Returns the Count of Distinct Elements from an Sorted Array of Numbers,
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|public int uniqueCountOfElements(int nums)|
|if (nums == null) return 0;|
|int arrSize = nums.length;|
|if (arrSize == 0) return 0;|
|int finalCount = 1; // Count the first element by default|
|for(int count = 0; count < arrSize; count++)|
|if (count != arrSize - 1 && nums[count] != nums[count+1]) finalCount++;|
Apr 12, 2021
Thank You @meekg33k for setting this up. Very well appreciated.
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Hello @deedee47, thank you for participating in Week 1 of Algorithm Fridays and congratulations, you are the winner of the $20 award
Your solution was selected because it is most memory-optimal and it is robust - takes care of edge cases, especially the edge case where the input array has a
nullvalue. You were also able to interpret the problem as returning the number of unique elements in an array thus avoiding unnecessary work in your code.
I have made a blog post here about the different solutions and about you as our winner.
We will contact you in less than 24 hours for your award.
Congratulations once again!