Leetcode 1266: Minimum Time Visiting All Points

Input: The input consists of an array of points, where each point is represented by a pair of integers [x, y].

Example: [[2, 2], [5, 5], [1, 3]]

Constraints:

• 1 <= points.length <= 100

• -1000 <= points[i][0], points[i][1] <= 1000

Output: The output is an integer representing the minimum time in seconds to visit all points in the given order.

Example: 6

Constraints:

• The time is measured in seconds, considering the movement options.

Goal: Determine the minimum time to visit all points in the given order.

Steps:

• 1. Calculate the time to move from the current point to the next using the maximum of the differences in x and y coordinates (taking diagonal moves into account).

• 2. Add the time for each pair of consecutive points to the total time.

• 3. Return the total time after visiting all points.

Goal: The constraints ensure that the solution works within the time and space limits.

Steps:

• 1 <= points.length <= 100

• -1000 <= points[i][0], points[i][1] <= 1000

Assumptions:

• The array contains at least one point.

• The input points are valid and within the given constraints.

• Input: [[2, 2], [5, 5], [1, 3]]

• Explanation: The total time to visit all points is 6 seconds: 3 seconds to go from [2, 2] to [5, 5] and 3 seconds to go from [5, 5] to [1, 3].

Approach: Use the concept of Manhattan distance but consider diagonal movements to minimize the time.

Observations:

• We can calculate the time for each move between points by considering the largest difference in coordinates (either horizontally or vertically).

• Diagonal moves can be used to reduce the total time.

• To find the minimum time, calculate the maximum of the horizontal and vertical distances between two consecutive points.

Steps:

• 1. Iterate through the list of points.

• 2. For each pair of consecutive points, calculate the time to move between them by taking the maximum of the absolute differences in x and y coordinates.

• 3. Add the time for each movement to the total time.

• 4. Return the total time after iterating through all points.

Empty Inputs:

• The input array will always contain at least one point as per the constraints.

Large Inputs:

• Ensure that the solution handles the maximum number of points efficiently.

Special Values:

• All points may be located at the same position, which would result in zero time.

Constraints:

• The input is valid and within the problem's constraints.

int minTimeToVisitAllPoints(vector<vector<int>>& points) {
    int ans = 0;
    for(int i = 1; i < points.size(); i++) {
        ans += max(abs(points[i][1] - points[i - 1][1]), abs(points[i][0] - points[i - 1][0]));
    }
    return ans;
}

1 : Function Definition

int minTimeToVisitAllPoints(vector<vector<int>>& points) {

This defines the function `minTimeToVisitAllPoints`, which takes a 2D vector `points` representing a list of coordinates and returns the minimum time required to visit all points.

2 : Variable Initialization

    int ans = 0;

This initializes an integer variable `ans` to 0, which will accumulate the total time needed to visit all the points.

3 : Loop (Iterate Through Points)

    for(int i = 1; i < points.size(); i++) {

This loop iterates through the points starting from the second point (index 1) and compares each point with the previous one.

4 : Distance Calculation

        ans += max(abs(points[i][1] - points[i - 1][1]), abs(points[i][0] - points[i - 1][0]));

This calculates the time it takes to move from the previous point to the current point. The time is the maximum of the vertical and horizontal distance between the points.

5 : Return Statement

    return ans;

This returns the total accumulated time `ans`, which represents the minimum time required to visit all points.

Best Case: O(n), where n is the number of points.

Average Case: O(n) for each iteration over the points.

Worst Case: O(n), where n is the number of points.

Description: The time complexity is linear because we iterate over the list of points once.

Best Case: O(1), as no additional space is used other than for variables.

Worst Case: O(1), since only a constant amount of extra space is needed.

Description: The space complexity is constant since only a few variables are used to calculate the total time.

Leetcode 1266: Minimum Time Visiting All Points

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