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;
}

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

How to improve my design, approach, code? please advise
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gudii9Asked:
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Gerwin Jansen, EE MVETopic Advisor Commented:
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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Amitkumar PSr. 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;
}

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gudii9Author 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?
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Gerwin Jansen, EE MVETopic Advisor Commented:
Nothing, work out an example and see what happens ;) Change to even numbers to see the difference.
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mccarlIT 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.
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gudii9Author 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;
		}

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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	

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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.
*/
}

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value-->[0, 0, 0, 1, 1, 1, 1]
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gudii9Author Commented:
any improvements, refinement or optimization to my code?
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gudii9Author 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;
}

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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;

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gudii9Author 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}
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