Tasks Details
easy
Find the minimal perimeter of any rectangle whose area equals N.
Task Score
100%
Correctness
100%
Performance
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].
Copyright 2009–2025 by Codility Limited. All Rights Reserved. Unauthorized copying, publication or disclosure prohibited.
Solution
Programming language used Java 8
Time spent on task 14 minutes
Notes
not defined yet
Code: 10:12:18 UTC,
java,
autosave
Code: 10:25:08 UTC,
java,
verify,
result: Passed
// 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 sqrtN = (int)Math.sqrt(N);
//System.out.println(sqrtN);
int B = 0;
int peri = Integer.MAX_VALUE;
for (int A=1; A<=sqrtN; A++) {
if (N%A == 0) {
B = N/A;
peri = Math.min(2*(A+B), peri);
}
}
return peri;
}
}
User test case 1:
[30]
User test case 2:
[20]
User test case 3:
[10]
User test case 4:
[1]
User test case 5:
[5]
User test case 6:
[9]
Analysis
expand all
User tests
1.
0.004 s
OK
function result: 22
function result: 22
1.
0.004 s
OK
function result: 18
function result: 18
1.
0.004 s
OK
function result: 14
function result: 14
1.
0.004 s
OK
function result: 4
function result: 4
1.
0.004 s
OK
function result: 12
function result: 12
1.
0.004 s
OK
function result: 12
function result: 12
Code: 10:25:17 UTC,
java,
verify,
result: Passed
// 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 sqrtN = (int)Math.sqrt(N);
//System.out.println(sqrtN);
int B = 0;
int peri = Integer.MAX_VALUE;
for (int A=1; A<=sqrtN; A++) {
if (N%A == 0) {
B = N/A;
peri = Math.min(2*(A+B), peri);
}
}
return peri;
}
}
User test case 1:
[30]
User test case 2:
[20]
User test case 3:
[10]
User test case 4:
[1]
User test case 5:
[5]
User test case 6:
[9]
Analysis
expand all
User tests
1.
0.004 s
OK
function result: 22
function result: 22
1.
0.008 s
OK
function result: 18
function result: 18
1.
0.008 s
OK
function result: 14
function result: 14
1.
0.004 s
OK
function result: 4
function result: 4
1.
0.004 s
OK
function result: 12
function result: 12
1.
0.008 s
OK
function result: 12
function result: 12
Code: 10:25:22 UTC,
java,
final,
score:
100
// 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 sqrtN = (int)Math.sqrt(N);
//System.out.println(sqrtN);
int B = 0;
int peri = Integer.MAX_VALUE;
for (int A=1; A<=sqrtN; A++) {
if (N%A == 0) {
B = N/A;
peri = Math.min(2*(A+B), peri);
}
}
return peri;
}
}
Analysis summary
The solution obtained perfect score.
Analysis
Detected time complexity:
O(sqrt(N))