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Leetcode 3000. Maximum Area of Longest Diagonal Rectangle

Problem Statement:

You are given a rectangle of dimensions n x m. You need to find the maximum area of a rectangle that can be cut out from this rectangle such that the cut-out rectangle has the same aspect ratio as the original rectangle.

Clarifying Questions:

1. Constraints: Are there any constraints on the values of n and m?
   • Let’s assume typical constraints up to (10^9).
2. Output: Should the result be an integer (if we’re to find max area) or a specific rectangle’s dimensions?
   • Let’s assume the problem asks for the maximum area.
3. Aspect Ratio: By “aspect ratio,” are we assuming it is the width to height or height to width?
   • Normally for “n x m,” “n” is the height and “m” is the width, so the aspect ratio will be height/width = n/m.

Strategy:

We observe that maintaining the aspect ratio of ( n/m ) means any rectangle we cut out of the given rectangle must match this ratio.

So, if our intention is to find the maximum possible area of a rectangle with the same aspect ratio, then it must be a portion of the original rectangle itself where both sides are scaled down but maintain the aspect ratio.

Given the constraints, the maximum area will actually be the area of the original rectangle itself because you can’t find a larger subrectangle maintaining the same ratio without scaling.

Code:

Here is how you’d implement this in C++:

#include <iostream>

int main() {
    long long n, m;
    std::cin >> n >> m;

    long long maxArea = n * m;
    std::cout << maxArea << std::endl;

    return 0;
}

Time Complexity:

• Time Complexity: The time complexity is (O(1)) because we’re simply multiplying two numbers and printing the result.
• Space Complexity: The space complexity is (O(1)) because we’re only storing two integers and the result.

Explanation:

• We take n and m as inputs which represent the dimensions of the given rectangle.
• The highest possible area maintaining the aspect ratio would be the full rectangle itself.
• Thus, we simply calculate ( n \times m ) and print this value.

This solution leverages simple arithmetic and basic I/O functionality to deliver the maximum area of a subrectangle cut out with the same aspect ratio as the original.

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