Closest Pair of Points - Problem

Given an array of points in a 2D plane, find the closest pair of points and return the minimum distance between them.

Each point is represented as [x, y] where x and y are integers.

The distance between two points (x1, y1) and (x2, y2) is calculated using the Euclidean distance formula: √((x2-x1)² + (y2-y1)²)

Your solution must run in O(n log n) time complexity using a divide and conquer approach.

Input & Output

Example 1 — Basic Case

$ Input: points = [[0,2],[1,4],[4,1],[5,3]]

› Output: 2.23606797749979

💡 Note: The closest pair is (0,2) and (1,4) with distance √((1-0)² + (4-2)²) = √(1+4) = √5 ≈ 2.236

Example 2 — Minimum Size

$ Input: points = [[1,1],[3,4]]

› Output: 3.605551275463989

💡 Note: Only two points: distance = √((3-1)² + (4-1)²) = √(4+9) = √13 ≈ 3.606

Example 3 — Same Distance

$ Input: points = [[0,0],[1,0],[0,1],[1,1]]

› Output: 1.0

💡 Note: Multiple pairs have distance 1.0: (0,0)-(1,0), (0,0)-(0,1), (1,0)-(1,1), (0,1)-(1,1)

Constraints

2 ≤ points.length ≤ 10⁴
-10⁴ ≤ x_i, y_i ≤ 10⁴
All points are distinct

Visualization

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Asked in

G Google 35 M Microsoft 28 A Amazon 22 F Facebook 18

The key insight is using divide and conquer to recursively split the 2D plane and only check boundary points that could improve the current best distance. The optimal approach sorts points by x-coordinate, divides recursively, and carefully examines the strip around the midline. Time: O(n log n), Space: O(n).

Common Approaches

✓ Brute Force

⏱️ Time: O(n²) Space: O(1)

Iterate through all possible pairs of points, calculate the Euclidean distance for each pair, and keep track of the minimum distance found.

Divide and Conquer

⏱️ Time: O(n log n) Space: O(n)

Sort points by x-coordinate, recursively divide the plane into halves, find closest pairs in each half, then check boundary region for potential closer pairs.

Brute Force — Algorithm Steps

Step 1: Initialize minimum distance to infinity
Step 2: Use nested loops to check every pair (i,j) where i < j
Step 3: Calculate distance between points[i] and points[j]
Step 4: Update minimum if current distance is smaller

Visualization

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Step-by-Step Walkthrough

Start with first point

Compare point 0 with all other points

Continue systematically

For each point, check distances to remaining points

Track minimum

Keep the smallest distance found so far

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#include <float.h>

double solution(int points[][2], int n) {
    if (n < 2) return 0.0;
    
    double minDist = DBL_MAX;
    
    for (int i = 0; i < n; i++) {
        for (int j = i + 1; j < n; j++) {
            int x1 = points[i][0], y1 = points[i][1];
            int x2 = points[j][0], y2 = points[j][1];
            double dist = sqrt((x2 - x1) * (x2 - x1) + (y2 - y1) * (y2 - y1));
            if (dist < minDist) {
                minDist = dist;
            }
        }
    }
    
    return minDist;
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    line[strcspn(line, "\n")] = 0;
    
    // Parse JSON array manually
    int points[1000][2];
    int n = 0;
    
    char *ptr = line + 1; // Skip first '['
    while (*ptr && *ptr != ']') {
        if (*ptr == '[') {
            ptr++;
            points[n][0] = strtol(ptr, &ptr, 10);
            if (*ptr == ',') ptr++;
            points[n][1] = strtol(ptr, &ptr, 10);
            n++;
            while (*ptr && *ptr != ']') ptr++;
            if (*ptr == ']') ptr++;
            if (*ptr == ',') ptr++;
        } else {
            ptr++;
        }
    }
    
    double result = solution(points, n);
    printf("%f\n", result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

Nested loops check all n×(n-1)/2 pairs of points

⚠ Quadratic Growth

Space Complexity

O(1)

Only uses a few variables to track minimum distance

✓ Linear Space

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Closest Pair of Points - Problem

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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