Codlility---MinPerimeterRectangle

Posted 2020-09-22 Hugh

Task description

An integer N is given, representing the area of some rectangle.

The area of a rectangle whose sides are of length A and B is A * B, and theperimeter is 2 * (A + B).

The goal is to find the minimal perimeter of any rectangle whose area equals N. The sides of this rectangle should be only integers.

For example, given integer N = 30, rectangles of area 30 are:

• (1, 30), with a perimeter of 62,
• (2, 15), with a perimeter of 34,
• (3, 10), with a perimeter of 26,
• (5, 6), with a perimeter of 22.

Write a function:

class Solution { public int solution(int N); }

that, given an integer N, returns the minimal perimeter of any rectangle whose area is exactly equal to N.

For example, given an integer N = 30, the function should return 22, as explained above.

Assume that:

• N is an integer within the range [1..1,000,000,000].

Complexity:

• expected worst-case time complexity is O(sqrt(N));
• expected worst-case space complexity is O(1).

 

 

Solution

 

Programming language used: Java

Code: 00:48:19 UTC, java, final, score:  100

show code in pop-up

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// you can also use imports, for example:
// import java.util.*;

// you can write to stdout for debugging purposes, e.g.
// System.out.println("this is a debug message");

class Solution {
    public int solution(int N) {
        // write your code in Java SE 8
        int max_perimeter = 2*(1+N);
        for(int i=2; i < Math.sqrt(N)+1; i++) {
            if(N%i == 0) {
                max_perimeter = Math.min(max_perimeter, 2*(i+N/i));
            }
        }
        return max_perimeter;
    }
}


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