tenRun challenge

Hi,

I am working on below challenge

http://codingbat.com/prob/p199484

Psedo code description of approach :
1. create new array of given array size
2. loop given array in for loop
3. check each element value not greater than 10
4. if yes multiply by 10
5 if no give same value from given array to new array

I wrote my code as below

public int[] tenRun(int[] nums) {
  
  
  int len =nums.length;
  int[] arr=new int[len];
  for(int i=0;i<len;i++){
    if(nums[i]<10){
    arr[i]=nums[i]*10;
    }
    else{
        arr[i]=nums[i];
    }
  }
  return arr;
}

Open in new window





I am  not passing all tests
Expected      Run            
tenRun([2, 10, 3, 4, 20, 5]) → [2, 10, 10, 10, 20, 20]      [20, 10, 30, 40, 20, 50]      X      
tenRun([10, 1, 20, 2]) → [10, 10, 20, 20]      [10, 10, 20, 20]      OK      
tenRun([10, 1, 9, 20]) → [10, 10, 10, 20]      [10, 10, 90, 20]      X      
tenRun([1, 2, 50, 1]) → [1, 2, 50, 50]      [10, 20, 50, 10]      X      
tenRun([1, 20, 50, 1]) → [1, 20, 50, 50]      [10, 20, 50, 10]      X      
tenRun([10, 10]) → [10, 10]      [10, 10]      OK      
tenRun([10, 2]) → [10, 10]      [10, 20]      X      
tenRun([0, 2]) → [0, 0]      [0, 20]      X      
tenRun([1, 2]) → [1, 2]      [10, 20]      X      
tenRun([1]) → [1]      [10]      X      
tenRun([]) → []      []      OK      
other tests
X      
How to improve my design, approach, code? please advise
LVL 7
gudii9Asked:
Who is Participating?
 
d-glitchCommented:
Go through the array one element at a time.
Do nothing until you find a multiple of 10.  You need a flag.
Once you find one you will use to replace later elements.  Remember it.
The rest of the array elements are either replaced or used to update replacer.
0
 
d-glitchCommented:
CREATE, GREATER THAN, and MULTIPLY  are not mentioned in the Challenge.
Work on the pseudo code, not the code.

How important is the pseudo code?
Do you realize you are actually failing all of the tests?
The only cases that pass are those where the input string is already in final form.

You can not improve your English comprehension by studying Java.
You might be able to improve your Java skills by slowing down and working on one question at a time.
But I've said that several times already and you don't seem to agree.
0
 
gudii9Author Commented:
tenRun([2, 10, 3, 4, 20, 5]) → [2, 10, 10, 10, 20, 20]
i thought above should be as below according to challenge description?
tenRun([2, 10, 3, 4, 20, 5]) → [2, 10, 10, 10, 20, 10]
please advise
0
Ultimate Tool Kit for Technology Solution Provider

Broken down into practical pointers and step-by-step instructions, the IT Service Excellence Tool Kit delivers expert advice for technology solution providers. Get your free copy now.

 
rrzCommented:
Are you familiar with the remainder operator? It is also called modulo or modulus.
You can use it in this challenge to determine whether a number is a multiple of ten .  
https://docs.oracle.com/javase/tutorial/java/nutsandbolts/op1.html   
http://www.javaranch.com/drive/modulo.html   
https://www.google.com/?gws_rd=ssl#q=modulus+operator+java
0
 
d-glitchCommented:
>>  I thought above should be as below according to challenge description?
      tenRun([2, 10, 3, 4, 20, 5]) → [2, 10, 10, 10, 20, 10]

Why do you think that?  What does the challenge say?

What do you think the results of the following should be:
      tenRun([2, 20, 3, 4, 10, 5]) → ???
0
 
gudii9Author Commented:
For each multiple of 10 in the given array, change all the values following it to be that multiple of 10, until encountering another multiple of 10. So {2, 10, 3, 4, 20, 5} yields {2, 10, 10, 10, 20, 20}.//however 3,4 became 10.10 i thought 5 should become 10
tenRun([2, 10, 3, 4, 20, 5]) → [2, 10, 10, 10, 20, 20]
tenRun([10, 1, 20, 2]) → [10, 10, 20, 20]
tenRun([10, 1, 9, 20]) → [10, 10, 10, 20]
tenRun([2, 10, 3, 4, 20, 5]) → [2, 10, 10, 10, 20, 20]
i thought above should be as below according to challenge description?
tenRun([2, 10, 3, 4, 20, 5]) → [2, 10, 10, 10, 20, 10]


What do you think the results of the following should be:
      tenRun([2, 20, 3, 4, 10, 5]) → ???tenRun([2, 20, 10, 10, 10, 10])
0
 
d-glitchCommented:
>>  However 3,4 became 10.10 i thought 5 should become 10
      What multiple of 10 does the 5 follow???


