Two Pointers were applied in this challenge, the first pointer[j] which is used to iterate through the given array of numbs and second pointer [i] which is used in placing a unique value to the position after it.
Procedure
Return 0, if nums.length = 0; which means there will be no elements in the array.
Start j with the value of 1.
Using the for loop, set i = 0; i < nums.length - 1 (we subtract 1 because once we get to the last value of array, we will not have any value to compare with).
On each iteration, there is comparison between the current value of nums[i] with the next value of nums[i + 1].
Using the if statement, compare[i] with [i + 1], if they are not equal, set nums of [j] equals to nums of [i + 1] and increment j.
Return j.
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