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Repair Windows errors before they cause bigger problemsFix Now →Scan for outdated or missing drivers - takes under a minuteDriver Scan →Clear out junk files and repair common Windows errorsFree Scan →Use a read pointer to scan the sorted list and a write pointer to build the unique-value prefix. The function below keeps one copy of each value in place and returns the length k of the valid prefix; it does not physically shorten the list.
In-place solution for one copy of each value
This is the contract in LeetCode problem 26: the input is sorted in non-decreasing order, and the first k positions must contain the unique values in sorted order. The remaining positions can be ignored.
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def remove_duplicates(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 writeFor example, given [1, 1, 2, 2, 3], the function returns 3, and the first three positions contain [1, 2, 3]. The list may still have five positions; only the prefix of length k is part of the result.
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Repair common Windows errors and clear accumulated junk for a smoother, more stable PC - no reinstall needed.Free scan · no reinstallHow the read and write pointers work
readvisits each input position from left to right.writemarks the next position where a new, unique value should go.- Because equal values are adjacent in a sorted list, comparing the current value with
nums[write - 1]is enough to detect whether it has already been retained. - When the values differ, the current value is copied to
nums[write], andwriteadvances.
At the end, write is the number of retained values. The algorithm scans the list once, taking O(n) time and O(1) auxiliary space for an ordinary mutable Python list.
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What the return value means
The returned value k is the length of the answer, not a new list. The problem specification says, “The first k elements of nums should contain the unique numbers in sorted order.” The tail after that prefix is unspecified and should not be treated as part of the result.
If your own code needs a physically shorter list, delete the unused tail after calling the function:
k = remove_duplicates(nums)
del nums[k:]That deletion is an additional choice for your Python API; it is not required by the in-place prefix contract.
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Edge cases
- An empty list returns
0. This is a useful extension for Python callers, even though the reference problem lists nonempty inputs. - A singleton list returns
1. - An all-equal list returns
1. - An already-unique list returns its original length.
When to use itertools.groupby instead
itertools.groupby groups consecutive elements with equal keys and assumes the input is already sorted on that key, as described in the Python Functional Programming HOWTO. It can produce a new list of unique values concisely:
from itertools import groupby
unique = [key for key, _ in groupby(nums)]This creates a separate output list rather than rewriting the input prefix. Use it when a new collection is what you need; use the pointer method when the required result is an in-place prefix and a returned length.
Independent reader supportYour contribution helps us test, update, and keep practical guides available for everyone.Do not confuse this with the at-most-two variation
LeetCode problem 80 asks for a different result: retain each value at most twice. Its keep condition compares a candidate with the value two positions behind the write pointer, once two values have already been retained. That is not the condition for problem 26, which keeps exactly one copy of each value.
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