Find the Shortest Distance Between Two Points in C++

Suppose we have a list of coordinates where each element is of the form [x, y], representing Euclidean coordinates. We have to find the smallest squared distance (x₁ - x₂) ² + (y₁ - y₂) ² between any two provided coordinates.

So, if the input is like coordinates = {{1, 2},{1, 4},{3, 5}}, then the output will be 4.

To solve this, we will follow these steps −

• Define one map ytorightmostx

• sort the array coordinates

• ret := infinity

• for each p in cordinates −

  • it = return the value where (p[1] - sqrt(ret)) is in ytorightmostx or the smallest value greater than it from ytorightmostx

  • while it is not equal to last element of ytorightmostx, do −

    • yd := first - p[1] of it

    • if yd > 0 and yd * yd >= ret, then −

      • Come out from the loop

    • nxt = it

    • increase nxt by 1

    • ret := minimum of (ret, dist(p[0], p[1], first value of it, second value of it)

    • xd := second value of it - p[0]

    • if xd * xd >= ret, then −

      • delete it from ytorightmostx

    • it := nxt

  • ytorightmostx[p[1]] := p[0]

• return ret

• Define a function dist(), this will take xl, yl, xr, yr.

  • xd := xl - xr

  • yd := yl - yr

  • return xd * xd + yd * yd

Example 

Let us see the following implementation to get a better understanding −

 Live Demo

#include <bits/stdc++.h>
using namespace std;
long long dist(long long xl, long long yl, long long xr, long long yr) {
   long long xd = xl - xr;
   long long yd = yl - yr;
   return xd * xd + yd * yd;
}
int solve(vector<vector<int>>& coordinates) {
   map<long long, long long> ytorightmostx;
   sort(coordinates.begin(), coordinates.end());
   long long ret = 1e18;
   for (auto& p : coordinates) {
      auto it = ytorightmostx.lower_bound(p[1] - sqrt(ret));
      while (it != ytorightmostx.end()) {
         long long yd = it->first - p[1];
         if (yd > 0 && yd * yd >= ret) {
            break;
         }
         auto nxt = it;
         nxt++;
         ret = min(ret, dist(p[0], p[1], it->second, it->first));
         long long xd = (it->second - p[0]);
         if (xd * xd >= ret) {
            ytorightmostx.erase(it);
         }
         it = nxt;
      }
      ytorightmostx[p[1]] = p[0];
   }
   return ret;
}
int main(){
   vector<vector<int>> coord = {{1, 2},{1, 4},{3, 5}};
   cout << solve(coord) << endl;
   return 0;
}

Input

{{1, 2},{1, 4},{3, 5}}

Output

4