fix34 challenge

Hi,

I am working on below challenge

http://codingbat.com/prob/p159339

I have not understood the description properly
Array-3 > fix34
prev  |  next  |  chance
Return an array that contains exactly the same numbers as the given array, but rearranged so that every 3 is immediately followed by a 4. Do not move the 3's, but every other number may move. The array contains the same number of 3's and 4's, every 3 has a number after it that is not a 3 or 4, and a 3 appears in the array before any 4.

fix34([1, 3, 1, 4]) → [1, 3, 4, 1]
fix34([1, 3, 1, 4, 4, 3, 1]) → [1, 3, 4, 1, 1, 3, 4]
fix34([3, 2, 2, 4]) → [3, 4, 2, 2]

please advise
LVL 7
gudii9Asked:
Who is Participating?

[Product update] Infrastructure Analysis Tool is now available with Business Accounts.Learn More

x
I wear a lot of hats...

"The solutions and answers provided on Experts Exchange have been extremely helpful to me over the last few years. I wear a lot of hats - Developer, Database Administrator, Help Desk, etc., so I know a lot of things but not a lot about one thing. Experts Exchange gives me answers from people who do know a lot about one thing, in a easy to use platform." -Todd S.

Gerwin Jansen, EE MVETopic Advisor Commented:
You need to put the numbers 4 that exist in the given array after the numbers 3.

In the given array, numbers 3 are not followed by a 4.

So:

1,3,5,4 will become 1,3,4,5
0
gudii9Author Commented:
how
fix34([1, 3, 1, 4, 4, 3, 1]) becomes  [1, 3, 4, 1, 1, 3, 4] ?
or just move 4 next 3 without moving 3?
what happens to non 4 and non 3 numbers? where we have to move them?
0
Gerwin Jansen, EE MVETopic Advisor Commented:
Don't move 3
Move 4 to right side of a 3
You may move other numbers
0

Experts Exchange Solution brought to you by

Your issues matter to us.

Facing a tech roadblock? Get the help and guidance you need from experienced professionals who care. Ask your question anytime, anywhere, with no hassle.

Start your 7-day free trial
Big Business Goals? Which KPIs Will Help You

The most successful MSPs rely on metrics – known as key performance indicators (KPIs) – for making informed decisions that help their businesses thrive, rather than just survive. This eBook provides an overview of the most important KPIs used by top MSPs.

gudii9Author Commented:
not getting idea on best approach?
0
Gerwin Jansen, EE MVETopic Advisor Commented:
don't understand what you mean, explain
0
nociSoftware EngineerCommented:
Moving does include exchanging  numbers.... Only 3's are fixed.... and there is a hint in the question as well.
There will be a 3 before any 4 anyway..., Also for every three there will be a 4....
so it can be done in a single pass through the array.

so
fix34([1, 3, 1, 4, 4, 3, 1]) 
becomes [ 1, 3, 4, 1, 4, 3, 1]
               -----
becomes  [1, 3, 4, 1, 1, 3, 4] ?
                     --    --

Open in new window

0
gudii9Author Commented:
public int[] fix34(int[] nums) {

	    for (int z = 0; z < nums.length; z++)
	        if (nums[z] == 3) {
	            int buffer = nums[z + 1];
	            nums[z + 1] = 4;
	            for (int j = z + 2; j < nums.length; j++)
	                if (nums[j] == 4) 
	                	nums[j] = buffer;
	        }
	    return nums;
	
}

Open in new window

Expected      Run            
fix34([1, 3, 1, 4]) → [1, 3, 4, 1]      [1, 3, 4, 1]      OK      
fix34([1, 3, 1, 4, 4, 3, 1]) → [1, 3, 4, 1, 1, 3, 4]      [1, 3, 4, 1, 1, 3, 4]      OK      
fix34([3, 2, 2, 4]) → [3, 4, 2, 2]      [3, 4, 2, 2]      OK      
fix34([3, 2, 3, 2, 4, 4]) → [3, 4, 3, 4, 2, 2]      [3, 4, 3, 4, 2, 2]      OK      
fix34([2, 3, 2, 3, 2, 4, 4]) → [2, 3, 4, 3, 4, 2, 2]      [2, 3, 4, 3, 4, 2, 2]      OK      
fix34([3, 1, 4]) → [3, 4, 1]      [3, 4, 1]      OK      
fix34([3, 4, 1]) → [3, 4, 1]      [3, 4, 1]      OK      
fix34([1, 1, 1]) → [1, 1, 1]      [1, 1, 1]      OK      
fix34([1]) → [1]      [1]      OK      
fix34([]) → []      []      OK      
fix34([7, 3, 7, 7, 4]) → [7, 3, 4, 7, 7]      [7, 3, 4, 7, 7]      OK      
fix34([3, 1, 4, 3, 1, 4]) → [3, 4, 1, 3, 4, 1]      [3, 4, 1, 3, 4, 1]      OK      
fix34([3, 1, 1, 3, 4, 4]) → [3, 4, 1, 3, 4, 1]      [3, 4, 1, 3, 4, 1]      OK      
other tests
OK      

