# 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.

`// 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);` `}` `// 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);` `}` ` ` `// 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 && 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;` `}` |

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)

**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

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