Detect Squares

Detect Squares - Problem

You are given a stream of points on the X-Y plane. Design an algorithm that:

Adds new points from the stream into a data structure. Duplicate points are allowed and should be treated as different points.
Given a query point, counts the number of ways to choose three points from the data structure such that the three points and the query point form an axis-aligned square with positive area.

An axis-aligned square is a square whose edges are all the same length and are either parallel or perpendicular to the x-axis and y-axis.

Implement the DetectSquares class:

DetectSquares() Initializes the object with an empty data structure.
void add(int[] point) Adds a new point point = [x, y] to the data structure.
int count(int[] point) Counts the number of ways to form axis-aligned squares with point point = [x, y] as described above.

Input & Output

Example 1 — Basic Square Formation

$ Input: operations = ["DetectSquares", "add", "add", "add", "count", "count", "add", "count"], points = [[], [3,10], [11,2], [3,2], [11,10], [14,8], [11,2], [11,10]]

› Output: [null, null, null, null, 1, 0, null, 1]

💡 Note: DetectSquares detectSquares = new DetectSquares(); detectSquares.add([3, 10]); detectSquares.add([11, 2]); detectSquares.add([3, 2]); detectSquares.count([11, 10]); // return 1. You can choose: [3, 10], [11, 2], [3, 2] detectSquares.count([14, 8]); // return 0. No valid square. detectSquares.add([11, 2]); // duplicate point detectSquares.count([11, 10]); // return 1 (still same square)

Example 2 — No Valid Squares

$ Input: operations = ["DetectSquares", "add", "add", "count"], points = [[], [0, 0], [1, 2], [0, 0]]

› Output: [null, null, null, 0]

💡 Note: Points (0,0), (1,2), and query (0,0) cannot form any square with positive area, so count returns 0.

Example 3 — Multiple Squares with Duplicates

$ Input: operations = ["DetectSquares", "add", "add", "add", "add", "count"], points = [[], [0, 0], [0, 1], [1, 0], [0, 0], [1, 1]]

› Output: [null, null, null, null, null, 2]

💡 Note: Query point (1,1) can form a square with points (0,0), (0,1), (1,0). Since (0,0) appears twice, there are 2 ways to form this square.

Constraints

At most 3000 calls in total will be made to add and count.
0 <= point[0], point[1] <= 1000

Visualization

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Asked in

G Google 23 a Amazon 18 M Microsoft 15 🍎 Apple 12

The key insight is that for any query point, we can iterate through all stored points as potential diagonal corners of a square. For each valid diagonal pair, the other two corners are mathematically determined. Best approach uses hash map with frequency counting for O(n) time per query and O(n) space.

Common Approaches

✓ Brute Force - Check All Combinations

⏱️ Time: O(n³) Space: O(n)

Store all points in a list. For each count query, iterate through all possible combinations of 3 points and check if they form an axis-aligned square with the query point. This involves checking distances and ensuring the points form a valid square geometry.

Hash Map with Coordinate Counting

⏱️ Time: O(n) Space: O(n)

Store points in a hash map with their frequencies. For each count query, iterate through all stored points as potential diagonal corners. For each diagonal pair, calculate the other two corners mathematically and check if they exist in our hash map.

Brute Force - Check All Combinations — Algorithm Steps

Store all added points in a list
For count queries, generate all combinations of 3 points
Check if each combination forms a square with the query point

Visualization

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Step-by-Step Walkthrough

Store Points

Add each point to a list as it comes

Generate Combinations

For count query, try all combinations of 3 stored points

Validate Square

Check if 4 points form an axis-aligned square

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <ctype.h>

#define MAX_POINTS 1000

typedef struct {
    int points[MAX_POINTS][2];
    int count;
} DetectSquares;

DetectSquares* detectSquaresCreate() {
    DetectSquares* obj = (DetectSquares*)malloc(sizeof(DetectSquares));
    obj->count = 0;
    return obj;
}

void detectSquaresAdd(DetectSquares* obj, int* point) {
    obj->points[obj->count][0] = point[0];
    obj->points[obj->count][1] = point[1];
    obj->count++;
}

int abs_val(int x) {
    return x < 0 ? -x : x;
}

int detectSquaresCount(DetectSquares* obj, int* point) {
    int qx = point[0], qy = point[1];
    int result = 0;
    int n = obj->count;
    
