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countClumps  challenge

Posted on 2016-08-14
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Last Modified: 2016-08-17
Hi,

I am working on below challenge

http://codingbat.com/prob/p193817


Say that a "clump" in an array is a series of 2 or more adjacent elements of the same value. Return the number of clumps in the given array.

countClumps([1, 2, 2, 3, 4, 4]) → 2
countClumps([1, 1, 2, 1, 1]) → 2
countClumps([1, 1, 1, 1, 1]) → 1

i am not clear on the description of the challenge. please advise
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Question by:gudii9
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Expert Comment

by:Gerwin Jansen, EE MVE
ID: 41756126
gudii9 - where are you from, what is your native language?
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Accepted Solution

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Gerwin Jansen, EE MVE earned 500 total points
ID: 41756128
You have to count blocks of the same numbers.

1 1 1 2 3 3 4 5 contains 2 blocks (numbers 1 and numbers 3)
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by:gudii9
ID: 41756451
countClumps([1, 2, 2, 3, 4, 4]) → 2//this is clear as it has one 2's clump and one other 4's clump
countClumps([1, 1, 2, 1, 1]) → 2//this is clear
countClumps([1, 1, 1, 1, 1]) → 1//how it is 1?
is it because it has one clump of 5 1's in it?
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LVL 38

Expert Comment

by:Gerwin Jansen, EE MVE
ID: 41756540
Yes, one clump of 1's
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Author Comment

by:gudii9
ID: 41758651
public int countClumps(int[] nums) {
  int counter=0;
  boolean flag=false;
  for(int i=0;i<nums.length-1;i++){
    if(nums[i]==nums[i+1]&&!flag){
      flag=true;
      counter++;
    }
    else if(nums[i]!=nums[i+1]){
      flag=false;
    }
  }
  return counter;
}

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above passes all tests.
basically checking i with i+1  and making flag to true and repeat in if loop then in else check for inequality and make flag false there
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Author Comment

by:gudii9
ID: 41759619
just thinking any other possible simple solutions/approaches for this? or any improvements, modifications, optimization to my code above?
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Expert Comment

by:Gerwin Jansen, EE MVE
ID: 41759689
Can't think of an easier way, if you want a different way then you have to write different pseudo code first. But it works and passes the tests so you've learned some here right?
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Author Comment

by:gudii9
ID: 41760276
yes definitely
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