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Minimum Time Visiting All Points - Leetcode Solution

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LeetCode:  Minimum Time Visiting All Points Leetcode Solution

Difficulty: Easy

Topics: math array  

Problem Statement: Given a list of points in the plane, find the minimum time it takes to visit all the points in the list where you can move in any cardinal direction (N,S,E or W) and in a single unit of time.

Solution: The problem requires us to find the time it takes to move from one point to another and to do so with the minimum time. To solve this problem, we can use the concept of the Manhattan distance. The Manhattan distance between two points (x1, y1) and (x2, y2) is defined as the sum of the absolute differences of their coordinates:

Manhattan Distance = |x2 - x1| + |y2 - y1|

Therefore, the time it takes to move from point A to point B is equal to the maximum of the absolute difference of their x-coordinates and their y-coordinates i.e. the maximum value between |x2 - x1| and |y2 - y1|. This is because the time taken to move horizontally and vertically is equal.

To solve this problem, we can iterate over all the given points in the list and calculate the time it takes to move from the current point to the next point. We can then add this time to total time. We can repeat this for all the points in the list.

Algorithm:

  1. Define a variable "total_time" initialized as 0.
  2. Iterate over all the given points in the list, starting from the second point.
  3. For each point, calculate the time it takes to move from the current point to the next point using the Manhattan distance formula and store it in a variable "time". The time taken to move between two points is equal to the maximum value between |x2 - x1| and |y2 - y1|.
  4. Add "time" to "total_time".
  5. Return "total_time".

Pseudo code:

function minTimeToVisitAllPoints(points):
    let total_time = 0
    for i from 1 to points.length-1:
        let x_diff = abs(points[i][0]-points[i-1][0])
        let y_diff = abs(points[i][1]-points[i-1][1])
        total_time += max(x_diff, y_diff)

    return total_time

Time Complexity: The time complexity of this algorithm is O(n) because we are iterating over all the given points and performing constant operations on each point. Thus, the time complexity of this algorithm is linear in the number of points.

Space Complexity: The space complexity of this algorithm is O(1) because we are not using any extra space apart from the input variable "points".

Conclusion: We have learned how to solve the Minimum Time Visiting All Points problem on LeetCode using the concept of Manhattan distance. We have defined an algorithm, provided pseudo-code, calculated the time and space complexities of the algorithm, and explained the entire solution. This solution is optimal in terms of time complexity and requires minimal space.

Minimum Time Visiting All Points Solution Code

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