# withoutTen challenge

Hi,

I am working on below challenge

http://codingbat.com/prob/p196976

My psuedo code of the logical approach is
1 loop through given array
2 check and see if all elements are not 10 if yes then return same array
3.if given element is 10 replace it with 0
4. display the array with all zeros at the front and non zero numbers at the end

I wrote my code as below
``````public int[] withoutTen(int[] nums) {
int[] result=null;
for(int i=0;i<nums.length;i++){
if(nums[i]!=10){
result=nums;
}

else if(nums[i]==10){
nums[i]=0;
}
}
return result;
}
``````

I am not passing all tests
Expected      Run
withoutTen([1, 10, 10, 2]) → [1, 2, 0, 0]      [1, 0, 0, 2]      X
withoutTen([10, 2, 10]) → [2, 0, 0]      [0, 2, 0]      X
withoutTen([1, 99, 10]) → [1, 99, 0]      [1, 99, 0]      OK
withoutTen([10, 13, 10, 14]) → [13, 14, 0, 0]      [0, 13, 0, 14]      X
withoutTen([10, 13, 10, 14, 10]) → [13, 14, 0, 0, 0]      [0, 13, 0, 14, 0]      X
withoutTen([10, 10, 3]) → [3, 0, 0]      [0, 0, 3]      X
withoutTen([1]) → [1]      [1]      OK
withoutTen([13, 1]) → [13, 1]      [13, 1]      OK
withoutTen([10]) → [0]      null      X
withoutTen([]) → []      null      X
other tests
X

LVL 7
###### Who is Participating?

Commented:
You might think about creating an array to return with the same length as nums, keeping in mind that such an array will contain all zeroes. You could then populate it with the values from nums that are not ten.
0

You should keep track of the amount of non-10 numbers you've found. Place the non-10 numbers in the result array and for each 10 found put a 0 at the end of the result array (minus the index of the non-10 numbers found).
0

Author Commented:
i will try
0

OK
0

IT Business Systems Analyst / Software DeveloperCommented:
Or another approach, knowing that all elements of an int array will already be initialised to 0, you just have to loop the original array and place any non-10 numbers into the result array. The key point though is you need one variable, call it "i", to keep track of the index in the original array but you also need another variable to keep track of where you are putting the numbers into the result array, because those two indexes will be different, so you need two variables.
0

Author Commented:
``````public int[] withoutTen(int[] nums) {
int[] result=null;
for(int i=0;i<nums.length;i++){
if(nums[i]!=10){
result=nums;
}

else if(nums[i]==10){
nums[i]=0;
}
}
//return result;

int len=nums.length;
int[] arr=new int[len];
int front=0;
int back=len-1;
for(int i=0;i<nums.length;i++){
if(nums[i]%2==1){
arr[front]=nums[i];
front++;

}
else if(nums[i]%2==0){
arr[back]=nums[i];
back--;
}
}
return arr;

}
``````
Expected      Run
withoutTen([1, 10, 10, 2]) → [1, 2, 0, 0]      [1, 2, 0, 0]      OK
withoutTen([10, 2, 10]) → [2, 0, 0]      [0, 2, 0]      X
withoutTen([1, 99, 10]) → [1, 99, 0]      [1, 99, 0]      OK
withoutTen([10, 13, 10, 14]) → [13, 14, 0, 0]      [13, 14, 0, 0]      OK
withoutTen([10, 13, 10, 14, 10]) → [13, 14, 0, 0, 0]      [13, 0, 14, 0, 0]      X
withoutTen([10, 10, 3]) → [3, 0, 0]      [3, 0, 0]      OK
withoutTen([1]) → [1]      [1]      OK
withoutTen([13, 1]) → [13, 1]      [13, 1]      OK
withoutTen([10]) → [0]      [0]      OK
withoutTen([]) → []      []      OK
other tests
X

i integrated my solution with evenodd challenge solution and failing couple of tests. how to fox them. please advise
0

IT Business Systems Analyst / Software DeveloperCommented:
Ok, working with the code that you have, first you should really take out the first loop that you have (lines 2 -> 12), because all that does is change any 10's to 0's. This sounds ok at first but the problem is that you would lose track of which 0's were real zero's and which were there because they were initially 10. Now there might not be any test case that would highlight that problem, but it is best to be sure.

Now, what your left with is a loop that copies odd numbers to the front of the array and copies even numbers to the back of the array.

But what you want (if using this code) is to copy non-10 numbers to the front of the array and puts a zero in for a 10 to the back of the array.

So, yes, there are similarities there and you can use the code that you have, but the changes required could be worked out with the bold words I have written above, ie. you have lines in your code that check if the number is odd or even (lines 21, 26), so you have to change those so that they check for non-10 and 10 numbers.

