Fix45 challenge

gudii9
gudii9 used Ask the Experts™
on
Hi,

I am working on below challenge.
http://codingbat.com/prob/p125819

My psedo code of approach is
1. iterate given arry
2. find position of 4
3. set next position to 5 by keeping orignal value in buffere
4. create other for loop from  index position where we found 4 plus two and if that value is 5 replace with buffere
5. return modified array

public int[] fix45(int[] nums) {
  

	    for (int z = 0; z < nums.length; z++)
	        if (nums[z] == 4) {
	            int buffer = nums[z + 1];
	            nums[z + 1] = 5;
	            for (int j = z + 2; j < nums.length; j++)
	                if (nums[j] == 5) 
	                	nums[j] = buffer;
	        }
	    return nums;
	

}

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i am failing below tests. please advise
Expected      Run            
fix45([5, 4, 9, 4, 9, 5]) → [9, 4, 5, 4, 5, 9]      [5, 4, 5, 4, 5, 9]      X      
fix45([1, 4, 1, 5]) → [1, 4, 5, 1]      [1, 4, 5, 1]      OK      
fix45([1, 4, 1, 5, 5, 4, 1]) → [1, 4, 5, 1, 1, 4, 5]      [1, 4, 5, 1, 1, 4, 5]      OK      
fix45([4, 9, 4, 9, 5, 5, 4, 9, 5]) → [4, 5, 4, 5, 9, 9, 4, 5, 9]      [4, 5, 4, 5, 9, 9, 4, 5, 9]      OK      
fix45([5, 5, 4, 1, 4, 1]) → [1, 1, 4, 5, 4, 5]      [5, 5, 4, 5, 4, 5]      X      
fix45([4, 2, 2, 5]) → [4, 5, 2, 2]      [4, 5, 2, 2]      OK      
fix45([4, 2, 4, 2, 5, 5]) → [4, 5, 4, 5, 2, 2]      [4, 5, 4, 5, 2, 2]      OK      
fix45([4, 2, 4, 5, 5]) → [4, 5, 4, 5, 2]      [4, 5, 4, 5, 2]      OK      
fix45([1, 1, 1]) → [1, 1, 1]      [1, 1, 1]      OK      
fix45([4, 5]) → [4, 5]      [4, 5]      OK      
fix45([5, 4, 1]) → [1, 4, 5]      [5, 4, 5]      X      
fix45([]) → []      []      OK      
fix45([5, 4, 5, 4, 1]) → [1, 4, 5, 4, 5]      [5, 4, 5, 4, 5]      X      
fix45([4, 5, 4, 1, 5]) → [4, 5, 4, 5, 1]      [4, 5, 4, 5, 1]      OK      
fix45([3, 4, 5]) → [3, 4, 5]      [3, 4, 5]      OK      
fix45([4, 1, 5]) → [4, 5, 1]      [4, 5, 1]      OK      
fix45([5, 4, 1]) → [1, 4, 5]      [5, 4, 5]      X      
fix45([2, 4, 2, 5]) → [2, 4, 5, 2]      [2, 4, 5, 2]      OK      
other tests
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What do all the cases you are failing have in common?

Author

Commented:
one extra 5
one extra 5
That's what's wrong with your result.

But what do the input arrays have in common?
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Author

Commented:
same number of 4's and 5's

Author

Commented:
also every 4 has a number after it that is not a 4
same number of 4's and 5's

also every 4 has a number after it that is not a 4
Both of those must be true for every case according to the problem definition.

What do the input arrays that you get wrong have in common that is different from the ones you get correct?   <== This is what debugging is all about.

How is this challenge harder than the fix34?  What is the difference?
As a hint:  Note that the challenge uses the words "this" and "version" twice.

Author

Commented:
they have 5 at beginning and failing

Below passes all tests

public int[] fix45(int[] nums) {
		  for (int x = 0; x < nums.length; x++) {
		    if (nums[x] == 4) {
		      for (int y = 0; y < nums.length; y++) {
		        if (nums[y] == 5) {
		          if (y > 0 && nums[y-1] != 4) {
		            int buffer = nums[x+1];
		            nums[x+1] = 5;
		            nums[y] = buffer;
		          } 
		          else if (y == 0) {
		            int buff = nums[x+1];
		            nums[x+1] = 5;
		            nums[y] = buff;
		          }
		                     
