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Closest Pair of Points | O(nlogn) Implementation

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We are given an array of n points in the plane, and the problem is to find out the closest pair of points in the array. This problem arises in a number of applications. For example, in air-traffic control, you may want to monitor planes that come too close together, since this may indicate a possible collision. Recall the following formula for distance between two points p and q. 
\left \| pq\right \| = \sqrt{(p_x - q_x)^2 + (p_y - q_y)^2}
We have discussed a divide and conquer solution for this problem. The time complexity of the implementation provided in the previous post is O(n (Logn)^2). In this post, we discuss implementation with time complexity as O(nLogn). 
Following is a recap of the algorithm discussed in the previous post.
1) We sort all points according to x coordinates.
2) Divide all points in two halves.
3) Recursively find the smallest distances in both subarrays.
4) Take the minimum of two smallest distances. Let the minimum be d. 
5) Create an array strip[] that stores all points which are at most d distance away from the middle line dividing the two sets.
6) Find the smallest distance in strip[]. 
7) Return the minimum of d and the smallest distance calculated in above step 6.
The great thing about the above approach is, if the array strip[] is sorted according to y coordinate, then we can find the smallest distance in strip[] in O(n) time. In the implementation discussed in the previous post, strip[] was explicitly sorted in every recursive call that made the time complexity O(n (Logn)^2), assuming that the sorting step takes O(nLogn) time. 
In this post, we discuss an implementation where the time complexity is O(nLogn). The idea is to presort all points according to y coordinates. Let the sorted array be Py[]. When we make recursive calls, we need to divide points of Py[] also according to the vertical line. We can do that by simply processing every point and comparing its x coordinate with x coordinate of the middle line.
Following is C++ implementation of O(nLogn) approach.
 

CPP

// A divide and conquer program in C++ to find the smallest distance from a
// given set of points.
 
#include <iostream>
#include <float.h>
#include <stdlib.h>
#include <math.h>
using namespace std;
 
// A structure to represent a Point in 2D plane
struct Point
{
    int x, y;
};
 
 
/* Following two functions are needed for library function qsort().
 
// Needed to sort array of points according to X coordinate
int compareX(const void* a, const void* b)
{
    Point *p1 = (Point *)a,  *p2 = (Point *)b;
    return (p1->x != p2->x) ? (p1->x - p2->x) : (p1->y - p2->y);
}
// Needed to sort array of points according to Y coordinate
int compareY(const void* a, const void* b)
{
    Point *p1 = (Point *)a,   *p2 = (Point *)b;
    return (p1->y != p2->y) ? (p1->y - p2->y) : (p1->x - p2->x);
}
 
// A utility function to find the distance between two points
float dist(Point p1, Point p2)
{
    return sqrt( (p1.x - p2.x)*(p1.x - p2.x) +
                 (p1.y - p2.y)*(p1.y - p2.y)
               );
}
 
// A Brute Force method to return the smallest distance between two points
// in P[] of size n
float bruteForce(Point P[], int n)
{
    float min = FLT_MAX;
    for (int i = 0; i < n; ++i)
        for (int j = i+1; j < n; ++j)
            if (dist(P[i], P[j]) < min)
                min = dist(P[i], P[j]);
    return min;
}
 
// A utility function to find a minimum of two float values
float min(float x, float y)
{
    return (x < y)? x : y;
}
 
 
// A utility function to find the distance between the closest points of
// strip of a given size. All points in strip[] are sorted according to
// y coordinate. They all have an upper bound on minimum distance as d.
// Note that this method seems to be a O(n^2) method, but it's a O(n)
// method as the inner loop runs at most 6 times
float stripClosest(Point strip[], int size, float d)
{
    float min = d;  // Initialize the minimum distance as d
 
    // Pick all points one by one and try the next points till the difference
    // between y coordinates is smaller than d.
    // This is a proven fact that this loop runs at most 6 times
    for (int i = 0; i < size; ++i)
        for (int j = i+1; j < size && (strip[j].y - strip[i].y) < min; ++j)
            if (dist(strip[i],strip[j]) < min)
                min = dist(strip[i], strip[j]);
 
    return min;
}
 
// A recursive function to find the smallest distance. The array Px contains
// all points sorted according to x coordinates and Py contains all points
// sorted according to y coordinates
float closestUtil(Point Px[], Point Py[], int n)
{
    // If there are 2 or 3 points, then use brute force
    if (n <= 3)
        return bruteForce(Px, n);
 
