Remove Element- LeetCode

Problem Statement

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.

Intuition

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

view raw removeElement.java hosted with ❤ by GitHub

Complexity analysis

• Time complexity : O(n). Both the pointers traverse the array once.
• Space complexity : O(1)

Explanation

When nums[i] equals to the given value, skip this element by incrementing i. As long as nums[i] ! = val, we copy nums[i] to nums[count] and increment both indexes at the same time. Repeat the process until i reaches the end of the array and the new length is count.

Feel free to post queries.