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# post4 challenge

Posted on 2016-08-08
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Hi,

I am working on below challenge
http://codingbat.com/prob/p164144

Psedo code description of approach :
1. create new arrray//not sure on size?
2. assign each element with new element arrays until not equal to 4 from backwars
3. if above true return till there
4 if false break for loop

I wrote my code as below

``````public int[] post4(int[] nums) {

int len =nums.length;
int[] arr=new int[len];
for(int i=len;i>0;i--){
if(nums[i]!=4){
arr[i]=nums[i];
}
else{
break;
}
}
return arr;

}
``````

I am not passing all tests due to edge cases

Expected      Run
post4([2, 4, 1, 2]) → [1, 2]      Exception:java.lang.ArrayIndexOutOfBoundsException: 4 (line number:9)      X
post4([4, 1, 4, 2]) → [2]      Exception:java.lang.ArrayIndexOutOfBoundsException: 4 (line number:9)      X
post4([4, 4, 1, 2, 3]) → [1, 2, 3]      Exception:java.lang.ArrayIndexOutOfBoundsException: 5 (line number:9)      X
post4([4, 2]) → [2]      Exception:java.lang.ArrayIndexOutOfBoundsException: 2 (line number:9)      X
post4([4, 4, 3]) → [3]      Exception:java.lang.ArrayIndexOutOfBoundsException: 3 (line number:9)      X
post4([4, 4]) → []      Exception:java.lang.ArrayIndexOutOfBoundsException: 2 (line number:9)      X
post4([4]) → []      Exception:java.lang.ArrayIndexOutOfBoundsException: 1 (line number:9)      X
post4([2, 4, 1, 4, 3, 2]) → [3, 2]      Exception:java.lang.ArrayIndexOutOfBoundsException: 6 (line number:9)      X
post4([4, 1, 4, 2, 2, 2]) → [2, 2, 2]      Exception:java.lang.ArrayIndexOutOfBoundsException: 6 (line number:9)      X
post4([3, 4, 3, 2]) → [3, 2]      Exception:java.lang.ArrayIndexOutOfBoundsException: 4 (line number:9)      X
other tests
X
Your progress graph for this problem
0
Question by:gudii9
• 15
• 5
• 3
• +3

LVL 35

Assisted Solution

mccarl earned 125 total points
ID: 41748243
Finish your "pre4" challenge before you worry about looking at this. Then you will have everything you should need to do this one!
2

LVL 30

Expert Comment

ID: 41748366
Hint: Read and try to understand the thrown exceptions. This is a good exercise to learn debugging. Debugging is extremely important when someone wants to become a good programmer.
0

LVL 16

Expert Comment

ID: 41749169
I agree with mccarl, and also mostly agree with Zoppo.
0

LVL 30

Expert Comment

ID: 41750030
Hm - why just mostly?
0

LVL 16

Expert Comment

ID: 41750121
Because endless and countless previous explanations of how these things work have not been able to penetrate the asker's thinking.
0

LVL 30

Expert Comment

ID: 41750153
:D
0

LVL 31

Assisted Solution

awking00 earned 125 total points
ID: 41753747
Pseudo Code
Find the last index of 4
Find the last index of the array
The difference will be the length of the new array
Then populate it starting with the last index of 4 plus 1
0

LVL 27

Expert Comment

ID: 41754092
mccarl showed  you a useful method in the API at
https://www.experts-exchange.com/questions/28962285/pre4-challenge.html#a41753818
Look at the API
http://docs.oracle.com/javase/8/docs/api/java/util/Arrays.html
directly below the copyOf method there is the copyOfRange method which you can use in this challenge.
0

