Welcome to Subscribe On Youtube

Formatted question description: https://leetcode.ca/all/2280.html

2280. Minimum Lines to Represent a Line Chart

  • Difficulty: Medium.
  • Related Topics: Array, Math, Geometry, Sorting, Number Theory.
  • Similar Questions: Max Points on a Line, Minimum Number of Lines to Cover Points.

Problem

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.

Solution (Java, C++, Python)

  • class Solution {
    public int minimumLines(int[][] stockPrices) {
        if (stockPrices.length == 1) {
            return 0;
        }
        Arrays.sort(stockPrices, (a, b) -> a[0] - b[0]);
        // multiply with 1.0 to make it double and multiply with 100 for making it big so that
        // difference won't come out to be very less and after division it become 0.
        // failing for one of the case without multiply 100
        double lastSlope =
                (stockPrices[1][1] - stockPrices[0][1])
                        * 100
                        / ((stockPrices[1][0] - stockPrices[0][0]) * 1.0);
        int ans = 1;
        for (int i = 2; i < stockPrices.length; i++) {
            double curSlope =
                    (stockPrices[i][1] - stockPrices[i - 1][1])
                            * 100
                            / ((stockPrices[i][0] - stockPrices[i - 1][0]) * 1.0);
            if (lastSlope != curSlope) {
                lastSlope = curSlope;
                ans++;
            }
        }
        return ans;
    }
}

############

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;
    }
}

  • 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

############

# 2280. Minimum Lines to Represent a Line Chart
# https://leetcode.com/problems/minimum-lines-to-represent-a-line-chart/

class Solution:
    def minimumLines(self, stock: List[List[int]]) -> int:
        n = len(stock)
        
        if n == 1: return 0
        
        stock.sort(key = lambda x: x[0])
        dx = stock[0][0] - stock[1][0]
        dy = stock[0][1] - stock[1][1]
            
        res = 1
        
        for index in range(2, n):
            x, y = stock[index]
        
            dx1 = stock[index - 1][0] - stock[index][0]
            dy1 = stock[index - 1][1] - stock[index][1]  
            
            if dx1 * dy != dx * dy1:
                res += 1
                dx = dx1
                dy = dy1
        
        return res


  • 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;
    }
};

  • 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
}

  • 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;
}

Explain:

nope.

Complexity:

  • Time complexity : O(n).
  • Space complexity : O(n).

All Problems

All Solutions