Link to home
Start Free TrialLog in
Avatar of gudii9
gudii9Flag for United States of America

asked on

ShiftLeft challenge

Hi,

I am working on below challenge

http://codingbat.com/prob/p105031

Psedo code description of approach :
1. create new array of given array lenght
2. assign each new array i th element with given array i-1 th element
3, return new array

I wrote my code as below

public int[] shiftLeft(int[] nums) {
  int len =nums.length;
  int[] arr=new int[len];
  for(int i=0;i<len;i++){
    arr[i]=nums[i-1];
  }
  return arr;
}

Open in new window






I am not passing all tests by failing edge cases
Expected      Run            
shiftLeft([6, 2, 5, 3]) → [2, 5, 3, 6]      Exception:java.lang.ArrayIndexOutOfBoundsException: -1 (line number:5)      X      
shiftLeft([1, 2]) → [2, 1]      Exception:java.lang.ArrayIndexOutOfBoundsException: -1 (line number:5)      X      
shiftLeft([1]) → [1]      Exception:java.lang.ArrayIndexOutOfBoundsException: -1 (line number:5)      X      
shiftLeft([]) → []      []      OK      
shiftLeft([1, 1, 2, 2, 4]) → [1, 2, 2, 4, 1]      Exception:java.lang.ArrayIndexOutOfBoundsException: -1 (line number:5)      X      
shiftLeft([1, 1, 1]) → [1, 1, 1]      Exception:java.lang.ArrayIndexOutOfBoundsException: -1 (line number:5)      X      
shiftLeft([1, 2, 3]) → [2, 3, 1]      Exception:java.lang.ArrayIndexOutOfBoundsException: -1 (line number:5)      X      
other tests
X      
Your progress graph for this problem


How to improve my design, approach, code? please advise
Avatar of CPColin
CPColin
Flag of United States of America image

What do you think is causing the exception you're seeing? They're all the same, so it must be something that's happening commonly.
Avatar of gudii9

ASKER

Edge case again as i am checking i-1. if i start for loop at 1 to avoid this issue then i have to separately handle i at 0 indexed element which is bit challenging
The definition of "edge case" normally refers to situations that happen very rarely and cause your code to fail. In this case, since you're failing almost every test the same way, the problem is, by definition, not an "edge case."

Terminology aside, what do you think is causing these exceptions? What in your code is making them happen?
Avatar of gudii9

ASKER

as
public int[] shiftLeft(int[] nums) {
  int len =nums.length;
  int[] arr=new int[len];
 // arr[len]=nums[0];
  for(int i=0;i<len-1;i++){
   
    arr[i]=nums[i+1];
  }
  return arr;
}

Open in new window

Expected      Run            
shiftLeft([6, 2, 5, 3]) → [2, 5, 3, 6]      [2, 5, 3, 0]      X      
shiftLeft([1, 2]) → [2, 1]      [2, 0]      X      
shiftLeft([1]) → [1]      [0]      X      
shiftLeft([]) → []      []      OK      
shiftLeft([1, 1, 2, 2, 4]) → [1, 2, 2, 4, 1]      [1, 2, 2, 4, 0]      X      
shiftLeft([1, 1, 1]) → [1, 1, 1]      [1, 1, 0]      X      
shiftLeft([1, 2, 3]) → [2, 3, 1]      [2, 3, 0]      X      
other tests
X

need to fix last element
Avatar of gudii9

ASKER

i thought edge case something happen at start and end of array
ASKER CERTIFIED SOLUTION
Avatar of CPColin
CPColin
Flag of United States of America image

Link to home
membership
This solution is only available to members.
To access this solution, you must be a member of Experts Exchange.
Start Free Trial
Avatar of gudii9

ASKER

took erasable slate board with erasable marker and also opened eclipse to debug these two are most effective tools for me. i wil solve soon
public class shiftLeft {

	public static void main(String[] args) {
	
		int[] ar={6, 2, 5, 3};
System.out.println("is=="+shiftLeft(ar));
	}

	
	public static int[] shiftLeft(int[] nums) {
		  int len =nums.length;
		  int[] arr=new int[len];
		 // arr[len]=nums[0];
		  for(int i=0;i<len-1;i++){
		   
		    arr[i]=nums[i+1];
		  }
		  return arr;
		}
 
}

Open in new window

Avatar of gudii9

ASKER

		  int len =nums.length;
		  int[] arr=new int[len];
		 // arr[len]=nums[0];
		  for(int i=0;i<len-1;i++){
		   
		    arr[i]=nums[i+1];
		  }
		  arr[len-1]=nums[0];
		  return arr;
		