LVL 7

Author Comment

ID: 41754523
``````public int[] post4(int[] nums) {
int len = nums.length;
int val=0;
for (int i = 0; i < len-1; i++) {
if (nums[i] == 4) {
val=i;
break;
}
}
int[] arr = new int[val];
for (int k = val; k < len-1; k++) {
arr[k] =nums[k];
}

return arr;
}
``````
Expected      Run
post4([2, 4, 1, 2]) → [1, 2]      Exception:java.lang.ArrayIndexOutOfBoundsException: 1 (line number:12)      X
post4([4, 1, 4, 2]) → [2]      Exception:java.lang.ArrayIndexOutOfBoundsException: 0 (line number:12)      X
post4([4, 4, 1, 2, 3]) → [1, 2, 3]      Exception:java.lang.ArrayIndexOutOfBoundsException: 0 (line number:12)      X
post4([4, 2]) → [2]      Exception:java.lang.ArrayIndexOutOfBoundsException: 0 (line number:12)      X
post4([4, 4, 3]) → [3]      Exception:java.lang.ArrayIndexOutOfBoundsException: 0 (line number:12)      X
post4([4, 4]) → []      Exception:java.lang.ArrayIndexOutOfBoundsException: 0 (line number:12)      X
post4([4]) → []      []      OK
post4([2, 4, 1, 4, 3, 2]) → [3, 2]      Exception:java.lang.ArrayIndexOutOfBoundsException: 1 (line number:12)      X
post4([4, 1, 4, 2, 2, 2]) → [2, 2, 2]      Exception:java.lang.ArrayIndexOutOfBoundsException: 0 (line number:12)      X
post4([3, 4, 3, 2]) → [3, 2]      Exception:java.lang.ArrayIndexOutOfBoundsException: 1 (line number:12)      X
other tests
X

0

LVL 7

Author Comment

ID: 41754532
``````import java.util.Arrays;

public class Post4 {

public static void main(String[] args) {
// TODO Auto-generated method stub
int[] arr = { 2, 4, 1, 2 };
System.out.println("value is" + Arrays.toString(post4(arr)));
}

public static int[] post4(int[] nums) {
int len = nums.length;
int val = 0;
for (int i = 0; i < len; i++) {
if (nums[i] == 4) {
val = i;
break;
}
}
int val2=len-(val+1);
int[] arr = new int[val2];
for (int k = val2; k < len; k++) {
arr[k] = nums[k];
}

return arr;
}

}
``````

i modified bit getting same error
Exception in thread "main" java.lang.ArrayIndexOutOfBoundsException: 2
at Post4.post4(Post4.java:23)
at Post4.main(Post4.java:8)
0

LVL 7

Author Comment

ID: 41754548
``````arr[k] = nums[k];
``````

compiler complaining about assigining 5 digit array on right into 3 digit array on left. not sure how to get last few digits after 4 into new array arr?
0

LVL 35

Expert Comment

ID: 41754549
Try..

arr[k-val] = nums[k];
0

LVL 7

Author Comment

ID: 41754550
``````import java.util.Arrays;

public class PreFour {

public static void main(String[] args) {
// TODO Auto-generated method stub
int[] ar = { 2,4,1,0,8 };
System.out.println("value===>" + Arrays.toString(pre4(ar)));

}

public static int[] pre4(int[] nums) {
/*		You don't always have to try and do things in one go. Just use two loops....
the first to find the index of the first 4, now you can create your new
array with the correct size,
and then use a second loop to copy the elements from the original to new array.*/
int len = nums.length;
int val=0;
for (int i = 0; i < len; i++) {
if (nums[i] == 4) {
val=i;
break;
}
}
int[] arr = new int[val];

for (int k = 0; k < val; k++) {
arr[k] =nums[k];
}
return arr;

}

}
``````

pre4 works ok where as on similar lines post4 not woring
``````import java.util.Arrays;

public class Post4 {

public static void main(String[] args) {
// TODO Auto-generated method stub
int[] arr = { 2, 4, 1, 0,8 };
System.out.println("value is" + Arrays.toString(post4(arr)));
}

public static int[] post4(int[] nums) {
int len = nums.length;
int val = 0;
for (int i = 0; i < len; i++) {
if (nums[i] == 4) {
val = i;
break;
}
}
int val2=len-(val);
int[] arr = new int[val2];
for (int k = (val2-1); k < len; k++) {
arr[k] = nums[k];
}

return arr;
}

/*public int[] pre4(int[] nums) {
int len = nums.length;
int val=0;
for (int i = 0; i < len; i++) {
if (nums[i] == 4) {
val=i;
break;
}
}
int[] arr = new int[val];
for (int k = 0; k < val; k++) {
arr[k] =nums[k];
}

return arr;
}
*/

}
``````

