Remove Duplicates from Sorted Array solution

    

  def removeDuplicates(nums):
  if not nums:
      return 0
  i = 0
  for j in range(1, len(nums)):
      if nums[j] != nums[i]:
          i += 1
          nums[i] = nums[j]
  return i + 1

  nums = [1, 1, 2, 2, 2, 3, 4, 5, 5]
  print(removeDuplicates(nums))

Explanation (Python):

In Python, we can use two pointers, i and j, to keep track of the current and next distinct elements.

Initialize i to 0 since the first element is always unique.

Iterate through the array with j starting from index 1.

If nums[j] is different from nums[i], it means we have found a new distinct element. Increment i and update nums[i] with nums[j].

Continue the iteration until j reaches the end of the array.

Finally, return i + 1, which represents the length of the modified array with duplicates removed.

         
  class Solution {
    public int removeDuplicates(int[] nums) {
        if (nums.length == 0) {
            return 0;
        }
        int i = 0;
        for (int j = 1; j < nums.length; j++) {
            if (nums[j] != nums[i]) {
                i++;
                nums[i] = nums[j];
            }
        }
        return i + 1;
    }

    public static void main(String[] args) {
        int[] nums = {1, 1, 2, 2, 2, 3, 4, 5, 5};
        Solution solution = new Solution();
        System.out.println(solution.removeDuplicates(nums));
    }
}

  

Explanation (Java):

In Java, we use the same two-pointer approach as in Python.

Check if the length of nums is 0. If so, return 0 since there are no elements to remove.

Initialize i to 0.

Iterate through the array with j starting from index 1.

If nums[j] is different from nums[i], increment i and update nums[i] with nums[j].

Continue the iteration until j reaches the end of the array.

Finally, return i + 1, which represents the length of the modified array with duplicates removed.

         
    #include <iostream>
    #include <vector>

    using namespace std;
    
    int removeDuplicates(vector& nums) {
        if (nums.empty()) {
            return 0;
        }
        int i = 0;
        for (int j = 1; j < nums.size(); j++) {
            if (nums[j] != nums[i]) {
                i++;
                nums[i] = nums[j];
            }
        }
        return i + 1;
    }
    
    int main() {
        vector nums = {1, 1, 2, 2, 2, 3, 4, 5, 5};
        cout << removeDuplicates(nums) << endl;
        return 0;
    }
    
    
        
        

Explanation (C++):

In C++, we also follow the same two-pointer approach as in Python and Java.

Check if the nums vector is empty. If so, return 0 since there are no elements to remove.

Initialize i to 0.

Iterate through the vector with j starting from index 1.

If nums[j] is different from nums[i], increment i and update nums[i] with nums[j].

Continue the iteration until j reaches the end of the vector.

Finally, return i + 1, which represents the length of the modified vector with duplicates removed.

I hope this explanation helps! Let me know if you have any further questions.

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