Given the length L and breadth B of a sheet of paper, the task is to find the maximum number of rectangles with given length l and breadth b that can be cut from this sheet of paper.
Examples:
Input: L = 5, B = 2, l = 14, b = 3
Output: 0
The sheet is smaller than the required rectangle. So, no rectangle of the given dimension can be cut from the sheet.
Input: L = 10, B = 7, l = 4, b = 3
Output: 4
Approach:
- Try to cut the rectangles horizontally i.e. length of the rectangle is aligned with the length of the sheet and breadth of the rectangle is aligned with the breadth of the sheet and store the count of rectangles possible in horizontal.
- Repeat the same with vertical alignment i.e. when length of the rectangle is aligned with the breadth of the sheet and breadth of the rectangle is aligned with the length of the sheet and store the result in vertical. Print max(horizontal, vertical) as the result.
Below is the implementation of the above approach:
C++
// CPP implementation of the approach#include<bits/stdc++.h>using namespace std;// Function to return the maximum rectangles possibleint maxRectangles(int L, int B, int l, int b){ int horizontal = 0, vertical = 0; // Cut rectangles horizontally if possible if (l <= L && b <= B) { // One rectangle is a single cell int columns = B / b; int rows = L / l; // Total rectangles = total cells horizontal = rows * columns; } // Cut rectangles vertically if possible if (l <= B && b <= L) { int columns = L / b; int rows = B / l; vertical = rows * columns; } // Return the maximum possible rectangles return max(horizontal, vertical);}// Driver codeint main(){ int L = 10, B = 7, l = 4, b = 3; cout << (maxRectangles(L, B, l, b)) << endl;}// This code is contributed by// Sanjit_Prasad |
Java
// Java implementation of the approachclass GFG { // Function to return the maximum rectangles possible static int maxRectangles(int L, int B, int l, int b) { int horizontal = 0, vertical = 0; // Cut rectangles horizontally if possible if (l <= L && b <= B) { // One rectangle is a single cell int columns = B / b; int rows = L / l; // Total rectangles = total cells horizontal = rows * columns; } // Cut rectangles vertically if possible if (l <= B && b <= L) { int columns = L / b; int rows = B / l; vertical = rows * columns; } // Return the maximum possible rectangles return Math.max(horizontal, vertical); } // Driver code public static void main(String[] args) { int L = 10, B = 7, l = 4, b = 3; System.out.print(maxRectangles(L, B, l, b)); }} |
Python3
# Python3 implementation of the approach # Function to return the maximum # rectangles possible def maxRectangles(L, B, l, b): horizontal, vertical = 0, 0 # Cut rectangles horizontally if possible if l <= L and b <= B: # One rectangle is a single cell columns = B // b rows = L // l # Total rectangles = total cells horizontal = rows * columns # Cut rectangles vertically if possible if l <= B and b <= L: columns = L // b rows = B // l vertical = rows * columns # Return the maximum possible rectangles return max(horizontal, vertical) # Driver code if __name__ == "__main__": L, B, l, b = 10, 7, 4, 3 print(maxRectangles(L, B, l, b)) # This code is contributed by Rituraj Jain |
C#
// C# implementation of the above approach using System;class GFG { // Function to return the // maximum rectangles possible static int maxRectangles(int L, int B, int l, int b) { int horizontal = 0, vertical = 0; // Cut rectangles horizontally if possible if (l <= L && b <= B) { // One rectangle is a single cell int columns = B / b; int rows = L / l; // Total rectangles = total cells horizontal = rows * columns; } // Cut rectangles vertically if possible if (l <= B && b <= L) { int columns = L / b; int rows = B / l; vertical = rows * columns; } // Return the maximum possible rectangles return Math.Max(horizontal, vertical); } // Driver code public static void Main() { int L = 10, B = 7, l = 4, b = 3; Console.WriteLine(maxRectangles(L, B, l, b)); } } // This code is contributed by Ryuga |
PHP
<?php// PHP implementation of the approach// Function to return the maximum // rectangles possiblefunction maxRectangles($L, $B, $l, $b){ $horizontal = 0; $vertical = 0; // Cut rectangles horizontally if possible if ($l <= $L && $b <= $B) { // One rectangle is a single cell $columns = (int)($B / $b); $rows = (int)($L / $l); // Total rectangles = total cells $horizontal = $rows * $columns; } // Cut rectangles vertically if possible if ($l <= $B && $b <= $L) { $columns = (int)($L / $b); $rows = (int)($B / $l); $vertical = $rows * $columns; } // Return the maximum possible rectangles return max($horizontal, $vertical);}// Driver code$L = 10;$B = 7;$l = 4;$b = 3;print(maxRectangles($L, $B, $l, $b));// This code is contributed by mits?> |
Javascript
<script>// Javascript implementation of the approach// Function to return the maximum // rectangles possiblefunction maxRectangles(L, B, l, b){ var horizontal = 0, vertical = 0; // Cut rectangles horizontally // if possible if (l <= L && b <= B) { // One rectangle is a single cell var columns = parseInt(B / b); var rows = parseInt(L / l); // Total rectangles = total cells horizontal = rows * columns; } // Cut rectangles vertically if possible if (l <= B && b <= L) { var columns = parseInt(L / b); var rows = parseInt(B / l); vertical = rows * columns; } // Return the maximum possible rectangles return Math.max(horizontal, vertical);}// Driver Codevar L = 10, B = 7, l = 4, b = 3;document.write(maxRectangles(L, B, l, b));// This code is contributed by kirti</script> |
Output:
4
Time Complexity: O(1), since there is only a basic arithmetic operation that takes constant time.
Auxiliary Space: O(1), since no extra space has been taken.
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