Today I have done leetcode's December LeetCoding Challenge with cpp.

December LeetCoding Challenge 3

Description

Minimum Time Visiting All Points

On a 2D plane, there are n points with integer coordinates points[i] = [xi, yi]. Return the minimum time in seconds to visit all the points in the order given by points.

You can move according to these rules:

• In 1 second, you can either:

  <ul>
  	<li>move vertically by one&nbsp;unit,</li>
  	<li>move horizontally by one unit, or</li>
  	<li>move diagonally <code>sqrt(2)</code> units (in other words, move one unit vertically then one unit horizontally in <code>1</code> second).</li>
  </ul>
  </li>
  <li>You have to visit the points in the same order as they appear in the array.</li>
  <li>You are allowed to pass through points that appear later in the order, but these do not count as visits.</li>
  

Example 1:

Input: points = [[1,1],[3,4],[-1,0]]
Output: 7
Explanation: One optimal path is [1,1] -> [2,2] -> [3,3] -> [3,4] -> [2,3] -> [1,2] -> [0,1] -> [-1,0]   
Time from [1,1] to [3,4] = 3 seconds 
Time from [3,4] to [-1,0] = 4 seconds
Total time = 7 seconds

Example 2:

Input: points = [[3,2],[-2,2]]
Output: 5

Constraints:

• points.length == n
• 1 <= n <= 100
• points[i].length == 2
• -1000 <= points[i][0], points[i][1] <= 1000

Solution

class Solution {
public:
  int minTimeToVisitAllPoints(vector<vector<int>>& points) {
    int answer = 0;
    for(int i = 1; i < points.size(); ++i) {
      answer += max(abs(points[i][1] - points[i - 1][1]), abs(points[i][0] - points[i - 1][0]));
    }

    return answer;
  }
};

// Accepted
// 122/122 cases passed (3 ms)
// Your runtime beats 95.33 % of cpp submissions
// Your memory usage beats 64.64 % of cpp submissions (10.4 MB)