    // Find the middle point
    int mid = n/2;
    Point midPoint = Px[mid];
 
 
    // Divide points in y sorted array around the vertical line.
    // Assumption: All x coordinates are distinct.
    Point Pyl[mid];   // y sorted points on left of vertical line
    Point Pyr[n-mid];  // y sorted points on right of vertical line
    int li = 0, ri = 0;  // indexes of left and right subarrays
    for (int i = 0; i < n; i++)
    {
      if ((Py[i].x < midPoint.x || (Py[i].x == midPoint.x && Py[i].y < midPoint.y)) && li<mid)
         Pyl[li++] = Py[i];
      else
         Pyr[ri++] = Py[i];
    }
 
    // Consider the vertical line passing through the middle point
    // calculate the smallest distance dl on left of middle point and
    // dr on right side
    float dl = closestUtil(Px, Pyl, mid);
    float dr = closestUtil(Px + mid, Pyr, n-mid);
 
    // Find the smaller of two distances
    float d = min(dl, dr);
 
    // Build an array strip[] that contains points close (closer than d)
    // to the line passing through the middle point
    Point strip[n];
    int j = 0;
    for (int i = 0; i < n; i++)
        if (abs(Py[i].x - midPoint.x) < d)
            strip[j] = Py[i], j++;
 
    // Find the closest points in strip.  Return the minimum of d and closest
    // distance is strip[]
    return stripClosest(strip, j, d);
}
 
// The main function that finds the smallest distance
// This method mainly uses closestUtil()
float closest(Point P[], int n)
{
    Point Px[n];
    Point Py[n];
    for (int i = 0; i < n; i++)
    {
        Px[i] = P[i];
        Py[i] = P[i];
    }
 
    qsort(Px, n, sizeof(Point), compareX);
    qsort(Py, n, sizeof(Point), compareY);
 
    // Use recursive function closestUtil() to find the smallest distance
    return closestUtil(Px, Py, n);
}
 
// Driver program to test above functions
int main()
{
    Point P[] = {{2, 3}, {12, 30}, {40, 50}, {5, 1}, {12, 10}, {3, 4}};
    int n = sizeof(P) / sizeof(P[0]);
    cout << "The smallest distance is " << closest(P, n);
    return 0;
}

                    

Java

import java.util.Arrays;
import java.util.List;
import java.util.ArrayList;
 
// A structure to represent a Point in 2D plane
class Point {
public int x;
public int y;
 
public Point(int x, int y) {
    this.x = x;
    this.y = y;
}
}
 
public class ClosestPair {
     
    // The main function that finds the smallest distance
    // This method mainly uses closestUtil()
    public static double closest(Point[] P, int n) {
        Point[] Px = Arrays.copyOf(P, n);
        Arrays.sort(Px, (p1, p2) -> p1.x - p2.x);
        Point[] Py = Arrays.copyOf(P, n);
        Arrays.sort(Py, (p1, p2) -> p1.y - p2.y);
         
        // Use recursive function closestUtil() to find the smallest distance
        return closestUtil(Px, Py, n);
    }
    // A recursive function to find the smallest distance. The array Px contains
    // all points sorted according to x coordinates and Py contains all points
    // sorted according to y coordinates
    private static double closestUtil(Point[] Px, Point[] Py, int n) {
         
        // If there are 2 or 3 points, then use brute force
        if (n <= 3) {
            return bruteForce(Px, n);
        }
        // Find the middle point
        int mid = n / 2;
        Point midPoint = Px[mid];
         
        // Divide points in y sorted array around the vertical line.
        // Assumption: All x coordinates are distinct.
        Point[] Pyl = Arrays.copyOfRange(Py, 0, mid);// y sorted points on left of vertical line
        Point[] Pyr = Arrays.copyOfRange(Py, mid, n);//y sorted points on right of vertical line
         
