How do you solve Remove Duplicates from Sorted Array II on LeetCode?

Remove Duplicates from Sorted Array II (LeetCode #80) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Use two pointers to efficiently manage the position of valid elements.

What is the brute force solution for Remove Duplicates from Sorted Array II?

The brute force approach for Remove Duplicates from Sorted Array II is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves iterating through the array and counting occurrences of each element. If an element appears more than twice, we can skip adding it to the result. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Remove Duplicates from Sorted Array II?

Remove Duplicates from Sorted Array II uses the following concepts: Array, Two Pointers. The recommended patterns to study are: Two Pointers, Array.

What are the interview tips for Remove Duplicates from Sorted Array II?

Explain your thought process clearly while coding. Consider edge cases before diving into the implementation. Practice coding the two-pointer technique to become comfortable with it.

What are common mistakes when solving Remove Duplicates from Sorted Array II?

Not handling edge cases like empty arrays. Confusing the index of the last valid element with the current element being processed.

#80

Remove Duplicates from Sorted Array II

Medium

ArrayTwo PointersTwo PointersArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

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Intuition

Time O(n)Space O(1)

The optimal solution uses a two-pointer technique to keep track of where to place the next valid number while iterating through the array. This avoids the need for extra space and reduces time complexity.

⚙️

Algorithm

5 steps

1Step 1: Initialize a pointer 'i' to track the position of the last valid element.
2Step 2: Iterate through the array with a second pointer 'j'.
3Step 3: For each element, check if it can be added (i.e., it appears at most twice).
4Step 4: If valid, increment 'i' and place the element at 'i'.
5Step 5: Return 'i + 1' as the new length of the modified array.

solution.py11 lines

1# Full working Python code
2
3def removeDuplicates(nums):
4    if not nums:
5        return 0
6    i = 0
7    for j in range(1, len(nums)):
8        if nums[j] != nums[i] or (i > 0 and nums[j] != nums[i - 1]):
9            i += 1
10            nums[i] = nums[j]
11    return i + 1

ℹ

Complexity note: The time complexity is O(n) because we only traverse the array once. The space complexity is O(1) since we are modifying the array in place without using any additional data structures.

1Use two pointers to efficiently manage the position of valid elements.
2Keep track of the last valid element to ensure duplicates are limited.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.