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has77  challenge

Posted on 2016-08-03
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Last Modified: 2016-08-08
Hi,

I am working on below challenge

http://codingbat.com/prob/p168357

Psedo code description of approach :
1. loop through given array
2. check if an(i th) element and i+1 th adjacent  are 7's or i, i+2 are 7's
3. if yes return true
4.if no return false

I wrote my code as below

public boolean has77(int[] nums) {
  boolean res=false;
  
  for(int i=0;i<nums.length-2;i++){
    
    if((nums[i]==7&&nums[i+1]==7)||(nums[i]==7&&nums[i+2]==7)){
      res=true;
      return res;
    }
  }
  return res;
}

Open in new window




I am not passing all tests
Expected      Run            
has77([1, 7, 7]) → true      false      X      
has77([1, 7, 1, 7]) → true      true      OK      
has77([1, 7, 1, 1, 7]) → false      false      OK      
has77([7, 7, 1, 1, 7]) → true      true      OK      
has77([2, 7, 2, 2, 7, 2]) → false      false      OK      
has77([2, 7, 2, 2, 7, 7]) → true      false      X      
has77([7, 2, 7, 2, 2, 7]) → true      true      OK      
has77([7, 2, 6, 2, 2, 7]) → false      false      OK      
has77([7, 7, 7]) → true      true      OK      
has77([7, 1, 7]) → true      true      OK      
has77([7, 1, 1]) → false      false      OK      
has77([1, 2]) → false      false      OK      
has77([1, 7]) → false      false      OK      
has77([7]) → false      false      OK      
other tests
X      
How to improve my design, approach, code? please advise
0
Comment
Question by:gudii9
  • 4
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9 Comments
 
LVL 35

Expert Comment

by:Terry Woods
ID: 41742259
You're missing the case where the last two numbers are 7's because you stop the loop too early for the first test to pick up that case.

I could put an extra test in the code to detect it, but you might like to solve that much yourself?
0
 
LVL 7

Author Comment

by:gudii9
ID: 41742443
an extra test in the code to detect it

where and how? i am also trying now
0
 
LVL 7

Author Comment

by:gudii9
ID: 41742451
public boolean has77(int[] nums) {
  boolean res=false;
  
  for(int i=0;i<nums.length-1;i++){
    
    if((nums[i]==7&&nums[i+1]==7)||(nums[i]==7&&nums[i+2]==7)){
      res=true;
      return res;
    }
  }
  return res;
}

Open in new window

if i put nums.length-1 in for loop failing tests

Expected      Run            
has77([1, 7, 7]) → true      true      OK      
has77([1, 7, 1, 7]) → true      true      OK      
has77([1, 7, 1, 1, 7]) → false      false      OK      
has77([7, 7, 1, 1, 7]) → true      true      OK      
has77([2, 7, 2, 2, 7, 2]) → false      Exception:java.lang.ArrayIndexOutOfBoundsException: 6 (line number:6)      X      
has77([2, 7, 2, 2, 7, 7]) → true      true      OK      
has77([7, 2, 7, 2, 2, 7]) → true      true      OK      
has77([7, 2, 6, 2, 2, 7]) → false      false      OK      
has77([7, 7, 7]) → true      true      OK      
has77([7, 1, 7]) → true      true      OK      
has77([7, 1, 1]) → false      false      OK      
has77([1, 2]) → false      false      OK      
has77([1, 7]) → false      false      OK      
has77([7]) → false      false      OK      
other tests
OK      
Correct for more than half the tests

Your progress graph for this problem

if i say i<nums.length failing more
Expected      Run            
has77([1, 7, 7]) → true      true      OK      
has77([1, 7, 1, 7]) → true      true      OK      
has77([1, 7, 1, 1, 7]) → false      Exception:java.lang.ArrayIndexOutOfBoundsException: 5 (line number:6)      X      
has77([7, 7, 1, 1, 7]) → true      true      OK      
has77([2, 7, 2, 2, 7, 2]) → false      Exception:java.lang.ArrayIndexOutOfBoundsException: 6 (line number:6)      X      
has77([2, 7, 2, 2, 7, 7]) → true      true      OK      
has77([7, 2, 7, 2, 2, 7]) → true      true      OK      
has77([7, 2, 6, 2, 2, 7]) → false      Exception:java.lang.ArrayIndexOutOfBoundsException: 6 (line number:6)      X      
has77([7, 7, 7]) → true      true      OK      
has77([7, 1, 7]) → true      true      OK      
has77([7, 1, 1]) → false      false      OK      
has77([1, 2]) → false      false      OK      
has77([1, 7]) → false      Exception:java.lang.ArrayIndexOutOfBoundsException: 2 (line number:6)      X      
has77([7]) → false      Exception:java.lang.ArrayIndexOutOfBoundsException: 1 (line number:6)      X      
other tests
X      
Correct for more than half the tests

Your progress graph for this problem


Forget It! -- delete my code for this problem
Progress graphs:
 Your progress graph for this problem

any tips on mastering these edge cases so that i can be right 100% all the time
0
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LVL 35

Accepted Solution

by:
Terry Woods earned 1000 total points
ID: 41743384
This is how I would write it to demonstrate cover of the missing case, though it's not necessarily the most efficient way to code it:

public boolean has77(int[] nums) {
  boolean res=false;
  
  for(int i=0;i<nums.length-2;i++){
    
    if((nums[i]==7&&nums[i+1]==7)||(nums[i+1]==7&&nums[i+2]==7)||(nums[i]==7&&nums[i+2]==7)){
      res=true;
      return res;
    }
  }
  return res;
}

Open in new window


The middle test is the one I've added.
0
 
LVL 32

Assisted Solution

by:awking00
awking00 earned 1000 total points
ID: 41744749
To possibly avoid the index out of bounds problem, you might consider starting at index 2 and searching backwards.
0
 
LVL 7

Author Comment

by:gudii9
ID: 41745115
public boolean has77(int[] nums) {
		boolean res = false;

		for (int i = 0; i < nums.length - 2; i++) {

			if ((nums[i] == 7 && nums[i + 1] == 7) || (nums[i] == 7 && nums[i + 2] == 7)
					|| ((nums[i + 1]) == 7 && (nums[i + 2] == 7))) {

				res = true;
				return res;
			}
		}
		return res;
	}

Open in new window



above passed all tests. one lesson i should take to myself is analyzing failing tests in this case {1,7,7} so i have to handle i+1 and i+2 case as well as above
0
 
LVL 7

Author Comment

by:gudii9
ID: 41745116
To possibly avoid the index out of bounds problem, you might consider starting at index 2 and searching backwards.


what do you mean by searching backwards?
can you please elaborate on that?
0
 
LVL 32

Expert Comment

by:awking00
ID: 41747231
>>what do you mean by searching backwards?
 can you please elaborate on that?
<<
You wrote -
>>for(int i=0;i<nums.length-1;i++){
    if((nums[i]==7&&nums[i+1]==7)||(nums[i]==7&&nums[i+2]==7)){
      res=true;<<
which caused ([2, 7, 2, 2, 7, 2]) to fail with indexOutOfBounds exception
because i is going from 0 to 5 so when it looks for a value at i + 2 when i = 4 or i + 1 when i = 5, there is no such index for that array.
However, if you increment i from 2 to 5 and first search for index 2 == 7 and (index 2-1 == 7 or index 2-2 == 7) up to index 5 == 7 and (index 5-1 == 7 or index 5-2 == 7), you will always be evaluating a valid index
0
 
LVL 32

Expert Comment

by:awking00
ID: 41747234
Needed to add another condition - where (index - 2 == 7 and index - 1 == 7)
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