Open in new window


all tests successful except one

Expected      Run            
shiftLeft([6, 2, 5, 3]) → [2, 5, 3, 6]      [2, 5, 3, 6]      OK      
shiftLeft([1, 2]) → [2, 1]      [2, 1]      OK      
shiftLeft([1]) → [1]      [1]      OK      
shiftLeft([]) → []      Exception:java.lang.ArrayIndexOutOfBoundsException: 0 (line number:9)      X      
shiftLeft([1, 1, 2, 2, 4]) → [1, 2, 2, 4, 1]      [1, 2, 2, 4, 1]      OK      
shiftLeft([1, 1, 1]) → [1, 1, 1]      [1, 1, 1]      OK      
shiftLeft([1, 2, 3]) → [2, 3, 1]      [2, 3, 1]      OK      
other tests
OK      

let me fix that also
Avatar of gudii9

ASKER

public int[] shiftLeft(int[] nums) {
		  int len =nums.length;
		  int[] arr=new int[len];
		 // arr[len]=nums[0];
		 if(len>0){
		  for(int i=0;i<len-1;i++){
		   
		    arr[i]=nums[i+1];
		  }
		  arr[len-1]=nums[0];
		 }else{
		   return arr;
		 }
		  return arr;
		}

Open in new window


above passing all tests. please advise any improvements/modifications
That looks good to me. If you want to do it without creating a separate array, you could store nums[0] in a variable before you do the looping, then stick it at the end after the looping is done. But since the challenge says you can do it either way, this will work fine!
Avatar of gudii9

ASKER

If you want to do it without creating a separate array, you could store nums[0] in a variable before you do the looping, then stick it at the end after the looping is done
how to do without creating new array?
not clear? please advise
The code you have in this comment is very close to what you would need. All you have to do is store the first element somewhere before you start looping and put it at the end after your loop finishes.
Avatar of gudii9

ASKER

The code you have in this comment is very close to what you would need. All you have to do is store the first element somewhere before you start looping and put it at the end after your loop finishes.
public int[] shiftLeft(int[] nums) {
  int len =nums.length;
  int[] arr=new int[len];
 // arr[len]=nums[0];
  for(int i=0;i<len-1;i++){
   
    arr[i]=nums[i+1];
  }
  return arr;
}

Open in new window



but i created new array right in above comment code?
So you did. My mistake. Just skip the line where you create a new array.
Avatar of gudii9

ASKER

public int[] shiftLeft(int[] nums) {
  int len =nums.length;
//  int[] arr=new int[len];
 // arr[len]=nums[0];
  for(int i=0;i<len-1;i++){
   
    arr[i]=nums[i+1];
  }
  return arr;
}

Open in new window


if i skip compilation error at assignment statement

 arr[i]=nums[i+1];

Open in new window

Well yeah, you need to update that line to use the right array on both sides of the = sign.
Avatar of gudii9

ASKER

how to use right array on both sides if i have only one array and i am not creating new array? confusing
Try it and see what happens.
Avatar of gudii9

ASKER

Right means opposite of left
i thought right means opposite of incorrect.
public int[] shiftLeft(int[] nums) {
  int len =nums.length;
//  int[] arr=new int[len];
 int x=nums[0];
  for(int i=0;i<len-1;i++){
   
    nums[i]=nums[i+1];
  }
  nums[len-1]=x;
  return nums;
}

Open in new window


above passes all tests except one fails
public int[] shiftLeft(int[] nums) {
  int len =nums.length;
//  int[] arr=new int[len];
 int x=nums[0];
  for(int i=0;i<len-1;i++){
   
    nums[i]=nums[i+1];
  }
  nums[len-1]=x;
  return nums;
}

Open in new window

how to fix that index out of bound for empty array?
Avatar of gudii9

ASKER

public int[] shiftLeft(int[] nums) {
  int len =nums.length;
//  int[] arr=new int[len];
if(len>0){
 int x=nums[0];
  for(int i=0;i<len-1;i++){
   
    nums[i]=nums[i+1];
  }
  nums[len-1]=x;
  return nums;
}

else
return nums;



}

Open in new window


i fixed as above. any improvements?
Right means opposite of left
i thought right means opposite of incorrect.

In this case, it meant both!

i fixed as above. any improvements?

Looks good to me! Glad you solved the problem with the empty array.