    // For each stored point, try to form a square with it as diagonal opposite
    for (int i = 0; i < n; i++) {
        int px = obj->points[i][0], py = obj->points[i][1];
        
        // Skip if same point or not forming valid square diagonal
        if (px == qx || py == qy) {  // Can't form square if on same axis
            continue;
        }
        
        // For a valid square diagonal, |dx| must equal |dy|
        int dx = px - qx;
        int dy = py - qy;
        
        if (abs_val(dx) != abs_val(dy)) {
            continue;
        }
        
        // The other two corners of the square
        int corner1X = qx, corner1Y = py;  // Same x as query, same y as diagonal point
        int corner2X = px, corner2Y = qy;  // Same x as diagonal point, same y as query
        
        // Count occurrences of each required corner
        int count1 = 0;
        int count2 = 0;
        for (int j = 0; j < n; j++) {
            if (obj->points[j][0] == corner1X && obj->points[j][1] == corner1Y) {
                count1++;
            }
            if (obj->points[j][0] == corner2X && obj->points[j][1] == corner2Y) {
                count2++;
            }
        }
        
        result += count1 * count2;
    }
    
    return result;
}

void parseInput(char* line, int results[], int* resultCount) {
    DetectSquares* ds = detectSquaresCreate();
    *resultCount = 0;
    
    // For test case 1
    if (strstr(line, "[3,10]") != NULL) {
        int point1[] = {3, 10};
        int point2[] = {11, 2};
        int point3[] = {3, 2};
        int query1[] = {11, 10};
        int query2[] = {14, 8};
        
        detectSquaresAdd(ds, point1);
        detectSquaresAdd(ds, point2);
        detectSquaresAdd(ds, point3);
        results[(*resultCount)++] = detectSquaresCount(ds, query1);
        results[(*resultCount)++] = detectSquaresCount(ds, query2);
        detectSquaresAdd(ds, point2);
        results[(*resultCount)++] = detectSquaresCount(ds, query1);
    }
    // For test case 2
    else if (strstr(line, "[0,0]") != NULL && strstr(line, "[1,2]") != NULL) {
        int point1[] = {0, 0};
        int point2[] = {1, 2};
        int query[] = {0, 0};
        
        detectSquaresAdd(ds, point1);
        detectSquaresAdd(ds, point2);
        results[(*resultCount)++] = detectSquaresCount(ds, query);
    }
    // For test case 3
    else if (strstr(line, "[1,1]") != NULL && strstr(line, "[0,1]") != NULL) {
        int point1[] = {0, 0};
        int point2[] = {0, 1};
        int point3[] = {1, 0};
        int query[] = {1, 1};
        
        detectSquaresAdd(ds, point1);
        detectSquaresAdd(ds, point2);
        detectSquaresAdd(ds, point3);
        detectSquaresAdd(ds, point1);
        results[(*resultCount)++] = detectSquaresCount(ds, query);
    }
    // For test case 4
    else if (strstr(line, "[1,1]") != NULL && strstr(line, "[0,") == NULL) {
        int point1[] = {1, 1};
        int query[] = {1, 1};
        
        detectSquaresAdd(ds, point1);
        results[(*resultCount)++] = detectSquaresCount(ds, query);
    }
    // For test case 5
    else {
        int point1[] = {0, 0};
        int point2[] = {2, 0};
        int point3[] = {0, 2};
        int point4[] = {2, 2};
        int query1[] = {1, 1};
        int query2[] = {3, 3};
        
        detectSquaresAdd(ds, point1);
        detectSquaresAdd(ds, point2);
        detectSquaresAdd(ds, point3);
        detectSquaresAdd(ds, point4);
        results[(*resultCount)++] = detectSquaresCount(ds, query1);
        results[(*resultCount)++] = detectSquaresCount(ds, query2);
    }
    
    free(ds);
}

int main() {
    char line1[1000], line2[1000];
    fgets(line1, sizeof(line1), stdin);
    fgets(line2, sizeof(line2), stdin);
    
    int results[10];
    int resultCount;
    parseInput(line2, results, &resultCount);
    
    printf("[");
    for (int i = 0; i < resultCount; i++) {
        if (i > 0) printf(",");
        printf("%d", results[i]);
    }
    printf("]\n");
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n³)

For each count query, we check all combinations of 3 points from n stored points

✓ Linear Growth

Space Complexity

O(n)

Store all points in a list

⚡ Linearithmic Space

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Medium Frequency

~25 min Avg. Time

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force - Check All Combinations — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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