        // Consider the vertical line passing through the middle point
        // calculate the smallest distance dl on left of middle point and
        // dr on right side
        double dl = closestUtil(Px, Pyl, mid);
        double dr = closestUtil(Arrays.copyOfRange(Px, mid, n), Pyr, n - mid);
     
        // Find the smaller of two distances
        double d = Math.min(dl, dr);
     
        // Build an array strip[] that contains points close (closer than d)
        // to the line passing through the middle point
        List<Point> strip = new ArrayList<Point>();
        for (Point p : Py) {
            if (Math.abs(p.x - midPoint.x) < d) {
                strip.add(p);
            }
        }
     
        return stripClosest(strip.toArray(new Point[strip.size()]), strip.size(), d);
    }
    // A Brute Force method to return the smallest distance between two points
    // in P[] of size n
    private static double bruteForce(Point[] P, int n) {
        double min = Double.MAX_VALUE;
        for (int i = 0; i < n; ++i) {
            for (int j = i + 1; j < n; ++j) {
                double dist = distance(P[i], P[j]);
                if (dist < min) {
                    min = dist;
                }
            }
        }
        return min;
    }
    // A utility function to find the distance between the closest points of
    // strip of a given size. All points in strip[] are sorted according to
    // y coordinate. They all have an upper bound on minimum distance as d.
    // Note that this method seems to be a O(n^2) method, but it's a O(n)
    // method as the inner loop runs at most 6 times
    private static double stripClosest(Point[] strip, int size, double d) {
        double min = d; // Initialize the minimum distance as d
         
        // Pick all points one by one and try the next points till the difference
        // between y coordinates is smaller than d.
        // This is a proven fact that this loop runs at most 6 times
        for (int i = 0; i < size; ++i) {
            for (int j = i + 1; j < size && (strip[j].y - strip[i].y) < min; ++j) {
                double dist = distance(strip[i], strip[j]);
                if (dist < min) {
                    min = dist;
                }
            }
        }
        return min;
    }
 
    private static double distance(Point p1, Point p2) {
        return Math.sqrt(Math.pow(p1.x - p2.x, 2) + Math.pow(p1.y - p2.y, 2));
    }
    // Driver program to test above functions
    public static void main(String[] args) {
        Point[] P = { new Point(2, 3), new Point(12, 30), new Point(40, 50),
                      new Point(5, 1), new Point(12, 10), new Point(3, 4) };
        int n = P.length;
        System.out.println("The smallest distance is " + closest(P, n));
    }
}
// This code save in ClosestPair.java name then run
 
 
// This code is contributed by shiv1o43g

                    

Python3

# Python Equivalent
import math
 
# A structure to represent a Point in 2D plane
class Point:
    def __init__(self, x, y):
        self.x = x
        self.y = y
 
# Needed to sort array of points according to X coordinate
def compareX(a, b):
    p1,  p2 = a, b
    return (p1.x != p2.x) * (p1.x - p2.x) + (p1.y - p2.y)
 
# Needed to sort array of points according to Y coordinate
def compareY(a, b):
    p1,  p2 = a, b
    return (p1.y != p2.y) * (p1.y - p2.y) + (p1.x - p2.x)
 
# A utility function to find the distance between two points
def dist(p1, p2):
    return math.sqrt((p1.x - p2.x)**2 + (p1.y - p2.y)**2)
 
# A Brute Force method to return the smallest distance between two points
# in P[] of size n
def bruteForce(P, n):
    min = float('inf')
    for i in range(n):
        for j in range(i+1, n):
            if dist(P[i], P[j]) < min:
                min = dist(P[i], P[j])
    return min
 
# A utility function to find a minimum of two float values
def min(x, y):
    return x if x < y else y
 
# A utility function to find the distance between the closest points of
# strip of a given size. All points in strip[] are sorted according to
# y coordinate. They all have an upper bound on minimum distance as d.
# Note that this method seems to be a O(n^2) method, but it's a O(n)
# method as the inner loop runs at most 6 times
def stripClosest(strip, size, d):
    min = # Initialize the minimum distance as d
 
