<gh_stars>10-100
'''

Description:

Given an array nums and a value val, remove all instances of that value in-place and return the new length.

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

The order of elements can be changed. It doesn't matter what you leave beyond the new length.

Example 1:

Given nums = [3,2,2,3], val = 3,

Your function should return length = 2, with the first two elements of nums being 2.

It doesn't matter what you leave beyond the returned length.



Example 2:

Given nums = [0,1,2,2,3,0,4,2], val = 2,

Your function should return length = 5, with the first five elements of nums containing 0, 1, 3, 0, and 4.

Note that the order of those five elements can be arbitrary.

It doesn't matter what values are set beyond the returned length.
Clarification:

Confused why the returned value is an integer but your answer is an array?

Note that the input array is passed in by reference, which means modification to the input array will be known to the caller as well.

Internally you can think of this:

// nums is passed in by reference. (i.e., without making a copy)
int len = removeElement(nums, val);

// any modification to nums in your function would be known by the caller.
// using the length returned by your function, it prints the first len elements.
for (int i = 0; i < len; i++) {
    print(nums[i]);
}

'''

from typing import List

class Solution:
    def removeElement(self, nums: List[int], val: int) -> int:
        
        size = len(nums)
        
        # two-pointers
        # others_idx: index for other elements
        # target_idx: index for target elements with val
        others_idx, target_idx = 0, size-1
        
        while others_idx <= target_idx:
            
            if nums[others_idx] == val:
                
                # swap target elements to the tail side
                nums[others_idx], nums[target_idx] = nums[target_idx], nums[others_idx]
                
                target_idx -= 1
                
            else:
                
                # increase the index when we meet others
                others_idx += 1
            
        
        # length of others is the answer
        return others_idx


# n : the length of input nums

## Time Complexity: O( n )
#
# The overhead in time is the cost of iteration, which is of O( n )

## Space Complexity: O( 1 )
#
# The overhead in space is the storage for two-pointers, which is of O( 1 )


import unittest

class Testing( unittest.TestCase ):

    def test_case_1( self ):

        result = Solution().removeElement( nums = [3,2,2,3], val = 3 )
        self.assertEqual(result, 2)

    
    def test_case_2( self ):

        result = Solution().removeElement( nums = [3,2,2,3], val = 3 )
        self.assertEqual(result, 2)
 





if __name__ == "__main__":

    unittest.main()
'''

Description:

Given an array nums and a value val, remove all instances of that value in-place and return the new length.

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

The order of elements can be changed. It doesn't matter what you leave beyond the new length.

Example 1:

Given nums = [3,2,2,3], val = 3,

Your function should return length = 2, with the first two elements of nums being 2.

It doesn't matter what you leave beyond the returned length.



Example 2:

Given nums = [0,1,2,2,3,0,4,2], val = 2,

Your function should return length = 5, with the first five elements of nums containing 0, 1, 3, 0, and 4.

Note that the order of those five elements can be arbitrary.

It doesn't matter what values are set beyond the returned length.
Clarification:

Confused why the returned value is an integer but your answer is an array?

Note that the input array is passed in by reference, which means modification to the input array will be known to the caller as well.

Internally you can think of this:

// nums is passed in by reference. (i.e., without making a copy)
int len = removeElement(nums, val);

// any modification to nums in your function would be known by the caller.
// using the length returned by your function, it prints the first len elements.
for (int i = 0; i < len; i++) {
    print(nums[i]);
}

'''

from typing import List

class Solution:
    def removeElement(self, nums: List[int], val: int) -> int:
        
        size = len(nums)
        
        # two-pointers
        # others_idx: index for other elements
        # target_idx: index for target elements with val
        others_idx, target_idx = 0, size-1
        
        while others_idx <= target_idx:
            
            if nums[others_idx] == val:
                
                # swap target elements to the tail side
                nums[others_idx], nums[target_idx] = nums[target_idx], nums[others_idx]
                
                target_idx -= 1
                
            else:
                
                # increase the index when we meet others
                others_idx += 1
            
        
        # length of others is the answer
        return others_idx


# n : the length of input nums

## Time Complexity: O( n )
#
# The overhead in time is the cost of iteration, which is of O( n )

## Space Complexity: O( 1 )
#
# The overhead in space is the storage for two-pointers, which is of O( 1 )


import unittest

class Testing( unittest.TestCase ):

    def test_case_1( self ):

        result = Solution().removeElement( nums = [3,2,2,3], val = 3 )
        self.assertEqual(result, 2)

    
    def test_case_2( self ):

        result = Solution().removeElement( nums = [3,2,2,3], val = 3 )
        self.assertEqual(result, 2)
 





if __name__ == "__main__":

    unittest.main()
Description:  Given an array nums and a value val, remove all instances of that value in-place and return the new length.  Do not allocate extra space for another array, you must do this by modifying the input array in-place with O(1) extra memory.  The order of elements can be changed. It doesn't matter what you leave beyond the new length.  Example 1:  Given nums = [3,2,2,3], val = 3,  Your function should return length = 2, with the first two elements of nums being 2.  It doesn't matter what you leave beyond the returned length.    Example 2:  Given nums = [0,1,2,2,3,0,4,2], val = 2,  Your function should return length = 5, with the first five elements of nums containing 0, 1, 3, 0, and 4.  Note that the order of those five elements can be arbitrary.  It doesn't matter what values are set beyond the returned length. Clarification:  Confused why the returned value is an integer but your answer is an array?  Note that the input array is passed in by reference, which means modification to the input array will be known to the caller as well.  Internally you can think of this:  // nums is passed in by reference. (i.e., without making a copy) int len = removeElement(nums, val);  // any modification to nums in your function would be known by the caller. // using the length returned by your function, it prints the first len elements. for (int i = 0; i < len; i++) {     print(nums[i]); } # two-pointers # others_idx: index for other elements # target_idx: index for target elements with val # swap target elements to the tail side # increase the index when we meet others # length of others is the answer # n : the length of input nums ## Time Complexity: O( n ) # # The overhead in time is the cost of iteration, which is of O( n ) ## Space Complexity: O( 1 ) # # The overhead in space is the storage for two-pointers, which is of O( 1 )