Remove Element

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Remove Element
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Remove Element

Remove Element is a Leetcode problem that presents this challenge:

Given an integer array nums and an integer val, remove all occurrences of val in nums in-place. The order of the elements may be changed. Then return the number of elements in nums which are not equal to val.

If you ignore the in-place requirement, you could do this with an ArrayList:

ArrayList<Integer> list = new ArrayList<>();
for(int i = 0; i < nums.length; i++){
  list.add(nums[i]);
}

int removed = 0;
while(list.remove(Integer.valueOf(val))){
  removed++;
}

for(int i = 0; i < list.size(); i++){
  nums[i] = list.get(i);
}

return nums.length - removed;

To do it in-place (without creating any other data structures), you might loop over each element. When you find a match to be removed, you can use another for loop to shift every subsequent element left by one:

int notEqualCount = 0;
for (int i = 0; i < nums.length; i++) {

  int element = nums[i];

  if(element != val){
    notEqualCount++;
  } else {
    // Shift every subsequent element left by one
    for(int j = i; j < nums.length - 1; j++) {
      nums[j] = nums[j+1];
    }
    // Keep i at the same index to avoid skipping shifted elements
    i--;
  }
}

return notEqualCount;

This works, but it’s inefficient (O(n^2)) because it shifts the entire array left every time it finds a match.

Instead, you can use two pointers to process the array in a single pass:

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

    int leftIndex = 0;

    for (int rightIndex = 0; rightIndex < nums.length; rightIndex++) {
      int rightElement = nums[rightIndex];
      if (rightElement != val) {
        nums[leftIndex] = rightElement;
        leftIndex++;
      }
    }

    return leftIndex;
  }
}

Using two pointers is a common approach for processing arrays or strings, so make sure it’s in your back pocket for problems like this.

Previous:
Arrays and Strings

Arrays and Strings Examples

Valid Palindrome Valid Palindrome

Return whether a string is the same forward and backward.

Search Insert Position Search Insert Position

Return the index of an item, even if it's not in the array.

Remove Element Remove Element

Remove an element from an array in-place.

Add Your Own Example Add Your Own Example

Add your own example to Happy Coding.

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