Minimum Lines To Represent A Line Chart - Leetcode Solution

Problem statement: Given a list of points, where each point is represented by a tuple of two integers x and y, find the minimum number of lines required to represent a line chart of the points.

Solution: To solve this problem, we need to identify the maximum number of lines that can be drawn between the given points. For each pair of points, we can calculate the slope of the line that passes through them. If the slope is the same, then these points are on the same line. We can store the slope and the intercept for each line in a dictionary.

For each point, we can calculate the number of lines that pass through it by iterating over all possible pairs of points and counting the number of lines that have the same slope and intersect with the current point. We can store this count in a dictionary.

Finally, we can find the line that has the most number of points on it and subtract the count of that line from the total number of points to get the minimum number of lines required to represent the line chart.

Here is the Python code for this solution:

def gcd(x, y):
    if y == 0:
        return x
    return gcd(y, x % y)

def get_slope(p1, p2):
    dx = p1[0] - p2[0]
    dy = p1[1] - p2[1]
    if dx == 0:
        return (float('inf'), p1[0])
    g = gcd(dx, dy)
    return (dy // g, dx // g)

def minLines(points):
    if len(points) <= 1:
        return len(points)

    max_count = 0
    lines = {}
    point_counts = {}

    for i in range(len(points)):
        point_counts[i] = 0

    for i in range(len(points)):
        for j in range(i + 1, len(points)):
            slope = get_slope(points[i], points[j])
            if slope in lines:
                lines[slope].add(i)
                lines[slope].add(j)
            else:
                lines[slope] = set([i, j])

    for slope in lines:
        count = len(lines[slope])
        for point in lines[slope]:
            point_counts[point] += count

        if count > max_count:
            max_count = count

    return len(points) - max_count + 1

Time Complexity: The time complexity of this code is O(n^2 log n), where n is the number of points. This is because we are iterating over all pairs of points and performing a gcd operation to calculate the slope. The gcd operation has a time complexity of log n.

Space Complexity: The space complexity of this code is O(n^2), because we are storing all possible lines in a dictionary.

1