Given a square of size n. There are n2 small squares inside the square n of size 1 unit each, in which any one of the squares is colored. Our task is to cut the square n into two equal parts. The cutting line should not have any common points with the colored cell and the resulting two parts should be equal up to rotation. Print “YES” if it’s possible to cut the square with such conditions and “NO” otherwise.
Note: The value of n should always be an even positive number.
Examples:
Input : n = 4, x = 1, y = 1 Output : YES // n = 4 and 1 1 is the coordinate of the colored square Input : n = 2, x = 1, y = 1 Output : NO
In the first example, the painted square has a coordinate 1 x 1. So we have to cut the bigger square into two parts so that there should not be any common point with the colored cell. The bold line shown in the above picture cuts the square into two equal parts.
Below is the step by step algorithm to solve this problem:
1 . Initialize the size of the square and the position of the painted square.
2 . Dividing a square into two equal parts will only be possible if the cutting line passes through the center of our bigger square.
3 . Thus, if the painted square is linked anyway to the center of the bigger square, then it is not possible to cut the bigger square into two equal parts.
4 . So to check, divide the size of the bigger square into half and check if any dimension of a painted square is linked to it.
Below is the implementation of the above approach:
C++
// C++ program to illustrate// the above problem#include <bits/stdc++.h>using namespace std;// function to check if it's possible to// divide the square in two equal partsvoid halfsquare(int n, int x, int y){ int half = n / 2; // if the painted square is linked anyway // to the center of the square // then it's not possible if ((half == x || half == x - 1) && (half == y || half == y - 1)) cout << "NO" << endl; // else yes it's possible else cout << "YES" << endl;}// Driver codeint main(){ // initialize the size of the square int n = 100; // initialize the dimension of the painted square int x = 51, y = 100; halfsquare(n, x, y); return 0;} |
Java
// Java program to illustrate// the above problemimport java.io.*;class GFG { // function to check if it's possible to// divide the square in two equal partsstatic void halfsquare(int n, int x, int y){ int half = n / 2; // if the painted square is linked anyway // to the center of the square // then it's not possible if ((half == x || half == x - 1) && (half == y || half == y - 1)) System.out.println( "NO"); // else yes it's possible else System.out.println( "YES");}// Driver code public static void main (String[] args) { // initialize the size of the square int n = 100; // initialize the dimension of the painted square int x = 51, y = 100; halfsquare(n, x, y); }}// This code is contributed // by inder_verma.. |
Python 3
# Python 3 program to illustrate # the above problem # function to check if it's possible to # divide the square in two equal parts def halfsquare(n, x, y) : half = n // 2 # if the painted square is # linked anyway to the center # of the square then it's # not possible if ((half == x or half == x - 1) and (half == y or half == y - 1)) : print("NO") # else yes it's possible else : print("YES") # Driver code if __name__ == "__main__" : # initialize the size of the square n = 100 # initialize the dimension # of the painted square x, y = 51, 100 halfsquare(n, x, y)# This code is contributed by ANKITRAI1 |
C#
// C# program to illustrate// the above problemusing System;class GFG {// function to check if it's possible to// divide the square in two equal partsstatic void halfsquare(int n, int x, int y){ int half = n / 2; // if the painted square is linked anyway // to the center of the square // then it's not possible if ((half == x || half == x - 1) && (half == y || half == y - 1)) Console.WriteLine( "NO"); // else yes it's possible else Console.WriteLine( "YES");}// Driver code public static void Main () { // initialize the size of the square int n = 100; // initialize the dimension of the painted square int x = 51, y = 100; halfsquare(n, x, y); }}// This code is contributed // by anuj_67.. |
PHP
<?php// PHP program to illustrate// the above problem// function to check if it's// possible to divide the// square in two equal partsfunction halfsquare($n, $x, $y){ $half = $n / 2; // if the painted square is // linked anyway to the center // of the square then it's // not possible if (($half == $x || $half == $x - 1) && ($half == $y || $half == $y - 1)) echo "NO" ; // else yes it's possible else echo "YES" ;}// Driver code// initialize the size // of the square$n = 100;// initialize the dimension// of the painted square$x = 51; $y = 100;halfsquare($n, $x, $y);// This code is contributed // by anuj_67?> |
Javascript
<script>// Java script program to illustrate// the above problem// function to check if it's possible to// divide the square in two equal partsfunction halfsquare(n,x,y){ let half = n / 2; // if the painted square is linked anyway // to the center of the square // then it's not possible if ((half == x || half == x - 1) && (half == y || half == y - 1)) document.write( "NO"); // else yes it's possible else document.write( "YES");}// Driver code // initialize the size of the square let n = 100; // initialize the dimension of the painted square let x = 51, y = 100; halfsquare(n, x, y); // This code is contributed by sravan kumar Gottumukkala</script> |
Output:
YES
Time complexity: O(1) because it is performing constant operations
Auxiliary Space: O(1)
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