Remove Duplicates from Sorted Array

Problem Statement:

Given an integer array nums sorted in non-decreasing order, remove the duplicates in-place such that each unique element appears only once. The relative order of the elements should be kept the same. Then return the number of unique elements in nums.

Consider the number of unique elements of nums to be k, to get accepted, you need to do the following things:

• Change the array nums such that the first k elements of nums contain the unique elements in the order they were present in nums initially. The remaining elements of nums are not important as well as the size of nums.

• Return k.

• Example:

  Example 1:
  
  Input: nums = [1,1,2]
  Output: 2, nums = [1,2,_]
  Explanation: Your function should return k = 2, with the first two elements of nums being 1 and 2 respectively.
  It does not matter what you leave beyond the returned k (hence they are underscores).
  Example 2:
  
  Input: nums = [0,0,1,1,1,2,2,3,3,4]
  Output: 5, nums = [0,1,2,3,4,_,_,_,_,_]
  Explanation: Your function should return k = 5, with the first five elements of nums being 0, 1, 2, 3, and 4 respectively.
  It does not matter what you leave beyond the returned k (hence they are underscores).

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✅ Solution: Two Pointer Technique (In-Place Deduplication)

🔁 Return Only the Count of Unique Elements

class Solution {
public:
    int removeDuplicates(vector<int>& nums) {
        if (nums.empty()) return 0;  // Edge case: empty array

        int uniquePos = 1; // Position to place next unique element

        for (int scan = 1; scan < nums.size(); scan++) {
            // If current number is different from previous
            if (nums[scan] != nums[scan - 1]) {
                nums[uniquePos] = nums[scan]; // Overwrite duplicate
                uniquePos++;                  // Move to next insert position
            }
        }

        return uniquePos; // Number of unique elements
    }
};

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🔁 Return Vector Containing Only Unique Elements

class Solution {
public:
    vector<int> removeDuplicates(vector<int>& nums) {
        if (nums.empty()) return {}; // Edge case: empty array

        int uniquePos = 1; // Position to insert next unique element

        for (int scan = 1; scan < nums.size(); scan++) {
            if (nums[scan] != nums[scan - 1]) {
                nums[uniquePos] = nums[scan];
                uniquePos++;
            }
        }

        // Return only the unique portion of the array
        return vector<int>(nums.begin(), nums.begin() + uniquePos);
    }
};

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📝 How It Works

• Traverse the sorted array with a scanning pointer.
• Use another pointer uniquePos to track where the next unique number should go.
• Overwrite duplicates in-place to maintain space efficiency.
• Either return:
  • The count of unique values (for in-place APIs like LeetCode), or
  • The sliced result vector (nums[0:uniquePos]).

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🧩 Key Formula / Logic

if (nums[scan] != nums[scan - 1]) {
    nums[uniquePos] = nums[scan];
    uniquePos++;
}

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⏱️ Time & Space Complexity

Complexity	Value
Time	O(N)
Space	O(1) (in-place) / O(K) for returned vector

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⚠️ Edge Cases

• Empty input → returns 0 or empty vector.
• Fully unique array → returns as-is.
• All same elements → returns one element.

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💡 Other Approaches

Approach	Time	Space	Notes
set or unordered_set	O(N log N) or O(N)	O(N)	Not in-place, unordered
Manual new vector build	O(N)	O(N)	More readable, but uses space

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🔁 Related Problems

• Remove Duplicates from Sorted Array II
• Delete Duplicates in Linked List
• Remove Element
• Move Zeroes

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