asked on

Fix45 challenge

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

d-glitch

What do all the cases you are failing have in common?

gudii9

ASKER

one extra 5

d-glitch

one extra 5

That's what's wrong with your result.

But what do the input arrays have in common?

gudii9

ASKER

same number of 4's and 5's

gudii9

ASKER

also every 4 has a number after it that is not a 4

ASKER CERTIFIED SOLUTION

d-glitch

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gudii9

ASKER

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;
		}

Open in new window

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?

rrz

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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d-glitch

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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gudii9

ASKER

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;
}
}

Open in new window

gudii9

ASKER

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;
}

gudii9

ASKER

can i solve fix34 challenge with same above approach?

gudii9

ASKER

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.

rrz

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]

gudii9

ASKER

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?