not sure where is the issue?
0

LVL 7

Author Comment

ID: 41754551
``````import java.util.Arrays;

public class Post4 {

public static void main(String[] args) {
// TODO Auto-generated method stub
int[] arr = { 2, 4, 1, 0,8 };
System.out.println("value is" + Arrays.toString(post4(arr)));
}

public static int[] post4(int[] nums) {
int len = nums.length;
int val = 0;
for (int i = 0; i < len; i++) {
if (nums[i] == 4) {
val = i;
break;
}
}
int val2=len-(val);
int[] arr = new int[val2];
for (int k = (val2-1); k < len; k++) {
arr[k-val] = nums[k];
}

return arr;
}

/*public int[] pre4(int[] nums) {
int len = nums.length;
int val=0;
for (int i = 0; i < len; i++) {
if (nums[i] == 4) {
val=i;
break;
}
}
int[] arr = new int[val];
for (int k = 0; k < val; k++) {
arr[k] =nums[k];
}

return arr;
}
*/

}
``````

gives below

value is[0, 0, 0, 8]
0

LVL 7

Author Comment

ID: 41754553
``````arr[k-val] = nums[k];
``````

what is the rule both left and right array needs to be same size?
0

LVL 7

Author Comment

ID: 41754554
why k-val??
0

LVL 7

Author Comment

ID: 41754555
``````import java.util.Arrays;

public class Post4 {

public static void main(String[] args) {
// TODO Auto-generated method stub
int[] arr = { 2, 4, 1, 0,8 };
System.out.println("value is" + Arrays.toString(post4(arr)));
}

public static int[] post4(int[] nums) {
int len = nums.length;
int val = 0;
for (int i = 0; i < len; i++) {
if (nums[i] == 4) {
val = i;
break;
}
}
int val2=len-(val+1);
int[] arr = new int[val2];
for (int k = (val2-1); k < len; k++) {
arr[k-(val+1)] = nums[k];
}

return arr;
}

/*public int[] pre4(int[] nums) {
int len = nums.length;
int val=0;
for (int i = 0; i < len; i++) {
if (nums[i] == 4) {
val=i;
break;
}
}
int[] arr = new int[val];
for (int k = 0; k < val; k++) {
arr[k] =nums[k];
}

return arr;
}
*/

}
``````

i got correct result above

value is[1, 0, 8]

but coding bat failing
Expected      Run
post4([2, 4, 1, 2]) → [1, 2]      Exception:java.lang.ArrayIndexOutOfBoundsException: -1 (line number:13)      X
post4([4, 1, 4, 2]) → [2]      [0, 4, 2]      X
post4([4, 4, 1, 2, 3]) → [1, 2, 3]      [0, 0, 2, 3]      X
post4([4, 2]) → [2]      Exception:java.lang.ArrayIndexOutOfBoundsException: -1 (line number:13)      X
post4([4, 4, 3]) → [3]      [4, 3]      X
post4([4, 4]) → []      Exception:java.lang.ArrayIndexOutOfBoundsException: -1 (line number:13)      X
post4([4]) → []      Exception:java.lang.ArrayIndexOutOfBoundsException: -1 (line number:13)      X
post4([2, 4, 1, 4, 3, 2]) → [3, 2]      [0, 4, 3, 2]      X
post4([4, 1, 4, 2, 2, 2]) → [2, 2, 2]      [0, 0, 0, 2, 2]      X
post4([3, 4, 3, 2]) → [3, 2]      Exception:java.lang.ArrayIndexOutOfBoundsException: -1 (line number:13)      X
other tests
X
Your progress graph for this problem

0

LVL 7

Author Comment

ID: 41754556
``````public int[] post4(int[] nums) {
int len = nums.length;
int val = 0;
for (int i = 0; i < len; i++) {
if (nums[i] == 4) {
val = i;
break;
}
}
int val2=len-(val+1);
int[] arr = new int[val2];
for (int k = (val+1); k < len; k++) {
arr[k-(val+1)] = nums[k];
}

return arr;
}
``````

i am close now , failing few
Expected      Run
post4([2, 4, 1, 2]) → [1, 2]      [1, 2]      OK
post4([4, 1, 4, 2]) → [2]      [1, 4, 2]      X
post4([4, 4, 1, 2, 3]) → [1, 2, 3]      [4, 1, 2, 3]      X
post4([4, 2]) → [2]      [2]      OK
post4([4, 4, 3]) → [3]      [4, 3]      X
post4([4, 4]) → []      [4]      X
post4([4]) → []      []      OK
post4([2, 4, 1, 4, 3, 2]) → [3, 2]      [1, 4, 3, 2]      X
post4([4, 1, 4, 2, 2, 2]) → [2, 2, 2]      [1, 4, 2, 2, 2]      X
post4([3, 4, 3, 2]) → [3, 2]      [3, 2]      OK
other tests
X
Your progress graph for this problem
0