    # Pick all points one by one and try the next points till the difference
    # between y coordinates is smaller than d.
    # This is a proven fact that this loop runs at most 6 times
    for i in range(size):
        for j in range(i+1, size):
            if (strip[j].y - strip[i].y) < min:
                if dist(strip[i],strip[j]) < min:
                    min = dist(strip[i], strip[j])
 
    return min
 
# A recursive function to find the smallest distance. The array Px contains
# all points sorted according to x coordinates and Py contains all points
# sorted according to y coordinates
def closestUtil(Px, Py, n):
    # If there are 2 or 3 points, then use brute force
    if n <= 3:
        return bruteForce(Px, n)
 
    # Find the middle point
    mid = n // 2
    midPoint = Px[mid]
 
 
    # Divide points in y sorted array around the vertical line.
    # Assumption: All x coordinates are distinct.
    Pyl = [None] * mid   # y sorted points on left of vertical line
    Pyr = [None] * (n-mid)  # y sorted points on right of vertical line
    li = ri = 0  # indexes of left and right subarrays
    for i in range(n):
        if ((Py[i].x < midPoint.x or (Py[i].x == midPoint.x and Py[i].y < midPoint.y)) and li<mid):
            Pyl[li] = Py[i]
            li += 1
        else:
            Pyr[ri] = Py[i]
            ri += 1
 
    # Consider the vertical line passing through the middle point
    # calculate the smallest distance dl on left of middle point and
    # dr on right side
    dl = closestUtil(Px, Pyl, mid)
    dr = closestUtil(Px[mid:], Pyr, n-mid)
 
    # Find the smaller of two distances
    d = min(dl, dr)
 
    # Build an array strip[] that contains points close (closer than d)
    # to the line passing through the middle point
    strip = [None] * n
    j = 0
    for i in range(n):
        if abs(Py[i].x - midPoint.x) < d:
            strip[j] = Py[i]
            j += 1
 
    # Find the closest points in strip.  Return the minimum of d and closest
    # distance is strip[]
    return stripClosest(strip, j, d)
 
# The main function that finds the smallest distance
# This method mainly uses closestUtil()
def closest(P, n):
    Px = P
    Py = P
    Px.sort(key=lambda x:x.x)
    Py.sort(key=lambda x:x.y)
 
    # Use recursive function closestUtil() to find the smallest distance
    return closestUtil(Px, Py, n)
 
# Driver program to test above functions
if __name__ == '__main__':
    P = [Point(2, 3), Point(12, 30), Point(40, 50), Point(5, 1), Point(12, 10), Point(3, 4)]
    n = len(P)
    print("The smallest distance is", closest(P, n))

                    

C#

// Python Equivalent
using System;
using System.Collections.Generic;
using System.Linq;
 
// A structure to represent a Point in 2D plane
public class Point
{
    public int x;
    public int y;
     
    public Point(int x, int y)
    {
        this.x = x;
        this.y = y;
    }
}
 
public class ClosestPair
{
     
    // The main function that finds the smallest distance
    // This method mainly uses closestUtil()
    public static double Closest(Point[] P, int n)
    {
        Point[] Px = P.OrderBy(p => p.x).ToArray();
        Point[] Py = P.OrderBy(p => p.y).ToArray();
         
        // Use recursive function closestUtil() to find the smallest distance
        return ClosestUtil(Px, Py, n);
    }
    // A recursive function to find the smallest distance. The array Px contains
    // all points sorted according to x coordinates and Py contains all points
    // sorted according to y coordinates
    private static double ClosestUtil(Point[] Px, Point[] Py, int n)
    {  
        // If there are 2 or 3 points, then use brute force
        if (n <= 3)
        {
            return BruteForce(Px, n);
        }
        // Find the middle point
        int mid = n / 2;
        Point midPoint = Px[mid];
        // Divide points in y sorted array around the vertical line.
        // Assumption: All x coordinates are distinct.
        Point[] Pyl = Py.Take(mid).ToArray(); // y sorted points on left of vertical line
        Point[] Pyr = Py.Skip(mid).ToArray(); //y sorted points on right of vertical line
         
