Count of Squares that are parallel to the coordinate axis from the given set of N points

• Difficulty Level : Hard
• Last Updated : 27 Mar, 2020

Given an array of points points[] in a cartesian coordinate system, the task is to find the count of the squares that are parallel to the coordinate axis.

Examples:

Input:points[] = {(0, 0), (0, 2), (2, 0), (2, 2), (1, 1)}
Output: 1
Explanation:
As the points (0, 0), (0, 2), (2, 0), (2, 2) forms square which is parallel to the X-axis and Y-axis, Hence the count of such squares is 1.

Input:points[] = {(2, 0), (0, 2), (2, 2), (0, 0), (-2, 2), (-2, 0)}
Output: 2
Explanation:
As the points (0, 0), (0, 2), (2, 0), (2, 2) forms one square, whereas points (0, 0), (0, 2), (-2, 0), (-2, 2) forms other square which is parallel to the X-axis and Y-axis,
Hence the count of such squares is 2.

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: The idea is to choose two points from the array of points such that these two points are parallel to co-ordinate axis and then find other two points of the square with the help of the distance between the points. If those points exist in the array then, there is one such possible square.

Below is the implementation of the above approach:

C++

 // C++ implementation to find count of Squares// that are parallel to the coordinate axis// from the given set of N points  #include using namespace std;  #define sz(x) int(x.size())  // Function to get distance// between two pointsint get_dis(pair p1,            pair p2){    int a = abs(p1.first - p2.first);    int b = abs(p1.second - p2.second);    return ((a * a) + (b * b));}  // Function to check that points// forms a square and parallel to// the co-ordinate axisbool check(pair p1,           pair p2,           pair p3,           pair p4){      int d2 = get_dis(p1, p2);    int d3 = get_dis(p1, p3);    int d4 = get_dis(p1, p4);    if (d2 == d3        && 2 * d2 == d4        && 2 * get_dis(p2, p4) == get_dis(p2, p3)) {        return true;    }    if (d3 == d4        && 2 * d3 == d2        && 2 * get_dis(p3, p2) == get_dis(p3, p4)) {        return true;    }    if (d2 == d4        && 2 * d2 == d3        && 2 * get_dis(p2, p3) == get_dis(p2, p4)) {        return true;    }    return false;}  // Function to find all the squares which is// parallel to co-ordinate axisint count(map, int> hash,          vector > v, int n){    int ans = 0;    map, int> vis;      // Loop to choose two points    // from the array of points    for (int i = 0; i < n; i++) {        for (int j = 0; j < n; j++) {            if (i == j)                continue;            pair p1                = make_pair(v[i].first,                            v[j].second);            pair p2                = make_pair(v[j].first,                            v[i].second);            set > s;            s.insert(v[i]);            s.insert(v[j]);            s.insert(p1);            s.insert(p2);            if (sz(s) != 4)                continue;              // Condition to check if the            // other points are present in the map            if (hash.find(p1) != hash.end()                && hash.find(p2) != hash.end()) {                if ((!vis[v[i]] || !vis[v[j]]                     || !vis[p1] || !vis[p2])                    && (check(v[i], v[j], p1, p2))) {                      vis[v[i]] = 1;                    vis[v[j]] = 1;                    vis[p1] = 1;                    vis[p2] = 1;                    ans++;                }            }        }    }    cout << ans;    return ans;}  // Function to Count the number of squaresvoid countOfSquares(vector > v, int n){    ios_base::sync_with_stdio(0);    cin.tie(0);      map, int> hash;      // Declaring iterator to a vector    vector >::iterator ptr;      // Adding the points to hash    for (ptr = v.begin(); ptr < v.end(); ptr++)        hash[*ptr] = 1;      // Count the number of squares    count(hash, v, n);}  // Driver Codeint main(){      int n = 5;    vector > v;    v.push_back(make_pair(0, 0));    v.push_back(make_pair(0, 2));    v.push_back(make_pair(2, 0));    v.push_back(make_pair(2, 2));    v.push_back(make_pair(0, 1));      // Function call    countOfSquares(v, n);    return 0;}
Output:
1

Performance Analysis:

• Time Complexity: As in the above approach, there are two loops which takes O(N2) time, Hence the Time Complexity will be O(N2).
• Auxiliary Space Complexity: As in the above approach, there is extra space used, Hence the auxiliary space complexity will be O(N).

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