no23 challenge

Hi,

i am trying

http://codingbat.com/prob/p175689
public boolean no23(int[] nums) {
  
  
  
  
  int len=nums.length;
  
  for(int i=0;i<=len-1;i++)
  {
  if(nums[i]==2 | nums[i]==3)
  return false;
  
  else
  return true;
  }
  
  return false;
}

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i failed few tests. please advise
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gudii9Asked:
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ozoCommented:
return (nums[0]&~1)!=2&&(nums[1]|1)!=3;
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ozoCommented:
or
public boolean no23(int[] nums) {
  
  
  
  
  int len=nums.length;
  
  for(int i=0;i<=len-1;i++)
  {
  if(nums[i]==2 | nums[i]==3)
  return false;
  
  //else
  //return true;
  }
  
  return /*false*/true;
}

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LazyFolkCommented:
this works

public int[] makeEnds(int[] nums) {
  return new int[]{nums[0],nums[Math.max(0,nums.length-1)]};
}
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gudii9Author Commented:
 return /*false*/true;

i see above additional line compared to my code  as above what it means by above line?


also what is meaning of below

return (nums[0]&~1)!=2&&(nums[1]|1)!=3;

please advise
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gudii9Author Commented:
public boolean no23(int[] nums) {  
  
  
   return new int[]{nums[0],nums[Math.max(0,nums.length-1)]};
}

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above giving below error
Compile problems:


Error:      return new int[]{nums[0],nums[Math.max(0,nums.length-1)]};
             ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
Type mismatch: cannot convert from int[] to boolean


see Example Code to help with compile problems

please advise
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LazyFolkCommented:
Oh sorry, I did answer to another challenges

here is the correct answer :

public boolean no23(int[] nums) {
   return (nums[0]!=2 && nums[0]!=3  && nums[1]!=2 && nums[1]!=3);
}

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It will return true if :

         Number1 is not 2             (nums[0]!=2)
AND Number1 is not 3             (&& nums[0]!=3)
AND Number2 is not 2             (&& nums[1]!=2)
AND Number2 is not 3             (&& nums[1]!=3)
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ozoCommented:
return /*false*/true;
means
return true;

return (nums[0]&~1)!=2&&(nums[1]|1)!=3;
means
return (nums[0]!=2&&nums[0]!=3)&&(nums[1]!=2&&nums[1]!=3);

nums[0]&~1 == 2 when nums[0]==2 or nums[0]==3
nums[1]|1 == 3 when nums[1]==2 or nums[1]==3
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gudii9Author Commented:
return /*false*/true;
means
return true;

in above code
 /*false*/ false is surrounded with comment right?
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gudii9Author Commented:
return (nums[0]&~1)!=2&&(nums[1]|1)!=3;
means
return (nums[0]!=2&&nums[0]!=3)&&(nums[1]!=2&&nums[1]!=3);


is above ~ is java 8 syntax like lambda?

return (nums[0]!=2&&nums[0]!=3)&&(nums[1]!=2&&nums[1]!=3);

how is above with brakets different from below


   return (nums[0]!=2 && nums[0]!=3  && nums[1]!=2 && nums[1]!=3);

without any brackets?
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ozoCommented:
(nums[0]&~1)==2 when nums[0]==3

parenthesis are unnecessary in
 return nums[0]!=2 && nums[0]!=3 && nums[1]!=2 && nums[1]!=3;
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gudii9Author Commented:
return (nums[0]!=2&&nums[0]!=3)&&(nums[1]!=2&&nums[1]!=3);

nums[0]&~1 == 2 when nums[0]==2 or nums[0]==3
nums[1]|1 == 3 when nums[1]==2 or nums[1]==3

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above solution is not clear.
how it check no 2 or no 3 in the given array for different lenghts.

any simple examples on ~
please advise
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ozoCommented:
public boolean no23(int[] nums) {
  for( int i : nums ){
     if( i==2||i==3 ){ return false; }
  }
  return true;
}

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public boolean no23(int[] nums) {
  for( int i : nums ){
     if( (i&~1)==2 ){ return false; }
  }
  return true;
}

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public boolean no23(int[] nums) {
  for( int i : nums ){
     if( (i|1)==3 ){ return false; }
  }
  return true;
}

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public boolean no23(int[] nums) {
  for( int i : nums ){
     if( i/2==1 ){ return false; }
  }
  return true;
}

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gudii9Author Commented:
public boolean no23(int[] nums) {
  for( int i : nums ){
     if( i==2||i==3 ){ return false; }
  }
  return true;
}

Open in new window


above is clear.

Below lines not clear please advise


if( (i&~1)==2 ){ //what it means by i&~1 means ??

  if( (i|1)==3 ){ //what it means by i|1 ??

 if( i/2==1 ){ //how it checks no 2 as well as no 3.  is it due to division operator only 2, 3 divisible with 1 as result? if i take  i value as 4 then result is 2 which is not equal to 1 so true right??
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gudii9Author Commented:
please advise
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ozoCommented:
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gudii9Author Commented:
1 = 0001
2 = 0010
3 = 0011
 
2 0010
1 0001 or
  ----
3 0011
 
3 0011
1 0001 or
  ----
3 0011


Bitwise OR is clear but bitwise AND solution as below is not clear.
return (nums[0]&~1)!=2&&(nums[1]|1)!=3;
 Please advise
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gudii9Author Commented:
2    0010
~1 1110 AND
  -------------------
2  0010
 


3    0011
~1 1110 AND
  ----------------
2   0010




nums[1]|1
2 0010
1 0001 OR
-----------
 3 0011
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