Algorithm Fridays Week 1

algoFridaysWeek1.js

// Given a sorted array nums, write a function to remove the 
// duplicates from the array such that each element appears 
// only once. Your function should return the new length

const removeDuplicateArrayElements = (nums) => {
    const noDuplicatesArray = [];
   
    for(let i = 0; i < nums.length; i++){
        //compare adjacent elements of the array and check if they
        //are the same since the array is sorted already, skip if true.
        if(nums[i] === nums[i + 1]) continue;
        noDuplicatesArray.push(nums[i]);
    }

    return noDuplicatesArray.length;
}

console.log(removeDuplicateArrayElements([0,0,1,1,2,2,3,3,4,4,4,5,5,5,6])
);  //7

Nice Amaka. This is actually very good. However I feel your space complexity of O(n) can be reduced to O(1). What if rather than pushing into a new array, you define a counter and increment when nums[i] != nums[i+1]? Let me know what your think.

Yes I didn't think of that😫😫. Thank you so much, since the deadline has elapsed I will apply the feedback to subsequent problems. I am already excited about this program Algorithm Fridays as I hope to improve on my problem solving skills.💃🏻

Lol I like your style. Good

Thank you so much🙏🏻

Hello @Amaka202, thank you for participating in Week 1 of Algorithm Fridays.

I really like the logic you implemented by comparing adjacent elements in the array, awesome stuff! While your solution works and passes most of the test cases, it fails on one.

The assumption your solution makes is that the input array cannot have a null value because if I pass a null value to your function, it would break on line 8. Ideally, you always want to write code that is robust and takes care of edge cases and unexpected input. Basically adding a if (!nums) return 0 check would suffice.

Also, is there a way we could have solved this problem without having to create another array like you did on line 6? Because if you think about it, all you care about is the number of unique elements and doing that would reduce your space usage from O(N) to O(1). What do you think?

I have posted my solution here, do let me know what you think.

Hello @Amaka202, thank you for participating in Week 1 of Algorithm Fridays.

I really like the logic you implemented by comparing adjacent elements in the array, awesome stuff! While your solution works and passes most of the test cases, it fails on one.

The assumption your solution makes is that the input array cannot have a null value because if I pass a null value to your function, it would break on line 8. Ideally, you always want to write code that is robust and takes care of edge cases and unexpected input. Basically adding a if (!nums) return 0 check would suffice.

Also, is there a way we could have solved this problem without having to create another array like you did on line 6? Because if you think about it, all you care about is the number of unique elements and doing that would reduce your space usage from O(N) to O(1). What do you think?

I have posted my solution here, do let me know what you think.

Yup Thanks alot Uduak for the feedback, I could have definitely improved the code by not creating an extra array and using a counter instead. And also for the cases I overlooked, I have learned today to account for every possible case in my code. Thanks!

I will take out time to update the code for future reference.

Looking forward to this week's problem!💃🏻💃🏻

Nice Amaka. This is actually very good. However I feel your space complexity of O(n) can be reduced to O(1). What if rather than pushing into a new array, you define a counter and increment when nums[i] != nums[i+1]? Let me know what your think.

Yes I didn't think of that😫😫. Thank you so much, since the deadline has elapsed I will apply the feedback to subsequent problems. I am already excited about this program Algorithm Fridays as I hope to improve on my problem solving skills.💃🏻

Great stuff Amaka and Weldone!!