Test results - Codility

Tasks Details

easy

1. MinPerimeterRectangle

Find the minimal perimeter of any rectangle whose area equals N.

100%

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 the perimeter 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.

Write an efficient algorithm for the following assumptions:

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

Solution

Programming language used Java 8

Time spent on task 5 minutes

Notes

not defined yet

Code: 14:24:19 UTC, java, verify, result: Passed

1 2 3 4 5 6 7 8 9 10 11 12 13 14

import java.util.*;

class Solution {
    public int solution(int N) {
        int root = (int) Math.sqrt(N);
        for (int i=root; i<=N; i++) {
            if (N % i == 0) {
                return ((i + (N / i)) * 2);
            }    
        }
        
        return -1;
    }
}

Analysis

▶

example
example test

✔

▶

test_data1

✔

Code: 14:28:01 UTC, java, verify, result: Passed

1 2 3 4 5 6 7 8 9 10 11 12 13 14

import java.util.*;

class Solution {
    public int solution(int N) {
        int root = (int) Math.sqrt(N);
        for (int i=root; i>0; i--) {
            if (N % i == 0) {
                return ((i + (N / i)) * 2);
            }    
        }
        
        return -1;
    }
}

Analysis

▶

example
example test

✔

▶

test_data1

✔

Code: 14:28:10 UTC, java, verify, result: Passed

1 2 3 4 5 6 7 8 9 10 11 12 13 14

import java.util.*;

class Solution {
    public int solution(int N) {
        int root = (int) Math.sqrt(N);
        for (int i=root; i>0; i--) {
            if (N % i == 0) {
                return ((i + (N / i)) * 2);
            }    
        }
        
        return -1;
    }
}

Analysis

▶

example
example test

✔

▶

test_data1

✔

Code: 14:28:14 UTC, java, final, score: 100

1 2 3 4 5 6 7 8 9 10 11 12 13 14

import java.util.*;

class Solution {
    public int solution(int N) {
        int root = (int) Math.sqrt(N);
        for (int i=root; i>0; i--) {
            if (N % i == 0) {
                return ((i + (N / i)) * 2);
            }    
        }
        
        return -1;
    }
}

Analysis summary

The solution obtained perfect score.

Analysis

Detected time complexity:

O(sqrt(N))

▶

example
example test

✔

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extreme_min
N = 1 test

✔

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simple1
N = 36 test

✔

▶

simple2
N = 48 test

✔

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simple3
N = 101 test

✔

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small
N = 1,234 test

✔

▶

medium
N = 4,564,320 test

✔

▶

prime1
N = 15,486,451 test

✔

▶

square
N = 100,000,000 test

✔

▶

prime2
N = 982,451,653 test

✔

▶

extreme_max
N = 1,000,000,000 test

✔