Problem statement

Given an integer array nums sorted in non-decreasing order, remove the duplicates so each unique element appears once, keeping the order. Return the resulting array.

Example 1

Input: nums = [1,1,2]
Output: [1,2]

Example 2

Input: nums = [0,0,1,1,1,2,2,3,3,4]
Output: [0,1,2,3,4]

Constraints

• 1 <= nums.length <= 30
• -100 <= nums[i] <= 100
• nums is sorted in non-decreasing order

Hints for Remove Duplicates from Sorted Array

1. Hint 1

   Because the array is sorted, duplicates sit next to each other.

2. Hint 2

   Compare each element with the previous one — keep it only when it differs.

How to solve Remove Duplicates from Sorted Array

Because the array is sorted, equal values are adjacent — so a value is a duplicate exactly when it equals the one just kept. Use a write pointer for the deduplicated prefix and only advance it when a genuinely new value appears.

Approach

1. Keep the first element; set write = 1.
2. Scan from the second element onwards.
3. If the current value differs from nums[write - 1] — the last value kept — write it at nums[write] and advance.
4. Return the first write elements.

Why it works

Sortedness turns "have I seen this value?" into a single comparison against the previous kept value: any earlier occurrence of the same value must be adjacent to it, so nothing further back needs checking. That is why no hash set is needed and the extra space stays constant. The write pointer trails the read pointer, so values are never overwritten before being examined.

Complexity

• Time — O(n)
• Space — O(1)

Pitfalls

• Compare against the last kept element (nums[write - 1]), not the previous element in the original array.
• Seed write = 1, since the first element is always kept.
• This relies on the input being sorted; on unsorted data it silently keeps duplicates.
• A single-element array must be returned unchanged, which the seeding covers.

Reference solution

Python

from typing import List

def removeDuplicates(nums: List[int]) -> List[int]:
    out = []
    for x in nums:
        if not out or out[-1] != x:
            out.append(x)
    return out

Java

public static int[] removeDuplicates(int[] nums) {
    int write = 1;
    for (int i = 1; i < nums.length; i++) {
        if (nums[i] != nums[write - 1]) {
            nums[write] = nums[i];
            write++;
        }
    }
    return Arrays.copyOf(nums, write);
}

C++

vector<int> removeDuplicates(vector<int>& nums) {
    int write = 1;
    for (int i = 1; i < (int) nums.size(); i++) {
        if (nums[i] != nums[write - 1]) {
            nums[write] = nums[i];
            write++;
        }
    }
    return vector<int>(nums.begin(), nums.begin() + write);
}

JavaScript

var removeDuplicates = function(nums) {
    const out = [];
    for (const x of nums) {
        if (out.length === 0 || out[out.length - 1] !== x) out.push(x);
    }
    return out;
};

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