		        }
		      }
		    }
		  }
		  return nums;
		}

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Expected      Run            
fix45([5, 4, 9, 4, 9, 5]) → [9, 4, 5, 4, 5, 9]      [9, 4, 5, 4, 5, 9]      OK      
fix45([1, 4, 1, 5]) → [1, 4, 5, 1]      [1, 4, 5, 1]      OK      
fix45([1, 4, 1, 5, 5, 4, 1]) → [1, 4, 5, 1, 1, 4, 5]      [1, 4, 5, 1, 1, 4, 5]      OK      
fix45([4, 9, 4, 9, 5, 5, 4, 9, 5]) → [4, 5, 4, 5, 9, 9, 4, 5, 9]      [4, 5, 4, 5, 9, 9, 4, 5, 9]      OK      
fix45([5, 5, 4, 1, 4, 1]) → [1, 1, 4, 5, 4, 5]      [1, 1, 4, 5, 4, 5]      OK      
fix45([4, 2, 2, 5]) → [4, 5, 2, 2]      [4, 5, 2, 2]      OK      
fix45([4, 2, 4, 2, 5, 5]) → [4, 5, 4, 5, 2, 2]      [4, 5, 4, 5, 2, 2]      OK      
fix45([4, 2, 4, 5, 5]) → [4, 5, 4, 5, 2]      [4, 5, 4, 5, 2]      OK      
fix45([1, 1, 1]) → [1, 1, 1]      [1, 1, 1]      OK      
fix45([4, 5]) → [4, 5]      [4, 5]      OK      
fix45([5, 4, 1]) → [1, 4, 5]      [1, 4, 5]      OK      
fix45([]) → []      []      OK      
fix45([5, 4, 5, 4, 1]) → [1, 4, 5, 4, 5]      [1, 4, 5, 4, 5]      OK      
fix45([4, 5, 4, 1, 5]) → [4, 5, 4, 5, 1]      [4, 5, 4, 5, 1]      OK      
fix45([3, 4, 5]) → [3, 4, 5]      [3, 4, 5]      OK      
fix45([4, 1, 5]) → [4, 5, 1]      [4, 5, 1]      OK      
fix45([5, 4, 1]) → [1, 4, 5]      [1, 4, 5]      OK      
fix45([2, 4, 2, 5]) → [2, 4, 5, 2]      [2, 4, 5, 2]      OK      
other tests

package com.solution;

import java.util.Arrays;

public class Fix45 {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		int[] ar={1,4,1,5};
		System.out.println("fix45 -->"+Arrays.toString(fix45(ar)));

	}

	
	public static int[] fix45(int[] nums) {
		  for (int x = 0; x < nums.length; x++) {
		    if (nums[x] == 4) {
		      for (int y = 0; y < nums.length; y++) {
		        if (nums[y] == 5) {
		          if (y > 0 && nums[y-1] != 4) {
		            int buffer = nums[x+1];
		            nums[x+1] = 5;
		            nums[y] = buffer;
		          } 
		          else if (y == 0) {
		            int buff = nums[x+1];
		            nums[x+1] = 5;
		            nums[y] = buff;
		          }
		                     
		        }
		      }
		    }
		  }
		  return nums;
		}
}

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fix45 -->[1, 4, 5, 1]

any improvement/refinement/alternate approach?

Commented:
I don't know if my solution is any better than yours.  I create a new array. I fill it with the values from the input array. I return the new array.
public int[] fix45(int[] nums) {
          int[] arr = new int[nums.length];
  	    for (int z = 0; z < nums.length; z++){
	        if (nums[z] == 4) {
	          arr[z] = 4;
	          arr[z + 1] = 5;
	        }
        }
        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] != 4 && nums[k] != 5){
	               arr[j] = nums[k];
	               mark = k + 1;
	               break;
	           }
	          }
	        } 
        }
	      return arr;
}

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How does your pseudo code compare with this:
b = 0
for  i= 0 to length
     if (nums[i] == 4) 
          buff[b] = nums[i+1]
          nums[i+1] = 5
          b ++

b = 0
for i= 0 to length	
     if (nums[i] == 5)
          if (i == 0)
               nums[i] = buff[b]
               b++
          else if (nums[i-1] != 4)
               nums[i] = buff[b]
               b++

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Author

Commented:
what is the purpose of

  mark = k + 1;

public int[] fix45(int[] nums) {
          int[] arr = new int[nums.length];
            for (int z = 0; z < nums.length; z++){
              if (nums[z] == 4) {
                arr[z] = 4;
                arr[z + 1] = 5;
              }
        }
        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] != 4 && nums[k] != 5){
                     arr[j] = nums[k];
                     mark = k + 1;
                     break;
                 }
                }
              }
        }
            return arr;
}

package com.solution;

import java.util.Arrays;

public class Fix46_2 {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		int[] ar = { 1, 4, 1, 5 };
		System.out.println("fix45 -->" + Arrays.toString(fix45(ar)));

