Given an integer array `nums`

and an integer `val`

, remove all occurrences of `val`

in `nums`

**in-place**. The relative order of the elements may be changed.

Since it is impossible to change the length of the array in some languages, you must instead have the result be placed in the **first part** of the array `nums`

. More formally, if there are `k`

elements after removing the duplicates, then the first `k`

elements of `nums`

should hold the final result. It does not matter what you leave beyond the first `k`

elements.

Return `k`

* after placing the final result in the first *`k`

* slots of *`nums`

.

Do **not** allocate extra space for another array. You must do this by **modifying the input array in-place** with O(1) extra memory.

**Custom Judge:**

The judge will test your solution with the following code:

int[] nums = [...]; // Input array int val = ...; // Value to remove int[] expectedNums = [...]; // The expected answer with correct length. // It is sorted with no values equaling val. int k = removeElement(nums, val); // Calls your implementation assert k == expectedNums.length; sort(nums, 0, k); // Sort the first k elements of nums for (int i = 0; i < actualLength; i++) { assert nums[i] == expectedNums[i]; }

If all assertions pass, then your solution will be **accepted**.

**Example 1:**

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

**Example 2:**

Input:nums = [0,1,2,2,3,0,4,2], val = 2Output:5, nums = [0,1,4,0,3,_,_,_]Explanation:Your function should return k = 5, with the first five elements of nums containing 0, 0, 1, 3, and 4. Note that the five elements can be returned in any order. It does not matter what you leave beyond the returned k (hence they are underscores).

**Constraints:**

`0 <= nums.length <= 100`

`0 <= nums[i] <= 50`

`0 <= val <= 100`

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

If you know the concept of two pointers, and how you can keep track of the position where you wish to replace the element with all the occurrences of a different numbers other than the given value, then it is an easy question.

A detailed analysis is provided in the Video.

Time Complexity: O(n)

Space Complexity: O(1)