# EvenOdd challenge

Hi,

I am working on below challenge

http://codingbat.com/prob/p105771

My psuedo code of the logical approach is
1. loop throguh given array
2. find given elemt is even or odd.
3. if even group to front
4. if odd group to end
5. return array

I wrote my code as below

``````public int[] evenOdd(int[] nums) {
for(int i=0;i<nums.length;i++){

if(nums[i]%2==0){

}
else if(nums[i]%2==1){

}
}
return nums;
}
``````

I am not passing my tests as i just created skeleton code as i am not sure how to code this?

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Go through array once, odd number found -> exchange with first position in array. Increase first replace position when exchange is done.
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Sr. ConsultantCommented:
Agree with Gerwin.

``````public int[] evenOdd(int[] nums) {
for(int i=0;i<nums.length-1;i++){
for(int j=i+1;j<nums.length;j++){
if(nums[i]%2==0){
break;
} else {
if(nums[j]%2==0){
int tmp = nums[i];
nums[i] = nums[j];
nums[j] = tmp;
break;
}
}
}

}
return nums;
}
``````
0
Author Commented:
Go through array once, odd number found -> exchange with first position in array. Increase first replace position when exchange is done.

what i supposed to do when i see even number?
0
Nothing, work out an example and see what happens ;) Change to even numbers to see the difference.
0
IT Business Systems Analyst / Software DeveloperCommented:
Another possible solution, would be to create a new array of the same size, loop through the original array, if the number is even fill the result array from the front, but if it is odd, fill the result array from the back. It does require a bit extra memory to store an extra array plus you need 3 index variables... One for the main loop index, one to track where you are filling the result array from the front and one to track where you are filling from the back. But the advantage is that you only need to loop through the array once, whereas the other solutions loop through many times.
0
Author Commented:
i like above approach which is simple and passing all tests
``````public int[] evenOdd(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]%2==0){
arr[front]=nums[i];
front++;

}
else if(nums[i]%2==1){
arr[back]=nums[i];
back--;
}
}
return arr;
}
``````

``````Expected	Run
evenOdd([1, 0, 1, 0, 0, 1, 1]) → [0, 0, 0, 1, 1, 1, 1]	[0, 0, 0, 1, 1, 1, 1]	OK
evenOdd([3, 3, 2]) → [2, 3, 3]	[2, 3, 3]	OK
evenOdd([2, 2, 2]) → [2, 2, 2]	[2, 2, 2]	OK
evenOdd([3, 2, 2]) → [2, 2, 3]	[2, 2, 3]	OK
evenOdd([1, 1, 0, 1, 0]) → [0, 0, 1, 1, 1]	[0, 0, 1, 1, 1]	OK
evenOdd([1]) → [1]	[1]	OK
evenOdd([1, 2]) → [2, 1]	[2, 1]	OK
evenOdd([2, 1]) → [2, 1]	[2, 1]	OK
evenOdd([]) → []	[]	OK
other tests
OK
``````

``````import java.util.Arrays;

public class EvenOdd {

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

public static int[] evenOdd(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]%2==0){
arr[front]=nums[i];
front++;

}
else if(nums[i]%2==1){
arr[back]=nums[i];
back--;
}
}
return arr;
}

/*Another possible solution, would be to create a new array of the same size,
loop through the original array, if the number is even fill the result array from the
front, but if it is odd, fill the result array from the back. It does require a bit
extra memory to store an extra array plus you need 3 index variables... One for the main
loop index, one to track where you are filling the result array from the front and one to
track where you are filling from the back. But the advantage is that you only need to
loop through the array once, whereas the other solutions loop through many times.
*/
}
``````

value-->[0, 0, 0, 1, 1, 1, 1]
0
Author Commented:
any improvements, refinement or optimization to my code?
0
Author Commented:
``````public int[] evenOdd(int[] nums) {
for(int i=0;i<nums.length-1;i++){
for(int j=i+1;j<nums.length;j++){
if(nums[i]%2==0){
break;
} else {
if(nums[j]%2==0){
int tmp = nums[i];
nums[i] = nums[j];
nums[j] = tmp;
break;
}
}
}

}
return nums;
}
``````

above solution is not clear to me. why we are using two for loops one with counter i starting at 0other with counter j starting at 1?
and what is happening inside else loop for odd number case as below?
``````  int tmp = nums[i];
nums[i] = nums[j];
nums[j] = tmp;
``````
0
Author Commented:
i think i got it.

for every i which is even we are breaking.\

if i is odd checking consecutive elements is even then swapping them till end.

so {1,0.1.0.0,1,1} becomes {0,0.0.1.1.1.1}
0
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