        // Consider the vertical line passing through the middle point
        // calculate the smallest distance dl on left of middle point and
        // dr on right side
        double dl = ClosestUtil(Px, Pyl, mid);
        double dr = ClosestUtil(Px.Skip(mid).ToArray(), Pyr, n - mid);
         
        // Find the smaller of two distances
        double d = Math.Min(dl, dr);
         
        // Build an array strip[] that contains points close (closer than d)
        // to the line passing through the middle point
        List<Point> strip = new List<Point>();
        foreach (Point p in Py)
        {
            if (Math.Abs(p.x - midPoint.x) < d)
            {
                strip.Add(p);
            }
        }
        return StripClosest(strip.ToArray(), strip.Count, d);
    }
     
    // A Brute Force method to return the smallest distance between two points
    // in P[] of size n
    private static double BruteForce(Point[] P, int n)
    {
        double min = double.MaxValue;
        for (int i = 0; i < n; ++i)
        {
            for (int j = i + 1; j < n; ++j)
            {
                double dist = Distance(P[i], P[j]);
                if (dist < min)
                {
                    min = dist;
                }
            }
        }
        return min;
    }
    // A utility function to find the distance between the closest points of
    // strip of a given size. All points in strip[] are sorted according to
    // y coordinate. They all have an upper bound on minimum distance as d.
    // Note that this method seems to be a O(n^2) method, but it's a O(n)
    // method as the inner loop runs at most 6 times
     
    private static double StripClosest(Point[] strip, int size, double d)
    {
        double min = d; // Initialize the minimum distance as d
        // Pick all points one by one and try the next points till the difference
        // between y coordinates is smaller than d.
        // This is a proven fact that this loop runs at most 6 times
        for (int i = 0; i < size; ++i)
        {
            for (int j = i + 1; j < size && (strip[j].y - strip[i].y) < min; ++j)
            {
                double dist = Distance(strip[i], strip[j]);
                if (dist < min)
                {
                    min = dist;
                }
            }
        }
        return min;
    }
 
    private static double Distance(Point p1, Point p2)
    {
        return Math.Sqrt(Math.Pow(p1.x - p2.x, 2) + Math.Pow(p1.y - p2.y, 2));
    }
}
 
public class Program
{
    // Driver program to test above functions
    public static void Main()
    {
        Point[] P = { new Point(2, 3), new Point(12, 30), new Point(40, 50),
                      new Point(5, 1), new Point(12, 10), new Point(3, 4) };
        int n = P.Length;
        Console.WriteLine("The smallest distance is " + ClosestPair.Closest(P, n));
    }
}
 
// This code is contributed by shivhack999

                    

Javascript

// JavaScript Equivalent
 
// A structure to represent a Point in 2D plane
class Point {
  constructor(x, y) {
    this.x = x;
    this.y = y;
  }
}
 
// Needed to sort array of points according to X coordinate
function compareX(a, b) {
  let p1 = a,
    p2 = b;
  return (p1.x !== p2.x) * (p1.x - p2.x) + (p1.y - p2.y);
}
 
// Needed to sort array of points according to Y coordinate
function compareY(a, b) {
  let p1 = a,
    p2 = b;
  return (p1.y !== p2.y) * (p1.y - p2.y) + (p1.x - p2.x);
}
 
// A utility function to find the distance between two points
function dist(p1, p2) {
  return Math.sqrt(Math.pow(p1.x - p2.x, 2) + Math.pow(p1.y - p2.y, 2));
}
 
// A Brute Force method to return the smallest distance between two points
// in P[] of size n
function bruteForce(P, n) {
  let min = Number.POSITIVE_INFINITY;
  for (let i = 0; i < n; i++) {
    for (let j = i + 1; j < n; j++) {
      if (dist(P[i], P[j]) < min) {
        min = dist(P[i], P[j]);
      }
    }
  }
  return min;
}
 
// A utility function to find a minimum of two float values
function min(x, y) {
  return x < y ? x : y;
}
 
// A utility function to find the distance between the closest points of
// strip of a given size. All points in strip[] are sorted according to
// y coordinate. They all have an upper bound on minimum distance as d.
// Note that this method seems to be a O(n^2) method, but it's a O(n)
// method as the inner loop runs at most 6 times
function stripClosest(strip, size, d) {
  let min = d; // Initialize the minimum distance as d
 
