Given a positive integer N, the task is to find the maximum number of unique squares that can be formed with N arbitrary points in the coordinate plane.
Note: Any two squares that are not overlapping are considered unique.
Examples:
Input: N = 9
Output: 5
Explanation:
Consider the below square consisting of N points:
The squares ABEF, BCFE, DEHG, EFIH is one of the possible squares of size 1 which are non-overlapping with each other.
The square ACIG is also one of the possible squares of size 2.Input: N = 6
Output: 2
Approach: This problem can be solved based on the following observations:
- Observe that if N is a perfect square then the maximum number of squares will be formed when sqrt(N)*sqrt(N) points form a grid of sqrt(N)*sqrt(N) and all of them are equally spaces.
- But when N is not a perfect square, then it still forms a grid but with the greatest number which is a perfect square having a value less than N.
- The remaining coordinates can be placed around the edges of the grid which will lead to maximum possible squares.
Follow the below steps to solve the problem:
- Initialize a variable, say ans that stores the resultant count of squares formed.
- Find the maximum possible grid size as sqrt(N) and the count of all possible squares formed up to length len to the variable ans which can be calculated byÂ
. - Decrement the value of N by len*len.
- If the value of N is at least len, then all other squares can be formed by placing them in another cluster of points. Find the count of squares as calculated in Step 2 for the value of len.
- After completing the above steps, print the value of ans as the result.
Below is the implementation of the above approach:
C++
// C++ program for the above approachÂ
#include <bits/stdc++.h>using namespace std;Â
// Function to find the maximum number// of unique squares that can be formed// from the given N pointsint maximumUniqueSquares(int N){    // Stores the resultant count of    // squares formed    int ans = 0;Â
    // Base Case    if (N < 4) {        return 0;    }Â
    // Subtract the maximum possible    // grid size as sqrt(N)    int len = (sqrt(double(N)));    N -= len * len;Â
    // Find the total squares till now    // for the maximum grid    for (int i = 1; i < len; i++) {Â
        // A i*i grid contains (i-1)*(i-1)        // + (i-2)*(i-2) + ... + 1*1 squares        ans += i * i;    }Â
    // When N >= len then more squares    // will be counted    if (N >= len) {        N -= len;        for (int i = 1; i < len; i++) {            ans += i;        }    }Â
    for (int i = 1; i < N; i++) {        ans += i;    }Â
    // Return total count of squares    return ans;}Â
// Driver Codeint main(){Â Â Â Â int N = 9;Â Â Â Â cout << maximumUniqueSquares(N);Â
    return 0;} |
Java
// Java program for the above approachimport java.io.*;Â
class GFG {Â
// Function to find the maximum number// of unique squares that can be formed// from the given N pointsstatic int maximumUniqueSquares(int N){    // Stores the resultant count of    // squares formed    int ans = 0;Â
    // Base Case    if (N < 4) {        return 0;    }Â
    // Subtract the maximum possible    // grid size as sqrt(N)    int len = (int)(Math.sqrt(N));    N -= len * len;Â
    // Find the total squares till now    // for the maximum grid    for (int i = 1; i < len; i++) {Â
        // A i*i grid contains (i-1)*(i-1)        // + (i-2)*(i-2) + ... + 1*1 squares        ans += i * i;    }Â
    // When N >= len then more squares    // will be counted    if (N >= len) {        N -= len;        for (int i = 1; i < len; i++) {            ans += i;        }    }Â
    for (int i = 1; i < N; i++) {        ans += i;    }Â
    // Return total count of squares    return ans;}Â
// Driver Codepublic static void main (String[] args) {Â Â Â Â int N = 9;Â Â Â Â System.out.println( maximumUniqueSquares(N));Â
}}Â
// This code is contributed by shivanisinghss2110. |
Python3
# Python program for the above approachÂ
# for math functionimport mathÂ
# Function to find the maximum number# of unique squares that can be formed# from the given N pointsdef maximumUniqueSquares(N):       # Stores the resultant count of    # squares formed    ans = 0         # Base Case    if N < 4:        return 0Â
    # Subtract the maximum possible    # grid size as sqrt(N)    len = int(math.sqrt(N))Â
    N -= len * lenÂ
    # Find the total squares till now    # for the maximum grid    for i in range(1, len):Â
        # A i*i grid contains (i-1)*(i-1)        # + (i-2)*(i-2) + ... + 1*1 squares        ans += i * iÂ
    # When N >= len then more squares    # will be counted    if (N >= len):        N -= len        for i in range(1, len):            ans += iÂ
    for i in range(1, N):        ans += iÂ
    # Return total count of squares    return ansÂ
# Driver Codeif __name__ == "__main__":Â Â Â Â N = 9Â Â Â Â print(maximumUniqueSquares(N))Â Â Â Â Â Â Â Â Â # This code is contributed by rakeshsahni |
C#
// C# program for the above approachusing System;Â
public class GFG{Â
    // Function to find the maximum number    // of unique squares that can be formed    // from the given N points    static int maximumUniqueSquares(int N)    {               // Stores the resultant count of        // squares formed        int ans = 0;             // Base Case        if (N < 4) {            return 0;        }             // Subtract the maximum possible        // grid size as sqrt(N)        int len = (int)(Math.Sqrt(N));        N -= len * len;             // Find the total squares till now        // for the maximum grid        for (int i = 1; i < len; i++) {                 // A i*i grid contains (i-1)*(i-1)            // + (i-2)*(i-2) + ... + 1*1 squares            ans += i * i;        }             // When N >= len then more squares        // will be counted        if (N >= len) {            N -= len;            for (int i = 1; i < len; i++) {                ans += i;            }        }             for (int i = 1; i < N; i++) {            ans += i;        }             // Return total count of squares        return ans;    }         // Driver Code    public static void Main (string[] args)     {        int N = 9;        Console.WriteLine( maximumUniqueSquares(N));         }}Â
// This code is contributed by AnkThon |
Javascript
<script>Â
Â
// Javascript program for the above approachÂ
// Function to find the maximum number// of unique squares that can be formed// from the given N pointsfunction maximumUniqueSquares(N){    // Stores the resultant count of    // squares formed    var ans = 0;    var i;Â
    // Base Case    if (N < 4) {        return 0;    }Â
    // Subtract the maximum possible    // grid size as sqrt(N)    var len = Math.sqrt(N);    N -= len * len;Â
    // Find the total squares till now    // for the maximum grid    for (i = 1; i < len; i++) {Â
        // A i*i grid contains (i-1)*(i-1)        // + (i-2)*(i-2) + ... + 1*1 squares        ans += i * i;    }Â
    // When N >= len then more squares    // will be counted    if (N >= len) {        N -= len;        for (i = 1; i < len; i++) {            ans += i;        }    }Â
    for (i = 1; i < N; i++) {        ans += i;    }Â
    // Return total count of squares    return ans;}Â
// Driver Code    var N = 9;    document.write(maximumUniqueSquares(N));Â
// This code is contributed by SURENDRA_GANGWAR.</script> |
Output:Â
5
Â
Time Complexity: O(sqrt(N))Â
Auxiliary Space: O(1)
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