can someone tell me whats wrong with this one ??

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public class Solution { public static int min(int a, int b, int c, int d) { int min2; int e = min(a, b); if (e < c && e < b) min2 = e; else if (c < e && c < d) min2 = c; else min2 = d; return min2; } public static int min(int a, int b) { int min; if (a < b) min = a; else min = b; return min;

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Guadalupe Gagnon

Level 37 , Tampa, United States

30 March 2021, 17:36

It doesn't return the correct answer. I tried it with the numbers: 2, 2, 5, 3457 And it output that the minimum was 3457.

Aryan Babaie

Level 4 , Jülich, Germany

2 April 2021, 17:58

Thanks, but I know it doesn't. can you tell me what's wrong with it?

Guadalupe Gagnon

Level 37 , Tampa, United States

2 April 2021, 19:03

solution

It is the logic say that for the code: a = 2 b = 2 c = 5 d = 3457 The logic would work like this:

int e = min(a, b); // min(2,2) so 'e' would be 2
if (e < c && e < b) // if 2 < 5 (false) and 2 < 2 (true) whole statement false
else if (c < e && c < d) // if 5 < 2 (false) and 5 < 3457 (true) whole statement false
else
    min2 = d; // min2 gets set to 3457 because none of the above resulted in 'true'
return min2;

The min(int, int) method works fine. In the min(int, int, int, int) method you can use the min(int, int) method more than once. In your code you use it to figure out the min of 'a' and 'b'. You could then use it to figure out the min of 'c' and 'd'. Then you can use it a third time to figure out the min of those two numbers.

Aryan Babaie

Level 4 , Jülich, Germany

3 April 2021, 17:04

Thank you for taking your time on this and it helped a lot🤘