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Returns new count of elements in a Numbers array excluding occurrences of a specified value
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| public int countOfElementsExcludingVal(int[] nums, int val){ | |
| if(nums == null) return 0; | |
| int countWithoutVal = nums.length; | |
| for(int item : nums){ | |
| if (item == val) countWithoutVal--; | |
| } | |
| return countWithoutVal; | |
| } |
Hi @meekg33k, thank you for your comment.
I have gone through your solution, really insightful with the double-pointer approach. Thank you
No worries, glad you found it insightful!
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Hello @deedee47, thank you for participating in Week 2 of Algorithm Fridays.
This is a really decent solution; I like that you included edge case checks and you interpreted the problem differently to come up with a memory-optimal solution. Really neat!
Your solution focuses on looping through the array to check elements with values not equal to
val. Do you think there's a faster way we could loop through the array?I've posted my solution here. Do let me know what you think.