Question

Formatted question description: https://leetcode.ca/all/27.html

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

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.


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]);
}

@tag-array

Algorithm

A variable is needed for counting, and then the original array is traversed. If the current value is different from the given value, the current value will cover the position of the counting variable, and the counting variable will be incremented by 1.

Code

Java

  • 
    public class Remove_Element {
    
        // time: O(N)
        // space: O(N)
    	public class Solution {
    	    public int removeElement(int[] nums, int val) {
    	        if(nums == null || nums.length == 0) {
    	            return 0;
    	        }
    
    	        int i = 0; // slow pointer
    	        int j = 0; // fast pointer
    
    	        while(j < nums.length) {
    	            while(j < nums.length && nums[j] == val) {
    	                j++;
    	            }
    
    	            // protect index j+1
    	            if(j < nums.length) {
    	                nums[i] = nums[j];
    
    	                i++;
    	                j++;
    	            }
    	        }
    
    	        return i; // i is incremented, but is index. So same as length
    
    	    }
    	}
    
    }
    
  • // OJ: https://leetcode.com/problems/remove-element/
    // Time: O(N)
    // Space: O(1)
    class Solution {
    public:
        int removeElement(vector<int>& A, int val) {
            int j = 0, N = A.size();
            for (int i = 0; i < N; ++i) {
                if (A[i] != val) A[j++] = A[i];
            }
            return j;
        }
    };
    
  • class Solution(object):
      def removeElement(self, nums, val):
        """
        :type nums: List[int]
        :type val: int
        :rtype: int
        """
        slow = -1
        for i in range(0, len(nums)):
          if nums[i] != val:
            slow += 1
            nums[slow] = nums[i]
        return slow + 1
    
    

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