Minimum Time Visiting All Points in C++


Suppose there are some points given as an array. We have to find the minimum time in seconds to visit all points. There are some conditions.

  • In one second, it can move vertically, horizontally and diagonally
  • We have to visit the points in the same order as they appear in the array.

So if the points are [(1, 1), (3, 4), (-1, 0)], then output will be 7. If we check the sequence for the shortest route, the sequence will be (1, 1), (2, 2), (3, 3), (3, 4), (2, 3), (1, 2), (0, 1), (-1, 0)

To solve this we will just find the maximum of x coordinate difference of two consecutive points, and y coordinate difference of two consecutive points. The max values will be added together.

Example

Let us see the following implementation to get better understanding −

 Live Demo

#include <bits/stdc++.h>
using namespace std;
class Solution {
   public:
      int minTimeToVisitAllPoints(vector<vector<int>>& p) {
         int ans = 0;
         int n = p.size();
         for(int i = 1; i < n; i++){
            ans += max(abs(p[i][0] - p[i-1][0]), abs(p[i][1] - p[i-1] [1]));
         }
         return ans;
      }
};
main(){
   Solution ob;
   vector<vector<int>> c = {{1,1},{3,4},{-1,0}};
   cout << ob.minTimeToVisitAllPoints(c);
}

Input

[[1,1],[3,4],[-1,0]]

Output

7

Updated on: 29-Apr-2020

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