1
        return ans

Java Code
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class Solution {
    public int minimumLines(int[][] stockPrices) {
        Arrays.sort(stockPrices, (a, b) -> a[0] - b[0]);
        int dx = 0, dy = 1;
        int ans = 0;
        for (int i = 1; i < stockPrices.length; ++i) {
            int x = stockPrices[i - 1][0], y = stockPrices[i - 1][1];
            int x1 = stockPrices[i][0], y1 = stockPrices[i][1];
            int dx1 = x1 - x, dy1 = y1 - y;
            if (dy * dx1 != dx * dy1) {
                ++ans;
            }
            dx = dx1;
            dy = dy1;
        }
        return ans;
    }
}

C++ Code
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class Solution {
public:
    int minimumLines(vector<vector<int>>& stockPrices) {
        sort(stockPrices.begin(), stockPrices.end());
        int dx = 0, dy = 1;
        int ans = 0;
        for (int i = 1; i < stockPrices.size(); ++i) {
            int x = stockPrices[i - 1][0], y = stockPrices[i - 1][1];
            int x1 = stockPrices[i][0], y1 = stockPrices[i][1];
            int dx1 = x1 - x, dy1 = y1 - y;
            if ((long long) dy * dx1 != (long long) dx * dy1) ++ans;
            dx = dx1;
            dy = dy1;
        }
        return ans;
    }
};

Go Code
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func minimumLines(stockPrices [][]int) int {
	ans := 0
	sort.Slice(stockPrices, func(i, j int) bool { return stockPrices[i][0] < stockPrices[j][0] })
	for i, dx, dy := 1, 0, 1; i < len(stockPrices); i++ {
		x, y := stockPrices[i-1][0], stockPrices[i-1][1]
		x1, y1 := stockPrices[i][0], stockPrices[i][1]
		dx1, dy1 := x1-x, y1-y
		if dy*dx1 != dx*dy1 {
			ans++
		}
		dx, dy = dx1, dy1
	}
	return ans
}

TypeScript Code
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function minimumLines(stockPrices: number[][]): number {
    const n = stockPrices.length;
    stockPrices.sort((a, b) => a[0] - b[0]);
    let ans = 0;
    let pre = [BigInt(0), BigInt(0)];
    for (let i = 1; i < n; i++) {
        const [x1, y1] = stockPrices[i - 1];
        const [x2, y2] = stockPrices[i];
        const dx = BigInt(x2 - x1),
            dy = BigInt(y2 - y1);
        if (i == 1 || dx * pre[1] !== dy * pre[0]) ans++;
        pre = [dx, dy];
    }
    return ans;
}