Minimum Distance between Two Points
Last Updated :
23 Jul, 2025
You are given an array arr[] of n distinct points in a 2D plane, where each point is represented by its (x, y) coordinates. Find the minimum Euclidean distance between two distinct points.
Note: For two points A(px,qx) and B(py,qy) the distance Euclidean between them is:
Distance = \sqrt{(p_{x}-q_{x})^{2}+ (p_{y}-q_{y})^{2}}
Examples:
Input: arr[] = [[-1, -2], [0, 0], [1, 2], [2, 3]]
Output: 1.414214
Explanation: The smallest distance is between points (1, 2) and (2, 3), which is 1.414214.
Input: arr[] = [[-2, -2], [1, 2], [-1, 0], [3, 3]]
Output: 2.236068
Explanation: The smallest distance is between points (-2, -2) and (-1, 0), which is 2.236068.
[Naive Approach] Checking All Pairs - O(n^2) Time and O(1) Space
The idea is to check the distance between all pairs of points and store the minimum distance found.
C++
// C++ program to find closet point
#include <iostream>
#include <vector>
#include <cmath>
#include <iomanip>
using namespace std;
// Function to compute Euclidean distance between two points
double distance(const vector<double>& p1, const vector<double>& p2) {
return sqrt((p1[0] - p2[0]) * (p1[0] - p2[0]) +
(p1[1] - p2[1]) * (p1[1] - p2[1]));
}
// Function that returns the smallest distance
// between any pair of points
double minDistance(const vector<vector<double>>& points) {
int n = points.size();
double minDist = 1e9;
// Brute force to check all pairs
for (int i = 0; i < n; ++i) {
for (int j = i + 1; j < n; ++j) {
double dist = distance(points[i], points[j]);
if (dist < minDist) {
minDist = dist;
}
}
}
// Return the smallest distance
return minDist;
}
int main() {
vector<vector<double>> points = {{-1, -2}, {0, 0}, {1, 2}, {2, 3}};
double res = minDistance(points);
cout << fixed << setprecision(6) << res << endl;
return 0;
}
// Java program to find closest point
import java.util.ArrayList;
import java.util.List;
import java.lang.Math;
class GfG {
// Function to compute Euclidean distance between two points
static double distance(double[] p1, double[] p2) {
return Math.sqrt((p1[0] - p2[0]) * (p1[0] - p2[0]) +
(p1[1] - p2[1]) * (p1[1] - p2[1]));
}
// Function that returns the smallest distance
// between any pair of points
static double minDistance(List<double[]> points) {
int n = points.size();
double minDist = Double.MAX_VALUE;
// Brute force to check all pairs
for (int i = 0; i < n; ++i) {
for (int j = i + 1; j < n; ++j) {
double dist = distance(points.get(i), points.get(j));
if (dist < minDist) {
minDist = dist;
}
}
}
// Return the smallest distance
return minDist;
}
public static void main(String[] args) {
List<double[]> points = new ArrayList<>();
points.add(new double[]{-1, -2});
points.add(new double[]{0, 0});
points.add(new double[]{1, 2});
points.add(new double[]{2, 3});
double res = minDistance(points);
System.out.printf("%.6f\n", res);
}
}
# Python program to find closet point
import math
# Function to compute Euclidean distance between two points
def distance(p1, p2):
return math.sqrt((p1[0] - p2[0]) ** 2 + (p1[1] - p2[1]) ** 2)
# Function that returns the smallest distance
# between any pair of points
def minDistance(points):
n = len(points)
minDist = float('inf')
# Brute force to check all pairs
for i in range(n):
for j in range(i + 1, n):
dist = distance(points[i], points[j])
if dist < minDist:
minDist = dist
# Return the smallest distance
return minDist
if __name__ == "__main__":
points = [[-1, -2], [0, 0], [1, 2], [2, 3]]
res = minDistance(points)
print(f"{res:.6f}")
// C# program to find closest point
using System;
using System.Collections.Generic;
class GfG {
// Function to compute Euclidean distance between two points
static double distance(double[] p1, double[] p2) {
return Math.Sqrt((p1[0] - p2[0]) * (p1[0] - p2[0]) +
(p1[1] - p2[1]) * (p1[1] - p2[1]));
}
// Function that returns the smallest distance
// between any pair of points
static double minDistance(List<double[]> points) {
int n = points.Count;
double minDist = double.MaxValue;
for (int i = 0; i < n; ++i) {
for (int j = i + 1; j < n; ++j) {
double dist = distance(points[i], points[j]);
