Number of squares of side length required to cover an N*M rectangle
Given three numbers ,
,
. Find Number of squares of dimension
required to cover
rectangle.
Note:
- It’s allowed to cover the surface larger than the rectangle, but the rectangle has to be covered.
- It’s not allowed to break a square.
- The sides of squares should be parallel to the sides of the rectangle.
Examples:
Input: N = 6, M = 6, a = 4 Output: 4 Input: N = 2, M = 3, a = 1 Output: 6
Approach: An efficient approach is to make an observation and find a formula. The constraint that edges of each square must be parallel to the edges of the rectangle allows to analyze X and Y axes separately, that is, how many squares of length ‘a’ are needed to cover squares of length ‘m’ and ‘n’ and take the product of these two quantities. The number of small squares of side length ‘a’ required to cover ‘m’ sized square are ceil(m/a). Similarly, number of ‘a’ sized squares required to cover ‘n’ sized square are ceil(n/a).
So, the answer will be ceil(m/a)*ceil(n/a).
Below is the implementation of the above approach:
C++
// CPP program to find number of squares // of a*a required to cover n*m rectangle #include <bits/stdc++.h> using namespace std; // function to find number of squares // of a*a required to cover n*m rectangle int Squares( int n, int m, int a) { return ((m + a - 1) / a) * ((n + a - 1) / a); } // Driver code int main() { int n = 6, m = 6, a = 4; // function call cout << Squares(n, m, a); return 0; } |
C
// C program to find number of squares // of a*a required to cover n*m rectangle #include <stdio.h> // function to find number of squares // of a*a required to cover n*m rectangle int Squares( int n, int m, int a) { return ((m + a - 1) / a) * ((n + a - 1) / a); } // Driver code int main() { int n = 6, m = 6, a = 4; // function call printf ( "%d" ,Squares(n, m, a)); return 0; } // This code is contributed by kothavvsaakash. |
Java
// Java program to find number of squares // of a*a required to cover n*m rectangle import java.util.*; class solution { // function to find a number of squares // of a*a required to cover n*m rectangle static int Squares( int n, int m, int a) { return ((m + a - 1 ) / a) * ((n + a - 1 ) / a); } // Driver code public static void main(String arr[]) { int n = 6 , m = 6 , a = 4 ; // function call System.out.println(Squares(n, m, a)); } } //This code is contributed by Surendra_Gangwar |
Python 3
# Python 3 program to find number # of squares of a*a required to # cover n*m rectangle # function to find number of # squares of a*a required to # cover n*m rectangle def Squares(n, m, a): return (((m + a - 1 ) / / a) * ((n + a - 1 ) / / a)) # Driver code if __name__ = = "__main__" : n = 6 m = 6 a = 4 # function call print (Squares(n, m, a)) # This code is contributed # by ChitraNayal |
C#
// CSHARP program to find number of squares // of a*a required to cover n*m rectangle using System; class GFG { // function to find a number of squares // of a*a required to cover n*m rectangle static int Squares( int n, int m, int a) { return ((m + a - 1) / a) * ((n + a - 1) / a); } static void Main() { int n = 6, m = 6, a = 4; // function call Console.WriteLine(Squares(n, m, a)); } // This code is contributed by ANKITRAI1 } |
PHP
<?php // PHP program to find number of squares // of a*a required to cover n*m rectangle // function to find number of squares // of a*a required to cover n*m rectangle function Squares( $n , $m , $a ) { return ((int)(( $m + $a - 1) / $a )) * ((int)(( $n + $a - 1) / $a )); } // Driver code $n = 6; $m = 6; $a = 4; // function call echo Squares( $n , $m , $a ); // This code is contributed // by Akanksha Rai ?> |
Javascript
<script> // JavaScript program to find number of squares // of a*a required to cover n*m rectangle // function to find a number of squares // of a*a required to cover n*m rectangle function Squares(n, m, a) { return parseInt(((m + a - 1) / a)) * parseInt(((n + a - 1) / a)); } // Driver code var n = 6, m = 6, a = 4; // Function call document.write(Squares(n, m, a)); // This code is contributed by Ankita saini </script> |
4
Time Complexity: O(1), since there is no loop or recursion.
Auxiliary Space: O(1), since no extra space has been taken.
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