# Area of the largest Rectangle without a given point

Given the length **L** and breadth **B** of a rectangle and the position of a hole in the rectangle as **(X, Y)** coordinate, the task is to find the area of largest Rectangle within the given Rectangle such that it does not contain the hole.**Note:** The rectangle is placed at the origin by two of its side touching the Co-ordinate axis.**Examples:**

Input:L = 8, B = 8, X = 0, Y = 0Output:56Explanation:

Since the hole is at origin, i.e. (0, 0), the maximum area rectangle can be cut from either (0, 1) or (1, 0) by reducing the length or breadth of the rectangle by one.

Hence, the maximum area rectangle that can be formed is = 7 * 8 = 56

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Input:L = 1, B = 10, X = 0, Y = 3Output:6Explanation:

Since the hole is at (0, 3), the maximum area rectangle can be cutted from the point (0, 4) by reducing the breadth to 6 and keeping the length as 1.

Hence, the maximum area rectangle that can be formed is = 6 * 1 = 6

**Approach:** In order to avoid the hole, the rectangle can be cut from either above, below, left or right of the hole, as:

Position - Maximum area of rectangle ------------------------------------ Left - X * B Right - (L - X - 1) * B Above - L * Y Below - (B - Y - 1) * L

Therefore, the required area of the largest rectangle can be computed by comparing the area calculated by using the above positions. The position with the largest area will yield the result.

Below is the implementation of the above approach:

## C++

`// C++ implementation to find area of` `// largest Rectangle without hole` `// within a given Rectangle` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Function to find the maximum area` `// such that it does not contains any hole` `void` `maximumArea(` `int` `l, ` `int` `b,` ` ` `int` `x, ` `int` `y)` `{` ` ` `// Area for all the possible` ` ` `// positions of the cut` ` ` `int` `left, right, above, below;` ` ` `left = x * b;` ` ` `right = (l - x - 1) * b;` ` ` `above = l * y;` ` ` `below = (b - y - 1) * l;` ` ` `// Find the maximum area` ` ` `// among the above rectangles` ` ` `cout << max(max(left, right),` ` ` `max(above, below));` `}` `// Driver Code` `int` `main()` `{` ` ` `int` `L = 8, B = 8;` ` ` `int` `X = 0, Y = 0;` ` ` `// Function call` ` ` `maximumArea(l, b, x, y);` ` ` `return` `0;` `}` |

## Java

`// Java implementation to find area of` `// largest Rectangle without hole` `// within a given Rectangle` `import` `java.util.*;` `class` `GFG{` ` ` `// Function to find the maximum area` `// such that it does not contains any hole` `static` `void` `maximumArea(` `int` `l, ` `int` `b,` ` ` `int` `x, ` `int` `y)` `{` ` ` ` ` `// Area for all the possible` ` ` `// positions of the cut` ` ` `int` `left, right, above, below;` ` ` ` ` `left = x * b;` ` ` `right = (l - x - ` `1` `) * b;` ` ` `above = l * y;` ` ` `below = (b - y - ` `1` `) * l;` ` ` ` ` `// Find the maximum area` ` ` `// among the above rectangles` ` ` `System.out.print(Math.max(Math.max(left, right),` ` ` `Math.max(above, below)));` `}` ` ` `// Driver Code` `public` `static` `void` `main(String[] args)` `{` ` ` `int` `L = ` `8` `, B = ` `8` `;` ` ` `int` `X = ` `0` `, Y = ` `0` `;` ` ` ` ` `// Function call` ` ` `maximumArea(L, B, X, Y);` `}` `}` `// This code is contributed by Rajput-Ji` |

## Python3

`# Python3 implementation to find area of` `# largest Rectangle without hole` `# within a given Rectangle` `# Function to find the maximum area` `# such that it does not contains any hole` `def` `maximumArea(l, b,x, y):` ` ` `# Area for all the possible` ` ` `# positions of the cut` ` ` `left, right, above, below ` `=` `0` `, ` `0` `, ` `0` `, ` `0` ` ` `left ` `=` `x ` `*` `b` ` ` `right ` `=` `(l ` `-` `x ` `-` `1` `) ` `*` `b` ` ` `above ` `=` `l ` `*` `y` ` ` `below ` `=` `(b ` `-` `y ` `-` `1` `) ` `*` `l` ` ` `# Find the maximum area` ` ` `# among the above rectangles` ` ` `print` `(` `max` `(` `max` `(left, right),` `max` `(above, below)))` `# Driver Code` `l ` `=` `8` `b ` `=` `8` `x ` `=` `0` `y ` `=` `0` `# Function call` `maximumArea(l, b, x, y)` `# This code is contributed by mohit kumar 29` |

## C#

`// C# implementation to find area of` `// largest Rectangle without hole` `// within a given Rectangle` `using` `System;` `class` `GFG{` ` ` `// Function to find the maximum area` `// such that it does not contains any hole` `static` `void` `maximumArea(` `int` `l, ` `int` `b,` ` ` `int` `x, ` `int` `y)` `{` ` ` ` ` `// Area for all the possible` ` ` `// positions of the cut` ` ` `int` `left, right, above, below;` ` ` ` ` `left = x * b;` ` ` `right = (l - x - 1) * b;` ` ` `above = l * y;` ` ` `below = (b - y - 1) * l;` ` ` ` ` `// Find the maximum area` ` ` `// among the above rectangles` ` ` `Console.Write(Math.Max(Math.Max(left, right),` ` ` `Math.Max(above, below)));` `}` ` ` `// Driver Code` `public` `static` `void` `Main(String[] args)` `{` ` ` `int` `L = 8, B = 8;` ` ` `int` `X = 0, Y = 0;` ` ` ` ` `// Function call` ` ` `maximumArea(L, B, X, Y);` `}` `}` `// This code is contributed by 29AjayKumar` |

## Javascript

`<script>` ` ` `// JavaScript implementation to find area of` ` ` `// largest Rectangle without hole` ` ` `// within a given Rectangle` ` ` `// Function to find the maximum area` ` ` `// such that it does not contains any hole` ` ` `function` `maximumArea(l, b, x, y) {` ` ` `// Area for all the possible` ` ` `// positions of the cut` ` ` `var` `left = x * b;` ` ` `var` `right = (l - x - 1) * b;` ` ` `var` `above = l * y;` ` ` `var` `below = (b - y - 1) * l;` ` ` `// Find the maximum area` ` ` `// among the above rectangles` ` ` `document.write(Math.max(Math.max(left, right), Math.max(above, below)));` ` ` `}` ` ` `// Driver Code` ` ` `var` `L = 8,` ` ` `B = 8;` ` ` `var` `X = 0,` ` ` `Y = 0;` ` ` `// Function call` ` ` `maximumArea(L, B, X, Y);` ` ` `</script>` |

**Output:**

56

**Performance Analysis:**

**Time Complexity:**There is a simple computation that does not involve any iterations or recursions. Hence, the Time Complexity will be**O(1)**.**Auxiliary Space Complexity:**There is no extra space used. Hence, the auxiliary space complexity will be**O(1)**.