Also, the other difference in the bold numbers above is instead of copying the even (line 27) you have to put 0 in place of the 10.

Hopefully, with those hints above, you can change just the 3 lines of code required to get the code working properly.
0

Author Commented:
``````public int[] withoutTen(int[] nums) {
int len=nums.length;
int[] arr=new int[len];
int front=0;
int back=len-1;
for(int i=0;i<nums.length;i++){
if(nums[i]%10==1||nums[i]%10==2||nums[i]%10==3||nums[i]%10==4||nums[i]%10==5||nums[i]%10==6||nums[i]%10==7||nums[i]%10==8||nums[i]%10==9){
arr[front]=nums[i];
front++;

}
else {
arr[back]=nums[i];
back--;
}
if(arr[i]%10==0){
arr[i]=0;
}
}
return arr;

}
``````

Expected      Run
withoutTen([1, 10, 10, 2]) → [1, 2, 0, 0]      [1, 2, 0, 0]      OK
withoutTen([10, 2, 10]) → [2, 0, 0]      [2, 10, 0]      X
withoutTen([1, 99, 10]) → [1, 99, 0]      [1, 99, 0]      OK
withoutTen([10, 13, 10, 14]) → [13, 14, 0, 0]      [13, 14, 0, 0]      OK
withoutTen([10, 13, 10, 14, 10]) → [13, 14, 0, 0, 0]      [13, 14, 10, 0, 0]      X
withoutTen([10, 10, 3]) → [3, 0, 0]      [3, 0, 0]      OK
withoutTen([1]) → [1]      [1]      OK
withoutTen([13, 1]) → [13, 1]      [13, 1]      OK
withoutTen([10]) → [0]      [0]      OK
withoutTen([]) → []      []      OK
other tests
OK
0

Author Commented:
``````public int[] withoutTen(int[] nums) {
int len=nums.length;
int[] arr=new int[len];
int front=0;
int back=len-1;
for(int i=0;i<nums.length;i++){
if(nums[i]%10==1||nums[i]%10==2||nums[i]%10==3||nums[i]%10==4||nums[i]%10==5||nums[i]%10==6||nums[i]%10==7||nums[i]%10==8||nums[i]%10==9){
arr[front]=nums[i];
front++;

}
/* else {
arr[back]=nums[i];
back--;
}
if(arr[i]%10==0){
arr[i]=0;
}*/
}
return arr;

}
``````

that passed all tests
Expected      Run
withoutTen([1, 10, 10, 2]) → [1, 2, 0, 0]      [1, 2, 0, 0]      OK
withoutTen([10, 2, 10]) → [2, 0, 0]      [2, 0, 0]      OK
withoutTen([1, 99, 10]) → [1, 99, 0]      [1, 99, 0]      OK
withoutTen([10, 13, 10, 14]) → [13, 14, 0, 0]      [13, 14, 0, 0]      OK
withoutTen([10, 13, 10, 14, 10]) → [13, 14, 0, 0, 0]      [13, 14, 0, 0, 0]      OK
withoutTen([10, 10, 3]) → [3, 0, 0]      [3, 0, 0]      OK
withoutTen([1]) → [1]      [1]      OK
withoutTen([13, 1]) → [13, 1]      [13, 1]      OK
withoutTen([10]) → [0]      [0]      OK
withoutTen([]) → []      []      OK
other tests
OK

any imrovement or refinement to above code? otherwise will close it
0

Commented:
This
if(nums[i]%10==1||nums[i]%10==2||nums[i]%10==3||nums[i]%10==4||nums[i]%10==5||nums[i]%10==6||nums[i]%10==7||nums[i]%10==8||nums[i]%10==9)

can be shortened to
if nums[i] != 0
0

Author Commented:
you mean
``````nums[i]%10 != 0
``````
right?

above also passed all
``````public int[] withoutTen(int[] nums) {
int len=nums.length;
int[] arr=new int[len];
int front=0;
int back=len-1;
for(int i=0;i<nums.length;i++){
if(nums[i]%10 != 0){
arr[front]=nums[i];
front++;

}
/* else {
arr[back]=nums[i];
back--;
}
if(arr[i]%10==0){
arr[i]=0;
}*/
}
return arr;

}
``````
0

Commented:
you mean
1:nums[i]%10 != 0

Yes, that is exactly what I meant :-)
1

IT Business Systems Analyst / Software DeveloperCommented:
``````if (nums[i]%10 != 0) {
``````

The above is NOT what you want either. Since all the tests are passing there is obviously no test case for it, but if the input array had any 0, 20, 30, 40, 50, 60, ... etc. it would NOT do what you want it to do.

Just do this...

``````if (nums[i] != 10) {
``````
0
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