Understanding the Problem: The problem "Remove Element" asks us to remove all instances of a given value from an array in-place and return the new length of the array.

What We Need to Do:

1. Remove all occurrences of the given value from the array.
2. Return the new length of the array after removals.

Let's break down the solution :

class Solution {
public:
    int removeElement(vector<int>& nums, int val) {
        for(int i = 0; i < nums.size(); i++){
            if(nums[i] == val){
                nums.erase(nums.begin() + i);
                i--;
            }
        }
        return nums.size();
    }
};

The Key Lines Explained:

1. nums.erase(nums.begin() + i);
   • This line removes the element at index i from the vector.
   • nums.begin() points to the start of the vector.
   • nums.begin() + i points to the i-th element.
   • erase() removes the element at this position and shifts all subsequent elements to the left.
2. i--;
   • This line is crucial and easy to overlook.
   • After erasing an element, all elements to the right shift left by one position.
   • If we don't decrement i, we would skip the next element that just moved into the current position.
   • By decrementing i, we ensure that in the next iteration, we check the element that has shifted into the current position.

Why These Lines Are Important:

1. In-Place Removal: The erase() function modifies the vector in-place, which is a requirement of the problem.
2. No Element Skipping: The i-- ensures we don't accidentally skip any elements after a removal.

Let's explore a better approach using the two-pointer technique. This method achieves O(n) time complexity and is generally preferred for its efficiency.

USING TWO POINTER ! 

class Solution {
public:
    int removeElement(vector<int>& nums, int val) {
        int k = 0; // k is the write pointer
        
        for (int i = 0; i < nums.size(); i++) {
            if (nums[i] != val) {
                nums[k] = nums[i];
                k++;
            }
        }
        return k;
    }
};

Let's break down this more efficient solution:

1. We use two pointers:
   • i: the read pointer, which scans through the entire array.
   • k: the write pointer, which keeps track of where to place the next non-val element.
2. As we iterate through the array:
   • If the current element (nums[i]) is not equal to val, we copy it to the position indicated by k and then increment k.
   • If the current element is equal to val, we simply move on to the next element without incrementing k.
3. At the end, k represents both:
   • The number of elements not equal to val (i.e., the new length of the array).
   • The position where any remaining val elements begin in the array.

Key advantages of this approach:

1. Time Complexity: O(n), where n is the length of the array. We only pass through the array once.
2. Space Complexity: O(1), as we modify the array in-place without using extra space.
3. No need for expensive erase operations: We're simply overwriting elements instead of removing them and shifting the rest.
4. Maintains relative order of elements: The order of non-val elements remains the same.

Example walkthrough: Let's say nums = [3, 2, 2, 3] and val = 3:

1. i = 0, k = 0: nums[0] == 3, skip
2. i = 1, k = 0: nums[1] == 2, copy to nums[k], increment k
3. i = 2, k = 1: nums[2] == 2, copy to nums[k], increment k
4. i = 3, k = 2: nums[3] == 3, skip
5. Final array: [2, 2, 2, 3], return k = 2

The first k elements of the array are now the result, with all instances of val effectively removed.