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fix34  challenge

Posted on 2016-08-14
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Last Modified: 2016-08-24
Hi,

I am working on below challenge

http://codingbat.com/prob/p159339

I have not understood the description properly
Array-3 > fix34
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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
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Question by:gudii9
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9 Comments
 
LVL 37

Expert Comment

by:Gerwin Jansen
ID: 41756057
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
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LVL 7

Author Comment

by:gudii9
ID: 41756444
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?
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LVL 37

Accepted Solution

by:
Gerwin Jansen earned 250 total points
ID: 41756536
Don't move 3
Move 4 to right side of a 3
You may move other numbers
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LVL 7

Author Comment

by:gudii9
ID: 41760248
not getting idea on best approach?
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LVL 37

Expert Comment

by:Gerwin Jansen
ID: 41760812
don't understand what you mean, explain
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LVL 40

Assisted Solution

by:noci
noci earned 250 total points
ID: 41764258
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] ?
                     --    --

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LVL 7

Author Comment

by:gudii9
ID: 41766305
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;
	
}

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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;
	}

}

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value-->[1, 3, 4, 1, 1, 3, 4]
1
 
LVL 7

Author Comment

by:gudii9
ID: 41769255
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
 
LVL 7

Author Comment

by:gudii9
ID: 41769277
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.
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