27. Remove Element

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]);
}

My Solution

Source Code

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/**
 * @param {number[]} nums
 * @param {number} val
 * @return {number}
 */
var removeElement = function(nums, val) {
    let lag = 0;
    
    for(let x = 0; x < nums.length; x++){
            if(nums[x] === val){
                continue;
            }
            else{
                nums[lag] = nums[x];
                console.log(nums)
                lag++;
            }   
    }
    return lag;
};

Analysis

The idea here is to only keep track of the valid numbers as we iterate over the array. We have a lagging pointer that only advances when we see a valid number. When we find a new valid number, we override the lagging pointer with it. If we see an invalid number, we skip over it and don't do anything. As we are only iterating over the array once, the running time is O(n) where n is the length of the nums array.

My first idea was to simply switch the invalid number with whatever the next element in the array was. This actually worked well until an array had two invalid numbers in a row. Then I realized it would just be easier to look at the valid numbers instead of the invalid ones.