Coding interview (NeetCode)

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  1. LeetCode
    1. Algorithms and Data Structures for Beginners
      1. Lesson 2 - LeetCode Questions

LeetCode

Algorithms and Data Structures for Beginners

Lesson 2 - LeetCode Questions

  1. Easy - Remove Duplicates From Sorted Array <a href=https://leetcode.com/problems/remove-duplicates-from-sorted-array/description>(Q.26) </a>
    • original input: [0,0,1,1,1,2,2,3,3,4]
    • excepted output: [0,1,2,3,4]

Goal: Return k : int, the idx that will slice the original input up to a certain output

#Initally my list is : 
[0, 0, 1, 1, 1, 2, 2, 3, 3, 4]
[0, 1, 1, 1, 1, 2, 2, 3, 3, 4]
[0, 1, 2, 1, 1, 2, 2, 3, 3, 4]
[0, 1, 2, 3, 1, 2, 2, 3, 3, 4]
[0, 1, 2, 3, 4, 2, 2, 3, 3, 4]

My terminating idx is : 5
[0, 1, 2, 3, 4, 2, 2, 3, 3, 4]

Note: This algo will fail if the list is sorted non-increasingly

class Solution:
    def removeDuplicates(self, nums: List[int]) -> int:
        
        # Initialize two pointers.
        ptr1 = 0  # Slow pointer
        for ptr2 in range(1, len(nums)):  # Fast pointer
            if nums[ptr2-1] != nums[ptr2]: # Increment i only if a new unique element is found.
            # else keep looping
                ptr1 += 1
                # Move the unique element next to the last unique element found.
                nums[ptr1] = nums[ptr2]
        
        # Return the length of the array consisting of unique elements only.
        # Since i is an index, add 1 to represent the count.
        return ptr1 + 1
  1. Easy - Remove elements From Array given val<a href=https://leetcode.com/problems/remove-element>(Q.27) </a>
    • original input: [0,1,2,2,3,0,4,2]
    • val: 2
    • excepted output: [0,1,3,0,4]

Goal: Return k : int, the idx that will slice the original input up to a certain output

[0, 1, 2, 2, 3, 0, 4, 2] # Initally
[0, 1, 2, 2, 3, 0, 4, 2] # 
[0, 1, 2, 2, 3, 0, 4, 2]
[0, 1, 3, 2, 3, 0, 4, 2]
[0, 1, 3, 0, 3, 0, 4, 2]
[0, 1, 3, 0, 4, 0, 4, 2]
Return solution: 5

class Solution:
    def removeElement(self, nums: List[int], val: int) -> int:
        L_idx = 0
        for R_idx in range(len(nums)):
            if nums[R_idx] != val:
                nums[L_idx] = nums[R_idx]
                L_idx += 1
        return L_idx

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