Leetcode problem to remove instances of a value in an array

user_13204848 0
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The instructions were:

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.

This was my solution but it says my output is returning the first two elements of the array instead of the last two?

Input: [3,2,2,3] My output: [3,3] Expected: [2,2]

def removeElement(self, nums, val):
    count = 0;
    for i in nums:
        if nums[i] == val:
            temp = nums[count]
            nums[count] = nums[i]
            nums[i] = temp

            count=count+1
    nums = nums[count:]
    return len(nums)
arrays python
4 Answers

Check out this!

class Solution:
    def removeElement(self, nums: List[int], val: int) -> int:
        for i in nums[:]:
            if i == val:
                nums.remove(i)
        return len(nums)
user_99492830

There are a few problems here although you have the right idea for a linear algorithm.

for i in nums:, then nums[i] in the loop block is a common Python mistake. It doesn't iterate over the indices like you may think. It actually iterates over the elements! Use for i, element in enumerate(nums): or for i in range(len(nums)) to get indices. A good way to avoid this is to never use i for anything but indices. Calling it n, num or elem makes it easier to avoid confusion.

Another misunderstanding is nums = nums[count:]. This doesn't modify the caller's list data at all--it only reassigns the local variable nums to point to a copy of the caller's list, which isn't what you intended. The only way to modify the caller's list is by assigning something to each index.

A simple approach is to walk the list and move elements that don't match the target val to the front of the list. A pointer length trails the i variable and indicates the length of the result and where the next selected item will be put.

def removeElement(self, nums, val):
    length = 0

    for i in range(len(nums)):
        if nums[i] != val:
            nums[length] = nums[i]
            length += 1

    return length
user_62433520 
class Solution {
    public int removeElement(int[] nums, int val) {
        int len=nums.length;
        int index=0;
        int flag=0;
      
            
        for(int i=0;i<len;i++)
        {
            if(nums[i]!=val)
            {
                nums[index++]=nums[i];
                flag=1;
            }
            if(nums[i]==val)
                flag=1;
        }
        
        if(flag==1)
        return index;
        
        else
            return nums.length;
    }
}
user_143326950

It is simple, the question only needs the number of replacements. So it should be like this:

public static int RemoveElement(int[] nums, int val) {
 int replacementCounter = 0;
 for(int i=0; i < nums.Length ; i++){
   if(nums[i] != val){
     nums[replacementCounter] = nums[i];
     replacementCounter++;
   }
 }
 return replacementCounter;
}
user_21828080
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