LeetCode 26: Remove Duplicates from Sorted Array — Python Solution

LeetCode 26: Remove Duplicates from Sorted Array is an Easy array / string problem. This Python walkthrough develops a sorted write pointer solution, ties every code decision to a concrete invariant, and includes the regression check used before publication.

This independent guide paraphrases the task rather than reproducing LeetCode’s prompt. Use the official page for the exact statement, examples, constraints, and submission runner.

Recognizing the pattern

Array and string questions usually reward a precise index invariant. Decide which prefix or suffix is already final before mutating the next position.

For this problem specifically, keep one write position for the next distinct value. Because the input is sorted, comparing with the last written value is sufficient. The invariant worth writing beside the code is: The written prefix contains one copy of every distinct value in sorted order.

Step-by-step algorithm

1. Identify the input state consumed by removeDuplicates(nums) and initialize the data required by the sorted write pointer pattern.
2. Keep one write position for the next distinct value. Because the input is sorted, comparing with the last written value is sufficient.
3. After each update, verify the page’s central invariant: The written prefix contains one copy of every distinct value in sorted order.
4. Finish only after the boundary behavior is covered: Handle an empty array and arrays containing only one repeated value.

Python solution

class Solution:
    def removeDuplicates(self, nums):
        if not nums:
            return 0
        write = 1
        for read in range(1, len(nums)):
            if nums[read] != nums[write - 1]:
                nums[write] = nums[read]
                write += 1
        return write

Reading the implementation

The main entry point is removeDuplicates(nums). The named working state includes write; those variables make the sorted write pointer state visible instead of hiding it in incidental control flow.

A single main loop advances the algorithm, which is the key reason the traversal does not revisit already resolved input unnecessarily. Early returns stop as soon as the answer is forced, avoiding work that cannot change the result. In concrete terms, keep one write position for the next distinct value. Because the input is sorted, comparing with the last written value is sufficient.

Correctness argument

Initialization. The data structure starts with exactly the information known before any input element is processed.

Preservation. Keep one write position for the next distinct value. Because the input is sorted, comparing with the last written value is sufficient. Each update records the current item without invalidating earlier decisions; consequently, the written prefix contains one copy of every distinct value in sorted order.

Termination. The traversal consumes a finite input or finite state space. At the end, the invariant covers the complete input, which is precisely the condition required for the returned result.

Complexity and trade-offs

O(n) time and O(1) extra space. The auxiliary-space figure excludes the returned output unless the output itself is the structure being built.

A copied output buffer can simplify reasoning, but the in-place version reduces auxiliary memory when mutation is allowed. That comparison is useful in an interview because it explains why the final implementation is preferable, not merely that it passes.

Regression check

The published implementation belongs to a 100-problem suite that is compiled and exercised behaviorally before deployment.

One reference assertion from that suite is shown below. It targets the normal path while the edge conditions in the next section cover the failure-prone boundaries.

nums=[1,1,2]; k=Solution().removeDuplicates(nums); assert nums[:k] == [1,2]

Common mistakes and edge cases

• Problem-specific boundary: Handle an empty array and arrays containing only one repeated value.
• Pattern-level pitfall: Do not let a write operation destroy input that a later read still needs; write direction and boundary conventions matter.
• Invariant check: after every update, confirm that the written prefix contains one copy of every distinct value in sorted order.

Interview review checklist

• Explain why sorted write pointer matches the structure of this input.
• State the invariant in one sentence before tracing code: The written prefix contains one copy of every distinct value in sorted order.
• Derive O(n) time and O(1) extra space from how many times each element or state is visited.
• Test the boundary explicitly: Handle an empty array and arrays containing only one repeated value.

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