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# has22 challenge

Posted on 2016-08-01
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Hi,

I am working on below challenge

http://codingbat.com/prob/p121853
I wrote my code as below
``````public boolean has22(int[] nums) {

boolean result=false;

int sum=0;
for(int n:nums){
if(n==2)
result=true;

}
return result;

}
``````

I am not passing all tests

Expected      Run
has22([1, 2, 2]) → true      true      OK
has22([1, 2, 1, 2]) → false      true      X
has22([2, 1, 2]) → false      true      X
has22([2, 2, 1, 2]) → true      true      OK
has22([1, 3, 2]) → false      true      X
has22([1, 3, 2, 2]) → true      true      OK
has22([2, 3, 2, 2]) → true      true      OK
has22([4, 2, 4, 2, 2, 5]) → true      true      OK
has22([1, 2]) → false      true      X
has22([2, 2]) → true      true      OK
has22([2]) → false      true      X
has22([]) → false      false      OK
has22([3, 3, 2, 2]) → true      true      OK
has22([5, 2, 5, 2]) → false      true      X
other tests
X

0
Question by:gudii9
[X]
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LVL 14

Accepted Solution

CPColin earned 250 total points
ID: 41738045
You're supposed to return true only when there's more than one 2 in a row. Your code is returning true when there's a 2 anywhere.
0

LVL 16

Assisted Solution

krakatoa earned 250 total points
ID: 41738077
``````public boolean has22(int[] nums) {

for(int y=0;y<nums.length-1;y++){xxxxxxxxxxx){return true;}}
return false;
}
``````

... what do you insert instead of the xs??
0

LVL 7

Author Comment

ID: 41738189
let me check
0

LVL 16

Expert Comment

ID: 41738266
Is this snippet relevant or not in this question:

``````if(nums[y]==2&&nums[y+1]==2)
``````
0

LVL 7

Author Comment

ID: 41738385
``````public boolean has22(int[] nums) {

boolean result=false;

int sum=0;
int len=nums.length;
for(int i=0;i<len-1;i++){
if(nums[i]==2&&nums[i+1]==2)
result=true;

}
return result;

}
``````

above passed all tests? any improvements, suggestions, comments?
0

LVL 14

Expert Comment

ID: 41738425
You could return true as soon as you find two 2's next to each other and save a little processing time. (If you have an array that's a thousand elements long and the first two elements are 2's, you don't need to check the rest of the array, since you already know you'll be returning true.)
0

LVL 7

Author Comment

ID: 41738441
You could return true as soon as you find two 2's next to each other and save a little processing time.

how to do this?
0

LVL 14

Expert Comment

ID: 41738446
return true;
0

LVL 7

Author Comment

ID: 41738454
``````public boolean has22(int[] nums) {

// boolean result=false;

int sum=0;
int len=nums.length;
for(int i=0;i<len-1;i++){
if(nums[i]==2&&nums[i+1]==2)
return true;

}
// return result;
return false;

}
``````

i see . this is very powerful to use return here.
0

LVL 14

Expert Comment

ID: 41738455
Yep! You can't always break out of a loop early or return from a loop early, but it can be nice, when you can.
0

LVL 16

Expert Comment

ID: 41738788
i see . this is very powerful to use return here.

I ALREADY posted code with this return clause in it !

Did you even *look* at that ???

You ARE a troll, are you not ????
1

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