if (dist < minDist) {
minDist = dist;
}
}
}
// Return the smallest distance
return minDist;
}
static void Main() {
List<double[]> points = new List<double[]> {
new double[] {-1, -2},
new double[] {0, 0},
new double[] {1, 2},
new double[] {2, 3}
};
double res = minDistance(points);
Console.WriteLine(res.ToString("F6"));
}
}
// JavaScript program to find closest point
// Function to compute Euclidean distance between two points
function distance(p1, p2) {
return Math.sqrt((p1[0] - p2[0]) ** 2 + (p1[1] - p2[1]) ** 2);
}
// Function that returns the smallest distance
// between any pair of points
function minDistance(points) {
let n = points.length;
let minDist = Infinity;
for (let i = 0; i < n; ++i) {
for (let j = i + 1; j < n; ++j) {
let dist = distance(points[i], points[j]);
if (dist < minDist) {
minDist = dist;
}
}
}
// Return the smallest distance
return minDist;
}
// Driver Code
const points = [[-1, -2], [0, 0], [1, 2], [2, 3]];
const res = minDistance(points);
console.log(res.toFixed(6));
[Expected Approach] Using Divide and Conquer - O(n log(n)) Time and O(n) Space
The main idea is to use the divide and conquer algorithm, where the points are recursively divided into smaller groups. The minimum distance is calculated within each groups, and during the merging step, we will check for possible closer pairs across the dividing point. This approach reduces the number of comparisons, as it only focuses on relevant points near the divide.
Step By Step Implementation:
- Sort the points by x-coordinate.
- Recursively divide the points into left and right subarrays until each subarrays has one or two points:
- If one point, return infinity.
- If two points, return their distance.
- Find the minimum distances
dl and dr in the left and right subarrays, and set d as the smaller value between dl and dr.
- Build a strip: collect points whose x-distance from the midline is ≤
d.
- Sort the strip by y-coordinate.
- For each point in the strip, compare it with the next up to 7 points (having y-distance ≤
d) to check for closer pairs. - Update
d if a smaller distance is found in the strip. - Return the smallest distance found from the left half, right half, or across the strip.
Key Insights:
Let the dividing point be at coordinate (x, y). After recursively finding the minimum distance d from the left and right halves, we focus on points near the dividing point that could potentially form a closer pair.
We stores all points whose x-distance from the dividing point is ≤ d, i.e., points between x - d and x + d. So, all these points lie inside a vertical strip of width 2d along the x-axis.
- Why only consider points within
x - d to x + d?
- If two points lie farther than
d apart in the x-direction, their distance is already greater than d, so they can’t be closer.
Now, to reduce unnecessary comparisons, we sort these points by their y-coordinate. For each point in this strip, we only compare it with points that are within d units vertically (i.e., y-distance ≤ d).
- Why compare points with y distance ≤
d?
- Because any pair with a y-distance >
d will have a total distance > d, so we can safely skip those.
This means for any point in the strip[], we’re only checking points inside a rectangle of dimensions 2d × d (width 2d along x, height d along y).
For each point in the strip[], we check at most 7 points, as there are at most 7 points within this rectangle that could potentially have a smaller distance.
- Why at most 7 comparisons per point in strip[]?
Let’s look at this 2d × d rectangle more closely:
- Split it into two
d × d squares: one for points from the left half, one for the right half. - Each square is then divided into 4 smaller squares of size
(d/2 × d/2). - The diagonal of these smaller squares is less than
d/√2 which is less than d, so no two points can occupy the same small square (otherwise, they’d be closer than d, violating the earlier recursive result). - So, each
d × d square can have at most 4 points, totaling at most 8 points in the full 2d × d rectangle.
Since we’re only comparing the current point with the others, that means at most 7 valid comparisons.