LVL 7

Author Comment

ID: 41754557
failing more than one 4.

how to find last array and get elements after that to new array?
0

LVL 27

Accepted Solution

rrz earned 250 total points
ID: 41754558
Let's focus on your for loop.
``````		for (int i = 0; i < len-1; i++) {
if (nums[i] == 4) {
val=i;
break;
}
}
``````
That is exactly what you used in that other challenge called pre4.  But in that challenge we were searching  for the first 4.  In this challenge, we are searching for the last 4. We could find it in a number of ways. We could start our loop search at the end and step through backwards until we find a 4.  Another way is to create a variable to hold the value of  the index of the last 4. That second way is similar to what I did in my solution to the twoTwo challenge that we a few days ago.  In that challenge, I created a variable to remember what happen earlier in the looping.
So, how do you want to proceed?
Step through backwards?
Create a variable to remember where the last 4 is?
Or maybe someone has another way?
0

LVL 7

Author Comment

ID: 41754564
i think backwards rings bell in my mind as easy let me try that
0

LVL 7

Author Comment

ID: 41754565
``````public int[] post4(int[] nums) {
int len = nums.length;
int val = 0;
for (int i = len-1; i >=0; i--) {
if (nums[i] == 4) {
val = i;
break;
}
}
int val2=len-(val+1);
int[] arr = new int[val2];
for (int k = (val+1); k < len; k++) {
arr[k-(val+1)] = nums[k];
}

return arr;
}
``````
Expected      Run
post4([2, 4, 1, 2]) → [1, 2]      [1, 2]      OK
post4([4, 1, 4, 2]) → [2]      [2]      OK
post4([4, 4, 1, 2, 3]) → [1, 2, 3]      [1, 2, 3]      OK
post4([4, 2]) → [2]      [2]      OK
post4([4, 4, 3]) → [3]      [3]      OK
post4([4, 4]) → []      []      OK
post4([4]) → []      []      OK
post4([2, 4, 1, 4, 3, 2]) → [3, 2]      [3, 2]      OK
post4([4, 1, 4, 2, 2, 2]) → [2, 2, 2]      [2, 2, 2]      OK
post4([3, 4, 3, 2]) → [3, 2]      [3, 2]      OK
other tests
OK

All Correct
boom...backward approach worked fine

Another way is to create a variable to hold the value of  the index of the last 4. That second way is similar to what I did in my solution to the twoTwo challenge that we a few days ago.  In that challenge, I created a variable to remember what happen earlier in the looping.

let me refresh memory and try that approach also
0

LVL 27

Expert Comment

ID: 41754567
Your backwards stepping for loop is good.
0

LVL 7

Author Comment

ID: 41754568
if challenge say for example middle 4 then second approach only works right not forward not backward?
0

LVL 27

Expert Comment

ID: 41754569
I don't understand that last question. But let's stay focused on this challenge.
0

LVL 7

Author Comment

ID: 41755183
twotwo challenge is not yet clear to me. But i will close this question as i got proper approach on this.
0

LVL 7

Author Comment

ID: 41755753
That second way is similar to what I did in my solution to the twoTwo challenge that we a few days ago.  In that challenge, I created a variable to remember what happen earlier in the looping.

i am stil thinking, i was bit unclear on this way? can you please advise
0

LVL 27

Expert Comment

ID: 41755792
This is similar to what you did above here, except it doesn't use a break statement.  When the for loop completes the fourIndex variable is equal to the index of the last four.
``````public int[] post4(int[] nums) {
int fourIndex = 0;
for (int i = 0; i < nums.length; i++) {
if (nums[i] == 4) {
fourIndex = i;
}
}
return Arrays.copyOfRange(nums, fourIndex + 1, nums.length);
}
``````
0

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