Why not accept the solution if work correctly ?

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public class Solution { public static int min(int a, int b, int c) { //write your code here int result=0; if (a >= b) { if ( b >= c) result = c; else result = b; } else if (c >= b) result = a; else result = a; return result; } public static void main(String[] args) throws Exception { System.out.println(min(0, 2, 2)); System.out.println(min(-1, -2, -2)); System.out.println(min(-13, -15, 0)); System.out.println(min(-8, -9, -10)); }

package com.codegym.task.task02.task0216; /* Minimum of three numbers */ public class Solution { public static int min(int a, int b, int c) { //write your code here int result=0; if (a >= b) { if ( b >= c) result = c; else result = b; } else if (c >= b) result = a; else result = a; return result; } public static void main(String[] args) throws Exception { System.out.println(min(0, 2, 2)); System.out.println(min(-1, -2, -2)); System.out.println(min(-13, -15, 0)); System.out.println(min(-8, -9, -10)); } }

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Thomas

Nivel 41 , Bayreuth, Germany

12 julio 2021, 04:42

Cause the code is errornous. The first thing I see when I look at your code is two times the same return value in you if-elseif-else construct. To reach the problem you need to get a false from the if in line 12 (eg.: a = 2, b = 3) and a false from the elseif (eg. with c = 1). Now you're at line 21,22 and return a (and that's the minimum 2 for the input 2, 3, 1).

Robbindick Sandoval

Nivel 4 , Cali

14 julio 2021, 16:59

Hi Thomas tanks for you answer but at line 21 else if.....and line 22 its the final else in case of the all other validation fails.

Guadalupe Gagnon

Nivel 37 , Tampa, United States

14 julio 2021, 17:22

What Thomas said is correct. You need to fix the code so that it reaches the correct answer no matter what three numbers are considered. When the numbers are 2, 3, 1 your code says 2 is the minimum but the correct answer is 1.

Thomas

Nivel 41 , Bayreuth, Germany

15 julio 2021, 11:00

I see what you wrote but it still is wrong and I told you why I saw it in just one view if you write

} else if (c >= b)
    result = a;
else
    result = a;

then this all is just redundant code. You have a condition and you do A. If the condition isn't fullfilled you again do A. So you do A in all cases no matter what. Then you could shorten your code and just write.

} else
    result = a;

What I told you above is, that you missed one condition and that one you need to insert between the c >= b and the else (else if c < a to give the problem a name).

if (a >= b) {
    if (b >= c)
        result = c;
    else
        result = b;
} else if (c >= b) {
    result = a;
} else if (c < a) {
    result = c;
} else
    result = a;

If you fix that, then the task probably will verify. Now you could initialize result with a and remove all conditions returning a to get some shorter code. But the initial problem still remains. The code is hard to understand. And by now you should have seen that such constructs are very error prone (even though you're 100% sure all is working smooth as butter). Better would be to divide up the work into smaller parts. Eg. check if a is the min and return it in that case. Then take care of b and c.

// if a is the min, return it
if (a < b && a < c)
    return a;
// at this point we know now that a can't be the min anymore so just checking the min of b, c remains
return (b < c) ? b : c;

There are a 100 ways to solve this task but it's the best to take an easy to understand and maintain approach.

Robbindick Sandoval

Nivel 4 , Cali

15 julio 2021, 22:48

Thanks for your answer I fix the code and run good.

Robbindick Sandoval

Nivel 4 , Cali

15 julio 2021, 22:49

public class Solution { public static int min(int a, int b, int c) { //write your code here int result=0; if(a<=b && a<=c){ result = a; } else if(b<=a && b<=c){ result = b; } else if(c<=a && c<=b) { result = c; } return result; }