C++
// C++ program to find minimum distance between points
#include <iostream>
#include <vector>
#include <cmath>
#include <algorithm>
#include <iomanip>
using namespace std;
// Function to compute Euclidean distance between two points
double distance(const vector<double>& p1, const vector<double>& p2) {
return sqrt((p1[0] - p2[0]) * (p1[0] - p2[0]) +
(p1[1] - p2[1]) * (p1[1] - p2[1]));
}
// Comparison function to sort points by x-coordinate
bool compareX(const vector<double>& p1, const vector<double>& p2) {
return p1[0] < p2[0];
}
// Comparison function to sort points by y-coordinate
bool compareY(const vector<double>& p1, const vector<double>& p2) {
return p1[1] < p2[1];
}
// Function to find the minimum distance in the strip
double stripClosest(vector<vector<double>>& strip, double d) {
double minDist = d;
// Sort points in the strip by their y-coordinate
sort(strip.begin(), strip.end(), compareY);
// Compare each point in the strip
for (int i = 0; i < strip.size(); ++i) {
// At most 7 times this will run
for (int j = i + 1; j < strip.size() &&
(strip[j][1] - strip[i][1]) < minDist; ++j) {
minDist = min(minDist, distance(strip[i], strip[j]));
}
}
return minDist;
}
// Divide and conquer function to find the minimum distance
double minDistUtil(vector<vector<double>>& points,
int left, int right) {
// Base case brute force for 2 or fewer points
if (right - left <= 2) {
double minDist = 1e9;
for (int i = left; i < right; ++i) {
for (int j = i + 1; j < right; ++j) {
minDist = min(minDist, distance(points[i], points[j]));
}
}
return minDist;
}
// Find the midpoint
int mid = (left + right) / 2;
double midX = points[mid][0];
// Recursively find the minimum distances in
// the left and right halves
double dl = minDistUtil(points, left, mid);
double dr = minDistUtil(points, mid, right);
double d = min(dl, dr);
// Build the strip of points within distance d from the midline
int k = 0;
vector<vector<double>> strip;
for (int i = left; i < right; ++i) {
if (abs(points[i][0] - midX) < d) {
strip.push_back(points[i]);
}
}
// Find the minimum distance in the strip
double stripDist = stripClosest(strip, d);
return min(d, stripDist);
}
// Function to find the closest pair of points
double minDistance(vector<vector<double>>& points) {
int n = points.size();
// Sort points by x-coordinate
sort(points.begin(), points.end(), compareX);
return minDistUtil(points, 0, n);
}
int main() {
vector<vector<double>> points = {{-1, -2}, {0, 0}, {1, 2}, {2, 3}};
double res = minDistance(points);
// Output the result with 6 decimal places
cout << fixed << setprecision(6) << res << endl;
return 0;
}
// Java program to find minimum distance between points
import java.util.*;
import java.lang.Math;
public class GfG{
// Function to compute Euclidean distance between two points
static double distance(double[] p1, double[] p2) {
return Math.sqrt((p1[0] - p2[0]) * (p1[0] - p2[0]) +
(p1[1] - p2[1]) * (p1[1] - p2[1]));
}
// Comparison function to sort points by x-coordinate
static Comparator<double[]> compareX = new Comparator<double[]>() {
public int compare(double[] p1, double[] p2) {
return Double.compare(p1[0], p2[0]);
}
};
// Comparison function to sort points by y-coordinate
static Comparator<double[]> compareY = new Comparator<double[]>() {
public int compare(double[] p1, double[] p2) {
return Double.compare(p1[1], p2[1]);
}
};
// Function to find the minimum distance in the strip
static double stripClosest(double[][] strip, double d) {
double minDist = d;
// Sort points in the strip by their y-coordinate
Arrays.sort(strip, compareY);
// Compare each point in the strip
for (int i = 0; i < strip.length; i++) {
for (int j = i + 1; j < strip.length && (strip[j][1] - strip[i][1]) < minDist; j++) {
minDist = Math.min(minDist, distance(strip[i], strip[j]));
}
}
return minDist;
}
// Divide and conquer function to find the minimum distance
static double minDistUtil(double[][] points, int left, int right) {
// Base case brute force for 2 or fewer points
if (right - left <= 2) {
double minDist = Double.MAX_VALUE;
for (int i = left; i < right; i++) {
for (int j = i + 1; j < right; j++) {
minDist = Math.min(minDist, distance(points[i], points[j]));
}
}
return minDist;
}
// Find the midpoint
int mid = (left + right) / 2;
double midX = points[mid][0];
// Recursively find the minimum distances in
// the left and right halves
double dl = minDistUtil(points, left, mid);
double dr = minDistUtil(points, mid, right);
double d = Math.min(dl, dr);
// Build the strip of points within distance d from the midline
List<double[]> strip = new ArrayList<>();
for (int i = left; i < right; i++) {
if (Math.abs(points[i][0] - midX) < d) {