What do you think the results of the following should be:
      tenRun([2, 20, 3, 4, 10, 5]) → ???tenRun([2, 20, 10, 10, 10, 10])

No, that is not correct

For each multiple of 10 in the given array, change all the values following it to be that multiple of 10

Doing it by hand:
tenRun([2, 20, 3, 4, 10, 5]
     2  - no change
    20  - multiple of 10 - no change
     3 - change it 20 since it follows that multiple of 10
     4 - change it 20 since it follows that multiple of 10
    10  - multiple of 10 - no change
     5 - change it 10 since it follows that multiple of 10

Open in new window

 You don't change the values that are multiples of 10.
And you don't change any values before the first multiple of 10.
0
 
d-glitchCommented:
What do you think the results of the following should be:
      tenRun([2, 10, 3, 20, 5, 1001, 10, 2001]) → ???
0
 
gudii9Author Commented:
>>  However 3,4 became 10.10 i thought 5 should become 10
      What multiple of 10 does the 5 follow???//[b]not sure what it means..10*0 is 0 10*1 is 10 so 5 falls in between 0 and 5..when you say What multiple of 10 does the 5 follow i am not following it?[/b]


What do you think the results of the following should be:
      tenRun([2, 20, 3, 4, 10, 5]) → ???tenRun([2, 20, 10, 10, 10, 10]) 

No, that is not correct

For each multiple of 10 in the given array, change all the values following it to be that multiple of 10

Doing it by hand:
tenRun([2, 20, 3, 4, 10, 5]
     2  - no change
    20  - multiple of 10 - no change
     3 - change it 20 since it follows that multiple of 10
     4 - change it 20 since it follows that multiple of 10
    10  - multiple of 10 - no change
     5 - change it 10 since it follows that multiple of 10//[b][i]but challenge expecting to change it to 20 not 10[/i][/b]

Open in new window

0
 
d-glitchCommented:
This is the case we are talking about now.  It is one of the original test cases that you did not understand.
      tenRun([2, 10, 3, 4, 20, 5]

>>  However 3,4 became 10.10 i thought 5 should become 10
      What multiple of 10 does the 5 follow???

>> not sure what it means..10*0 is 0 10*1 is 10 so 5 falls in between 0 and 5..when you say
      What multiple of 10 does the 5 follow i am not following it?[/b]

falls between has nothing to do with following

multiple of  does not mean  multiply by

What multiple of 10 does the 5 follow in the array?
                                    is the same as
What multiple of 10 comes before 5 in the array?

The answer in this case is 20.  -- So change the 5 to 20.
0
 
gudii9Author Commented:

What multiple of 10 does the 5 follow in the array?
                                    is the same as
What multiple of 10 comes before 5 in the array?

The answer in this case is 20.  -- So change the 5 to 20.
still thinking above
0
 
d-glitchCommented:
This is the text of the Challenge, in case you have forgotten:
     For each multiple of 10 (like 0, 10, 20, 30, ...) in the given array
     change all the values following it (in the array) to be that multiple of 10
     until encountering another multiple of 10.
0
 
gudii9Author Commented:
tenRun([2, 10, 3, 20, 5, 1001, 10, 2001]) → ???    ([2, 10, 10, 20, 20, 20, 10, 10]) → is this answer correct?
1
 
gudii9Author Commented:
public int[] tenRun(int[] nums) {
  
  
  int len =nums.length;
  int[] arr=new int[len];
  for(int i=0;i<len;i++){
    if((nums[i]/10)==0){
    arr[i]=10*(nums[i]%10);
    }
    else{
        arr[i]=nums[i];
    }
  }
  return arr;
}

Open in new window



Expected      Run            
tenRun([2, 10, 3, 4, 20, 5]) → [2, 10, 10, 10, 20, 20]      [20, 10, 30, 40, 20, 50]      X      
tenRun([10, 1, 20, 2]) → [10, 10, 20, 20]      [10, 10, 20, 20]      OK      
tenRun([10, 1, 9, 20]) → [10, 10, 10, 20]      [10, 10, 90, 20]      X      
tenRun([1, 2, 50, 1]) → [1, 2, 50, 50]      [10, 20, 50, 10]      X      
tenRun([1, 20, 50, 1]) → [1, 20, 50, 50]      [10, 20, 50, 10]      X      
tenRun([10, 10]) → [10, 10]      [10, 10]      OK      
tenRun([10, 2]) → [10, 10]      [10, 20]      X      
tenRun([0, 2]) → [0, 0]      [0, 20]      X      
tenRun([1, 2]) → [1, 2]      [10, 20]      X      
tenRun([1]) → [1]      [10]      X      
tenRun([]) → []      []      OK      
other tests
X      
Your progress graph for this problem


i think now i understood challene trying to fix it and failing due to not able to start from first occurence of 10 multiple but starting from index 0
0
 
gudii9Author Commented:
how to start for loop from first occurrence of 10 multiple not from i=0 which resolve the issues here
0
 
d-glitchCommented:
You are still multiplying by 10.
That is wrong.