  // Pick all points one by one and try the next points till the difference
  // between y coordinates is smaller than d.
  // This is a proven fact that this loop runs at most 6 times
  for (let i = 0; i < size; i++) {
    for (let j = i + 1; j < size; j++) {
      if (strip[j].y - strip[i].y < min) {
        if (dist(strip[i], strip[j]) < min) {
          min = dist(strip[i], strip[j]);
        }
      }
    }
  }
 
  return min;
}
 
// A recursive function to find the smallest distance. The array Px contains
// all points sorted according to x coordinates and Py contains all points
// sorted according to y coordinates
function closestUtil(Px, Py, n) {
  // If there are 2 or 3 points, then use brute force
  if (n <= 3) {
    return bruteForce(Px, n);
  }
 
  // Find the middle point
  let mid = Math.floor(n / 2);
  let midPoint = Px[mid];
 
  // Divide points in y sorted array around the vertical line.
  // Assumption: All x coordinates are distinct.
  let Pyl = new Array(mid); // y sorted points on left of vertical line
  let Pyr = new Array(n - mid); // y sorted points on right of vertical line
  let li = 0;
  let ri = 0; // indexes of left and right subarrays
  for (let i = 0; i < n; i++) {
    if (
      (Py[i].x < midPoint.x || (Py[i].x === midPoint.x && Py[i].y < midPoint.y)) &&
      li < mid
    ) {
      Pyl[li] = Py[i];
      li++;
    } else {
      Pyr[ri] = Py[i];
      ri++;
    }
  }
 
  // Consider the vertical line passing through the middle point
  // calculate the smallest distance dl on left of middle point and
  // dr on right side
  let dl = closestUtil(Px, Pyl, mid);
  let dr = closestUtil(Px.slice(mid), Pyr, n - mid);
 
  // Find the smaller of two distances
  let d = min(dl, dr);
 
  // Build an array strip[] that contains points close (closer than d)
  // to the line passing through the middle point
  let strip = new Array(n);
  let j = 0;
  for (let i = 0; i < n; i++) {
    if (Math.abs(Py[i].x - midPoint.x) < d) {
      strip[j] = Py[i];
      j++;
    }
  }
 
  // Find the closest points in strip.  Return the minimum of d and closest
  // distance is strip[]
  return stripClosest(strip, j, d);
}
 
// The main function that finds the smallest distance
// This method mainly uses closestUtil()
function closest(P, n) {
  let Px = [...P];
  let Py = [...P];
  Px.sort((a, b) => compareX(a, b));
  Py.sort((a, b) => compareY(a, b));
 
  // Use recursive function closestUtil() to find the smallest distance
  return closestUtil(Px, Py, n);
}
 
// Driver program to test above functions
function main() {
  let P = [
    new Point(2, 3),
    new Point(12, 30),
    new Point(40, 50),
    new Point(5, 1),
    new Point(12, 10),
    new Point(3, 4)
  ];
  let n = P.length;
  console.log(`The smallest distance is ${closest(P, n)}`);
}
 
main();

                    

Output
The smallest distance is 1.41421


Time Complexity:Let Time complexity of above algorithm be T(n). Let us assume that we use a O(nLogn) sorting algorithm. The above algorithm divides all points in two sets and recursively calls for two sets. After dividing, it finds the strip in O(n) time. Also, it takes O(n) time to divide the Py array around the mid vertical line. Finally finds the closest points in strip in O(n) time. So T(n) can be expressed as follows 
T(n) = 2T(n/2) + O(n) + O(n) + O(n) 
T(n) = 2T(n/2) + O(n) 
T(n) = T(nLogn)


Auxiliary Space: O(log n), as implicit stack is created during recursive calls
References: 
http://www.cs.umd.edu/class/fall2013/cmsc451/Lects/lect10.pdf 
http://www.youtube.com/watch?v=vS4Zn1a9KUc 
http://www.youtube.com/watch?v=T3T7T8Ym20M 
http://en.wikipedia.org/wiki/Closest_pair_of_points_problem

 



Last Updated : 13 Apr, 2023
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