	}

	public static int[] fix45(int[] nums) {
		
        int[] arr = new int[nums.length];
	    for (int z = 0; z < nums.length; z++){
	        if (nums[z] == 4) {
	          arr[z] = 4;
	          arr[z + 1] = 5;
	        }
      }
      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] != 4 && nums[k] != 5){
	               arr[j] = nums[k];
	               mark = k + 1;
	               break;
	           }
	          }
	        } 
      }
	      return arr;
}
}

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Author

Commented:
i see using mark and other for loop you are trying to fill non 4 and non 5 values from imput nums asrray to output new arr array.

I found your solution of filling new array is more straight forward?
why are  we starting second for loop from mark rather than 0?
public int[] fix45(int[] nums) {
          int[] arr = new int[nums.length];
            for (int z = 0; z < nums.length; z++){
              if (nums[z] == 4) {
                arr[z] = 4;
                arr[z + 1] = 5;
              }
        }
        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] != 4 && nums[k] != 5){
                     arr[j] = nums[k];
                     mark = k + 1;
                     break;
                 }
                }
              }
        }
            return arr;
}

Author

Commented:
can i solve fix34 challenge with same above approach?

Author

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.

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.
I used the mark variable to know where to begin looking for the next number to use to fill into the arr array. The increment in the loop(k++) is still used.  
There probably is room for improvement. But, when you closed this question, I stopped working on this.  Anyway, maybe adding some debug statements, will shed some light on my code. See below,
import java.util.Arrays;
public class Fix{
	public static void main(String[] args) {
		int[] input = {2, 3, 4, 6, 4, 7, 5, 5, 4, 8, 5};
		System.out.println("  input -->" + Arrays.toString(input));
		fix45(input);
	}
	public static int[] fix45(int[] nums) {
        int[] arr = new int[nums.length];
  	    for (int z = 0; z < nums.length; z++){
	        if (nums[z] == 4) {
	          arr[z] = 4;
	          arr[z + 1] = 5;
	        }
        }
		System.out.println(" output -->" + Arrays.toString(arr) + " (after first for loop)");
        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] != 4 && nums[k] != 5){
	               arr[j] = nums[k];
				   mark = k + 1;
				   System.out.println(" j is " + j + ", k is " + k + ", mark is " + mark);
				   System.out.println(" output -->" + Arrays.toString(arr));
	               break;
	            }   
	          }
	        } 
        }
	      return arr;	
}
}

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The output is :    
  input -->[2, 3, 4, 6, 4, 7, 5, 5, 4, 8, 5]
 output -->[0, 0, 4, 5, 4, 5, 0, 0, 4, 5, 0] (after first for loop)
 j is 0, k is 0, mark is 1
 output -->[2, 0, 4, 5, 4, 5, 0, 0, 4, 5, 0]
 j is 1, k is 1, mark is 2
 output -->[2, 3, 4, 5, 4, 5, 0, 0, 4, 5, 0]
 j is 6, k is 3, mark is 4
 output -->[2, 3, 4, 5, 4, 5, 6, 0, 4, 5, 0]
 j is 7, k is 5, mark is 6
 output -->[2, 3, 4, 5, 4, 5, 6, 7, 4, 5, 0]
 j is 10, k is 9, mark is 10
 output -->[2, 3, 4, 5, 4, 5, 6, 7, 4, 5, 8]

Author

Commented:
 input -->[2, 3, 4, 6, 4, 7, 5, 5, 4, 8, 5]
 output -->[0, 0, 4, 5, 4, 5, 0, 0, 4, 5, 0] (after first for loop)
 j is 0, k is 0, mark is 1
 output -->[2, 0, 4, 5, 4, 5, 0, 0, 4, 5, 0]
 j is 1, k is 1, mark is 2
 output -->[2, 3, 4, 5, 4, 5, 0, 0, 4, 5, 0]
 j is 6, k is 3, mark is 4
 output -->[2, 3, 4, 5, 4, 5, 6, 0, 4, 5, 0]
 j is 7, k is 5, mark is 6
 output -->[2, 3, 4, 5, 4, 5, 6, 7, 4, 5, 0]
 j is 10, k is 9, mark is 10
 output -->[2, 3, 4, 5, 4, 5, 6, 7, 4, 5, 8]

above output is more clear now. To understand again today taking some time around 1 hour even though i saw this day before yesterday?

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