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Closest Pair of Points | O(nlogn) Implementation

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.

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  #include  #include  #include  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().    Refer: http://www.cplusplus.com/reference/clibrary/cstdlib/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

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 strip = new ArrayList();         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 = 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 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

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 strip = new List();         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