above passes all tests
package com.solution;

import java.util.Arrays;

public class Fix34 {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		int[] ar={1, 3, 1, 4, 4, 3, 1};
		System.out.println("value-->"+Arrays.toString(fix34(ar)));

	}
	
	public static int[] fix34(int[] nums) {
	    for (int z = 0; z < nums.length; z++)
	        if (nums[z] == 3) {
	            int buffer = nums[z + 1];
	            nums[z + 1] = 4;
	            for (int j = z + 2; j < nums.length; j++)
	                if (nums[j] == 4) 
	                	nums[j] = buffer;
	        }
	    return nums;
	}

}

Open in new window

value-->[1, 3, 4, 1, 1, 3, 4]
1
gudii9Author Commented:
public int[] fix34(int[] nums) {
		
        int[] arr = new int[nums.length];
	    for (int z = 0; z < nums.length; z++){
	        if (nums[z] == 3) {
	          arr[z] = 3;
	          arr[z + 1] = 4;
	        }
      }
      int mark = 0;
	      for (int j = 0; j < nums.length; j++){
	        if(arr[j] == 0){
	          for (int k = mark; k < nums.length; k++){
	           if(nums[k] != 3 && nums[k] != 4){
	               arr[j] = nums[k];
	               mark = k + 1;
	               break;
	           }
	          }
	        } 
      }
	      return arr;
}

Open in new window


Expected      Run            
fix34([1, 3, 1, 4]) → [1, 3, 4, 1]      [1, 3, 4, 1]      OK      
fix34([1, 3, 1, 4, 4, 3, 1]) → [1, 3, 4, 1, 1, 3, 4]      [1, 3, 4, 1, 1, 3, 4]      OK      
fix34([3, 2, 2, 4]) → [3, 4, 2, 2]      [3, 4, 2, 2]      OK      
fix34([3, 2, 3, 2, 4, 4]) → [3, 4, 3, 4, 2, 2]      [3, 4, 3, 4, 2, 2]      OK      
fix34([2, 3, 2, 3, 2, 4, 4]) → [2, 3, 4, 3, 4, 2, 2]      [2, 3, 4, 3, 4, 2, 2]      OK      
fix34([3, 1, 4]) → [3, 4, 1]      [3, 4, 1]      OK      
fix34([3, 4, 1]) → [3, 4, 1]      [3, 4, 1]      OK      
fix34([1, 1, 1]) → [1, 1, 1]      [1, 1, 1]      OK      
fix34([1]) → [1]      [1]      OK      
fix34([]) → []      []      OK      
fix34([7, 3, 7, 7, 4]) → [7, 3, 4, 7, 7]      [7, 3, 4, 7, 7]      OK      
fix34([3, 1, 4, 3, 1, 4]) → [3, 4, 1, 3, 4, 1]      [3, 4, 1, 3, 4, 1]      OK      
fix34([3, 1, 1, 3, 4, 4]) → [3, 4, 1, 3, 4, 1]      [3, 4, 1, 3, 4, 1]      OK      
other tests
OK      
above similar to fix45 challenge passed all tests where fill new array simply with values which seems to me more efficient and easy way
0
gudii9Author Commented:
i see you are using mark to increment inner for loop with k as you are breaking out of it prematurely for k++ to take place later.
0
It's more than this solution.Get answers and train to solve all your tech problems - anytime, anywhere.Try it for free Edge Out The Competitionfor your dream job with proven skills and certifications.Get started today Stand Outas the employee with proven skills.Start learning today for free Move Your Career Forwardwith certification training in the latest technologies.Start your trial today
Java

From novice to tech pro — start learning today.