strip.add(points[i]);
}
}
// Find the minimum distance in the strip
double stripDist = stripClosest(strip.toArray(new double[strip.size()][]), d);
return Math.min(d, stripDist);
}
// Function to find the closest pair of points
static double minDistance(double[][] points) {
int n = points.length;
// Sort points by x-coordinate
Arrays.sort(points, compareX);
return minDistUtil(points, 0, n);
}
public static void main(String[] args) {
double[][] points = {{-1, -2}, {0, 0}, {1, 2}, {2, 3}};
double res = minDistance(points);
// Output the result with 6 decimal places
System.out.printf("%.6f\n", res);
}
}
# Python program to find minimum distance between points
import math
# Function to compute Euclidean distance between two points
def distance(p1, p2):
return math.sqrt((p1[0] - p2[0]) ** 2 + (p1[1] - p2[1]) ** 2)
# Function to find the minimum distance in the strip
def stripClosest(strip, d):
min_dist = d
# Sort points in the strip by their y-coordinate
strip.sort(key=lambda point: point[1])
# Compare each point in the strip
for i in range(len(strip)):
for j in range(i + 1, len(strip)):
if (strip[j][1] - strip[i][1]) < min_dist:
min_dist = min(min_dist, distance(strip[i], strip[j]))
else:
break
return min_dist
# Divide and conquer function to find the minimum distance
def minDistUtil(points, left, right):
# Base case brute force for 2 or fewer points
if right - left <= 2:
min_dist = float('inf')
for i in range(left, right):
for j in range(i + 1, right):
min_dist = min(min_dist, distance(points[i], points[j]))
return min_dist
# Find the midpoint
mid = (left + right) // 2
mid_x = points[mid][0]
# Recursively find the minimum distances
# in the left and right halves
dl = minDistUtil(points, left, mid)
dr = minDistUtil(points, mid, right)
d = min(dl, dr)
# Build the strip of points within distance d from the midl
strip = []
for i in range(left, right):
if abs(points[i][0] - mid_x) < d:
strip.append(points[i])
# Find the minimum distance in the strip
stripDist = stripClosest(strip, d)
return min(d, stripDist)
# Function to find the closest pair of points
def minDistance(points):
n = len(points)
# Sort points by x-coordinate
points.sort(key=lambda point: point[0])
return minDistUtil(points, 0, n)
if __name__ == '__main__':
points = [[-1, -2], [0, 0], [1, 2], [2, 3]]
res = minDistance(points)
# Output the result with 6 decimal places
print(f'{res:.6f}')
// C# program to find minimum distance between points
using System;
using System.Linq;
using static System.Math;
using System.Collections.Generic;
class GfG {
// Function to compute Euclidean distance between two points
static double distance(double[] p1, double[] p2){
double dx = p1[0] - p2[0];
double dy = p1[1] - p2[1];
return Sqrt(dx * dx + dy * dy);
}
// Function to find the minimum distance in the strip
static double stripClosest(double[][] strip, double d){
double minDist = d;
// Sort points in the strip by their y-coordinate
Array.Sort(strip, (p1, p2) => p1[1].CompareTo(p2[1]));
for (int i = 0; i < strip.Length; ++i){
// The inner loop runs for at most 7 points
for (int j = i + 1; j < strip.Length &&
(strip[j][1] - strip[i][1]) < minDist; ++j){
minDist = Min(minDist, distance(strip[i], strip[j]));
}
}
return minDist;
}
// Divide and conquer function to find the minimum distance
static double minDistUtil(double[][] points, int left, int right){
// Base case there are 2 or fewer points
if (right - left <= 2){
if (right - left <= 0) return double.MaxValue;
return distance(points[left], points[right - 1]);
}
// Find the midpoint index
int midIndex = left + (right - left) / 2;
double midX = points[midIndex][0];
// Recursively find the minimum distances
// in the left and right halves
double dl = minDistUtil(points, left, midIndex);
double dr = minDistUtil(points, midIndex, right);
double d = Min(dl, dr);
// Build the strip of points whose x-coordinate
// is within distance 'd' from the mid
List<double[]> stripList = new List<double[]>();
for (int i = left; i < right; ++i){
if (Abs(points[i][0] - midX) < d){
stripList.Add(points[i]);
}
}
double[][] stripArray = stripList.ToArray();
// Find the minimum distance in the strip
double stripDist = stripClosest(stripArray, d);
return Min(d, stripDist);
}
// Function to find the closest pair of points
static double minDistance(double[,] points2D){
int n = points2D.GetLength(0);
double[][] points = new double[n][];
for (int i = 0; i < n; i++){
points[i] = new double[] { points2D[i, 0], points2D[i, 1] };
}
// Sort points by x-coordinate
Array.Sort(points, (p1, p2) => p1[0].CompareTo(p2[0]));
return minDistUtil(points, 0, n);
}
static void Main(){
double[,] points = {
{ -1, -2 },