You have to start at 0 to look for the first multiple of 10.

What is your pseudo code?
0
 
d-glitchCommented:
tenRun([2, 10, 3, 20, 5, 1001, 10, 2001]) → ???    ([2, 10, 10, 20, 20, 20, 10, 10]) → is this answer correct?

Yes, but your code is not even close.

Where is your pseudo code?
0
 
rrzCommented:
You should use the remainder operator to make your search. See my first comment above here.
0
 
gudii9Author Commented:
how to use moduls operator for this challenge?

Psedo code description of approach :
1. create new array of given array size
2. loop given array in for loop
3. check each element value modulus is 0(ie reminder)
4. if yes multiply subsequent elements by same previous element value?
5 if no give same value from given array to new array
0
 
rrzCommented:
how to use moduls operator for this challenge?
You can search for the condition  
nums[i]%10) == 0

Open in new window

that will be true for multiples of ten.
4. if yes multiply subsequent elements by same previous element value?
No. As d-glitch has posted
... MULTIPLY  are not mentioned in the Challenge.
Multiply is an arithmetic operation. If we multiply 3 by 4 the result is 12. The numbers that are multiples of ten are 10,  20 ,30, 40, and so on.
0
 
gudii9Author Commented:
public int[] tenRun(int[] nums) {
  
  
  int totalLen =nums.length;
  //int[] arr=new int[len];
  int firstVal=0;
  for(int i=0;i<totalLen;i++){
    if((nums[i]%10)==0){
    firstVal=i;
    break;
    }
  }
  
  
    for(int k=firstVal;k<totalLen;k++){
    if((nums[k]%10)==0){
    secondVal=k;
    break;
    }
  }
    
    
  return nums;
}

Open in new window


i tried above approach. I have to print  all values between firstval and secondval which i am not able to figure out how to proceed further esp if there is thirdval and fourthvale etc
0
 
rrzCommented:
Hint: instead of using the name firstVal use the name currentVal.
0
 
gudii9Author Commented:
let me try
0
 
gudii9Author Commented:
public int[] tenRun(int[] nums) {
  
  int totalLen =nums.length;
  //int[] arr=new int[len];
  int val=0;
  boolean tenFlag=false;
  for(int i=0;i<totalLen;i++){
    if((nums[i]%10)==0){
    val=nums[i];
    tenFlag=true;
    }
    else if((nums[i]%10)!=0&&tenFlag){
    nums[i]=val;
    
    }
  }
  return nums;
}

Open in new window



Expected

Run



tenRun([2, 10, 3, 4, 20, 5]) → [2, 10, 10, 10, 20, 20] [2, 10, 10, 10, 20, 20] OK  
tenRun([10, 1, 20, 2]) → [10, 10, 20, 20] [10, 10, 20, 20] OK  
tenRun([10, 1, 9, 20]) → [10, 10, 10, 20] [10, 10, 10, 20] OK  
tenRun([1, 2, 50, 1]) → [1, 2, 50, 50] [1, 2, 50, 50] OK  
tenRun([1, 20, 50, 1]) → [1, 20, 50, 50] [1, 20, 50, 50] OK  
tenRun([10, 10]) → [10, 10] [10, 10] OK  
tenRun([10, 2]) → [10, 10] [10, 10] OK  
tenRun([0, 2]) → [0, 0] [0, 0] OK  
tenRun([1, 2]) → [1, 2] [1, 2] OK  
tenRun([1]) → [1] [1] OK  
tenRun([]) → [] [] OK  

other tests
 OK
now I passed all tests
0
 
rrzCommented:
Here is my code.
public int[] tenRun(int[] nums) {
    int current = 1;
  for(int i=0;i<nums.length; i++){
    if((nums[i]%10) == 0){
        current = nums[i];
    }
    else{
        if(current != 1) nums[i] = current; 
    }
  }
  return nums;
}

Open in new window

1
 
d-glitchCommented:
Yes, you can combine the flag and the currVal variables for a more elegant solution.

You can come up with this sort of trick if you focus your thinking on algorithms and optimizations.  You will never come up with them if you keep asking for basic help.

You are currently working on a problem that can use this exact trick.  Good luck.
0
 
gudii9Author Commented:

You are currently working on a problem that can use this exact trick.

how to combine this?
0
 
d-glitchCommented:
Look at rrz's code until you understand it.  You may actually have to think about it.  
That is the point.

How many problems are you currently working on?
0
Question has a verified solution.

Are you are experiencing a similar issue? Get a personalized answer when you ask a related question.

Have a better answer? Share it in a comment.

All Courses

From novice to tech pro — start learning today.