Given a set of N points in the 2-D plane. The task is to find the minimum value of M such that a square centered at the origin with side 2*M contains at least floor(N/2) points inside or on it.
Examples:
Input : N = 4
Points are: {(1, 2), (-3, 4), (1, 78), (-3, -7)}
Output : 4
The square with end point (4, 4), (-4, -4), (4, -4), (-4, 4) will contain the points (1, 2) and (-3, 4).
Smallest Possible value of M such that the square has at least 2 points is 4.
Input : N = 3
Points are: {(1, 2), (-3, 4), (1, 78)}
Output : 2
Square contains the point (1, 2). {floor(3/2) = 1}
Approach:
- One major observation for any point (x, y), minimum M in which this point lies is max(abs(x), abs(y)).
- Using point 1. We can find the minimum value of M for all the points and store them in an array.
- Sort the array.
- Now, array[i] denotes minimum M such that if i points are required in the square of side 2*M. (as all the points below i have the minimum value of M less than or equal to i).
- Print the value of array[floor(n/2)].
Below is the implementation of the above approach:
C++
// C++ implementation of the above approach#include <bits/stdc++.h>using namespace std;// Function to Calculate Absolute Valueint mod(int x){ if (x >= 0) return x; return -x;}// Function to Calculate the Minimum value of Mvoid findSquare(int n){ int points[n][2] = { { 1, 2 }, { -3, 4 }, { 1, 78 }, { -3, -7 } }; int a[n]; // To store the minimum M for each // point in array for (int i = 0; i < n; i++) { int x, y; x = points[i][0]; y = points[i][1]; a[i] = max(mod(x), mod(y)); } // Sort the array sort(a, a + n); // Index at which atleast required point are // inside square of length 2*M int index = floor(n / 2) - 1; cout << "Minimum M required is: " << a[index] << endl;}// Driver Codeint main(){ int N; N = 4; findSquare(N); return 0;} |
Java
import java.util.*;// Java program to find next identical yearclass GFG{// Function to Calculate Absolute Valuestatic int mod(int x){ if (x >= 0) return x; return -x;}// Function to Calculate the Minimum value of Mstatic void findSquare(int n){ int points[][] = { { 1, 2 }, { -3, 4 }, { 1, 78 }, { -3, -7 } }; int []a = new int[n]; // To store the minimum M for each // point in array for (int i = 0; i < n; i++) { int x, y; x = points[i][0]; y = points[i][1]; a[i] = Math.max(mod(x), mod(y)); } // Sort the array Arrays.sort(a); // Index at which atleast required point are // inside square of length 2*M int index = (int) (Math.floor(n / 2) - 1); System.out.println("Minimum M required is: " + a[index]);}// Driver Codepublic static void main(String[] args) { int N; N = 4; findSquare(N);}}// This code contributed by Rajput-Ji |
Python3
# Python3 implementation of the# above approach # Function to Calculate the # Minimum value of M def findSquare(n): points = [[1, 2], [-3, 4], [1, 78], [-3, -7]] a = [None] * n # To store the minimum M # for each point in array for i in range(0, n): x = points[i][0] y = points[i][1] a[i] = max(abs(x), abs(y)) # Sort the array a.sort() # Index at which atleast required # point are inside square of length 2*M index = n // 2 - 1 print("Minimum M required is:", a[index]) # Driver Code if __name__ == "__main__": N = 4 findSquare(N) # This code is contributed # by Rituraj Jain |
C#
// C# program to find next identical year using System;class GFG { // Function to Calculate Absolute Value static int mod(int x) { if (x >= 0) return x; return -x; } // Function to Calculate the Minimum value of M static void findSquare(int n) { int [,]points = new int[4,2]{ { 1, 2 }, { -3, 4 }, { 1, 78 }, { -3, -7 } }; int []a = new int[n]; // To store the minimum M for each // point in array for (int i = 0; i < n; i++) { int x, y; x = points[i,0]; y = points[i,1]; a[i] = Math.Max(mod(x), mod(y)); } // Sort the array Array.Sort(a); // Index at which atleast required point are // inside square of length 2*M int index = (int) (n / 2 - 1); Console.WriteLine("Minimum M required is: " + a[index]); } // Driver Code public static void Main(String []args) { int N; N = 4; findSquare(N); } } // This code contributed by Arnab Kundu |
PHP
<?php// PHP implementation of the above approach// Function to Calculate Absolute Valuefunction mod($x){ if ($x >= 0) return $x; return -$x;}// Function to Calculate the// Minimum value of Mfunction findSquare($n){ $points = array(array( 1, 2 ), array( -3, 4 ), array( 1, 78 ), array( -3, -7 )); $a[$n] = array(); // To store the minimum M for each // point in array for ($i = 0; $i < $n; $i++) { $x; $y; $x = $points[$i][0]; $y = $points[$i][1]; $a[$i] = max(mod($x), mod($y)); } // Sort the array sort($a); // Index at which atleast required point // are inside square of length 2*M $index = floor($n / 2) - 1; echo "Minimum M required is: ", $a[$index], "\n";}// Driver Code$N = 4;findSquare($N);// This code is contributed by ajit.?> |
Javascript
<script> // Javascript program to find next identical year // Function to Calculate Absolute Value function mod(x) { if (x >= 0) return x; return -x; } // Function to Calculate the Minimum value of M function findSquare(n) { let points = [ [ 1, 2 ], [ -3, 4 ], [ 1, 78 ], [ -3, -7 ] ]; let a = new Array(n); // To store the minimum M for each // point in array for (let i = 0; i < n; i++) { let x, y; x = points[i][0]; y = points[i][1]; a[i] = Math.max(mod(x), mod(y)); } // Sort the array a.sort(function(a, b){return a - b}); // Index at which atleast required point are // inside square of length 2*M let index = (Math.floor(n / 2) - 1); document.write("Minimum M required is: " + a[index]); } let N; N = 4; findSquare(N); </script> |
Output:
Minimum M required is: 4
Time Complexity: O(n*log(n))
Auxiliary Space: O(n)
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