INTUITION:

• Since the array is sorted, duplicates are next to each other. We use two pointers—one (i) to track the last unique element and another (j) to find new unique ones.
• When a new unique element is found, we move it next to i. This way, duplicates get overwritten while keeping the order intact.

APPROACH:

• Initialize two pointers: i at index 0 and j at index 1.
• Traverse the array with j, comparing nums[i] with nums[j].
• If nums[j] is different from nums[i], increment i and update nums[i] with nums[j].
• After the loop, the number of unique elements is i + 1.

Here’s the code:

class Solution {
public:
    int removeDuplicates(vector<int>& nums) {
     int n=nums.size();
     int i=0;
     for(int j=1;j<n;j++)
     {
         if(nums[i]!=nums[j]){
         i++;
         nums[i]=nums[j];
         }
     }
     return i+1;
    }
};

KEYPOINTS:

• Since the array is sorted, all duplicates will be adjacent.
• Use two pointers (i and j) to traverse the array:
  • i tracks the position of the last unique element.
  • j iterates through the array to find the next unique element.

Feel free to share your approach and feedback in comment box.