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Use a read pointer to scan the sorted list and a write pointer to build the unique prefix in place. Return the write pointer as k: the first k elements hold one copy of each value, in sorted order. The remaining tail is not part of the result and does not need to be removed.
In-place solution for keeping one copy of each value
This implementation follows the contract for LeetCode 26: mutate the input list, keep one occurrence of each value, and return the valid prefix length.
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 write
For example, given [1, 1, 2, 2, 3], the function returns 3. The first three positions contain [1, 2, 3]; the contents after that prefix can be ignored.
How the read and write pointers work
read scans every input value
The loop starts at index 1 because the first value is already the first unique value. Since the input is sorted in non-decreasing order, duplicates appear next to one another. Each candidate can therefore be compared with the last value retained.
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write marks the next free position
When nums[read] differs from nums[write - 1], it is a new value. The function copies it to nums[write] and advances write. If the values match, it skips the duplicate. At the end, write is the number of unique values and the length of the valid prefix.
Understand the prefix-length contract
The problem specification says, “The first k elements of nums should contain the unique numbers in sorted order.” In other words, k tells the caller how much of the mutated list is valid; the task does not require Python to physically shorten the list.
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If your own code needs an actually shorter list, delete the tail after calling the function:
k = remove_duplicates(nums)
del nums[k:]
This resizing is an extra step for callers that need it, not part of the prefix-based return contract.
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Edge cases and complexity
- Empty list: returns
0. This is a useful extension for Python callers, although the reference problem specifies nonempty inputs. - One value: returns
1. - All values equal: returns
1. - Already unique: returns the original length.
The scan takes O(n) time and uses O(1) auxiliary space, assuming a mutable list with ordinary indexed access. The input must already be sorted; otherwise, equal values may not be adjacent and this comparison will not remove every duplicate.
Alternative when you want a new list
Python’s itertools.groupby groups consecutive elements with the same key; the Functional Programming HOWTO notes that it assumes input is already grouped or sorted by that key. It can construct a separate list of unique values:
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from itertools import groupby
unique = [key for key, _ in groupby(nums)]
This is concise when preserving the original list matters, but it allocates a new output list rather than rewriting the valid prefix in place.
Independent reader supportYour contribution helps us test, update, and keep practical guides available for everyone.Do not confuse the one-copy task with the at-most-two variation
LeetCode 80 is a related but different problem: it keeps each value at most twice. For that variation, keep an item when fewer than two items have been written, or when it differs from the value two positions behind the write pointer:
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def keep_at_most_two(nums):
write = 0
for value in nums:
if write < 2 or value != nums[write - 2]:
nums[write] = value
write += 1
return write
Use the first implementation when the requirement is exactly one copy of each value; changing the retention rule changes the answer.
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