{ 0, 0 },
{ 1, 2 },
{ 2, 3 },
{ 5, 1 },
{ 6, 3 },
{ 8, 0 },
{ 9, 2 }
};
double res = minDistance(points);
// Output the result with 6 decimal places
Console.WriteLine(res.ToString("F6"));
}
}
// JavaScript program to find minimum distance between points
// Function to compute Euclidean distance between two points
function distance(p1, p2) {
return Math.sqrt((p1[0] - p2[0]) ** 2 + (p1[1] - p2[1]) ** 2);
}
// Comparison function to sort points by x-coordinate
function compareX(p1, p2) {
return p1[0] - p2[0];
}
// Comparison function to sort points by y-coordinate
function compareY(p1, p2) {
return p1[1] - p2[1];
}
// Function to find the minimum distance in the strip
function stripClosest(strip, d) {
let minDist = d;
// Sort points in the strip by their y-coordinate
strip.sort(compareY);
// Compare each point in the strip
for (let i = 0; i < strip.length; i++) {
// At most 7 times this will run
for (let j = i + 1; j < strip.length &&
(strip[j][1] - strip[i][1]) < minDist; j++) {
minDist = Math.min(minDist, distance(strip[i], strip[j]));
}
}
return minDist;
}
// Divide and conquer function to find the minimum distance
function minDistUtil(points, left, right) {
// Base case brute force for 2 or fewer points
if (right - left <= 2) {
let minDist = Infinity;
for (let i = left; i < right; i++) {
for (let j = i + 1; j < right; j++) {
minDist = Math.min(minDist, distance(points[i], points[j]));
}
}
return minDist;
}
// Find the midpoint
let mid = Math.floor((left + right) / 2);
let midX = points[mid][0];
// Recursively find the minimum distances in the left and right halves
let dl = minDistUtil(points, left, mid);
let dr = minDistUtil(points, mid, right);
let d = Math.min(dl, dr);
// Build the strip of points within distance d from the midline
let strip = [];
for (let i = left; i < right; i++) {
if (Math.abs(points[i][0] - midX) < d) {
strip.push(points[i]);
}
}
// Find the minimum distance in the strip
let stripDist = stripClosest(strip, d);
return Math.min(d, stripDist);
}
// Function to find the closest pair of points
function minDistance(points) {
let n = points.length;
// Sort points by x-coordinate
points.sort(compareX);
return minDistUtil(points, 0, n);
}
// Driver Code
let points = [[-1, -2], [0, 0], [1, 2], [2, 3]];
let res = minDistance(points);
// Output the result with 6 decimal places
console.log(res.toFixed(6));
Applications:
- Used to detect planes that are too close to each other, preventing potential collisions.
- Used in motion planning to avoid obstacles by determining the closest obstacles or points in the robot's path.
- Helps in data classification and pattern recognition by finding nearest data points for training models.
- Optimizing the placement of devices by finding nearest neighbors for improved communication.
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Dynamic Programming or DPDynamic Programming is an algorithmic technique with the following properties.It is mainly an optimization over plain recursion. Wherever we see a recursive solution that has repeated calls for the same inputs, we can optimize it using Dynamic Programming. The idea is to simply store the results of
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Bitwise AlgorithmsBitwise algorithms in Data Structures and Algorithms (DSA) involve manipulating individual bits of binary representations of numbers to perform operations efficiently. These algorithms utilize bitwise operators like AND, OR, XOR, NOT, Left Shift, and Right Shift.BasicsIntroduction to Bitwise Algorit
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Advanced
Segment TreeSegment Tree is a data structure that allows efficient querying and updating of intervals or segments of an array. It is particularly useful for problems involving range queries, such as finding the sum, minimum, maximum, or any other operation over a specific range of elements in an array. The tree
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Pattern SearchingPattern searching algorithms are essential tools in computer science and data processing. These algorithms are designed to efficiently find a particular pattern within a larger set of data. Patten SearchingImportant Pattern Searching Algorithms:Naive String Matching : A Simple Algorithm that works i
2 min read
GeometryGeometry is a branch of mathematics that studies the properties, measurements, and relationships of points, lines, angles, surfaces, and solids. From basic lines and angles to complex structures, it helps us understand the world around us.Geometry for Students and BeginnersThis section covers key br
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