LeetCode #27 Remove Element Solution & Explanation

LeetCode Problem

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.

LeetCode #27 Custom Judge
LeetCode #27 Examples
LeetCode #27 Constraints

Solution

When I first saw this problem, I thought that was so easy, just a piece of cake !

C# Solution

First try

public class Solution {
    public int RemoveElement(int[] nums, int val) {

        //ANS1
        /*
        int []expectedNums  = nums.Where(s => s != val).ToArray();
        int l = expectedNums.Length;
        
        for (int i = 0; i <l ; i++) {
            nums[i] = expectedNums[i];
        }
        
        return l;
        */
        //ANS2
        
        int[]x = new int[nums.Length];
        
        int cnt = 0;
        foreach(int i in nums)
        {
            if(i!=val)
            {
                x[cnt] = i;
                cnt+=1;
            }
        }
        
        for (int i = 0; i <cnt ; i++) {
            nums[i] = x[i];
        }
        return cnt;
        
    }
}

Both of the solution had succussed, but when I look at the problem carefully, I found out one thing…

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

Oh my god, both of the solution had used extra space, let’s try another way.

Solution1

public class Solution {
    public int RemoveElement(int[] nums, int val) {

        int j = 0;
        for(int i=0;i<nums.Length;i++)
        {
            if(nums[i]!=val)
            {
                nums[j++] = nums[i];
            }
        }
        return j;
    }
}

Runtime : 134ms、128ms、155ms

Solution2

public class Solution {
    public int RemoveElement(int[] nums, int val) {

        int j = 0;
        foreach(int i in nums)
        {
            if(i!=val)
            {
                nums[j++] = i;
            }
        }
        
        return j;
    }
}

We can also use foreach to replace classic for iteration

Java Solution

Solution1

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

Runtime : 0ms、0ms、0ms

Python3 Solution

Solution1

class Solution:
    def removeElement(self, nums: List[int], val: int) -> int:
        j = 0
        for i in range(0,len(nums)):
            if(nums[i]!=val):
                nums[j] = nums[i]
                j+=1
        return j

Runtime : 44ms、36ms、47ms

JavaScript Solution

Solution1

/**
 * @param {number[]} nums
 * @param {number} val
 * @return {number}
 */
var removeElement = function(nums, val) {
    var j = 0;
    for(var i=0;i<nums.length;i++)
    {
        if(nums[i]!=val)
        {
            nums[j++] = nums[i];
        }
    }
    return j;
};

Runtime : 72ms、55ms、80ms


Submission Detail


Conclusion

🧡If my solution helps, that is my honor!

See Next Problem : Zyrastory-LeetCode #28 Find the Index of the First Occurrence in a String

The problem link : Remove Element – LeetCode

Latest Post

Leave a Reply

Your email address will not be published. Required fields are marked *