Remove Element

class Solution {
public:
    int deleteElement(vector<int>& elements, int target) {
        int index = 0;
        int length = elements.size();
        while (index < length) {
            if (elements[index] == target) {
                elements[index] = elements[length - 1];
                length--;
            } else {
                index++;
            }
        }
        return length;
    }
};

The provided C++ solution defines a method deleteElement that efficiently removes all instances of a specified target value from a given vector of integers elements. The function operates by using a two-pointer approach where:

• One pointer, index, iterates through the vector from the beginning.
• Instead of resizing the vector whenever a target element is found, the code swaps the target element with the last non-processed element in the vector and reduces the vector's effective size (length).

As a result, each encounter with the target does not require shifting elements, making the overall operation more efficient, especially for larger data sets.

The main features of this function are:

• It loops through the vector up to the current length, decrementing length whenever a target element is found.
• It performs an in-place modification by swapping elements, thus avoiding the creation of a new array.
• The function returns the new length of the vector after all target values have been removed.

The solution leverages efficiency in terms of space by modifying the input array directly without using additional storage, and in terms of time by reducing unnecessary movements of elements.

class Solution {
    public int deleteValue(int[] array, int target) {
        int k = 0;
        int size = array.length;
        while (k < size) {
            if (array[k] == target) {
                array[k] = array[size - 1];
                size--;
            } else {
                k++;
            }
        }
        return size;
    }
}

The Java solution presented focuses on removing a specified target value from an array, termed as deleteValue method in the Solution class. Follow this method to tackle the given problem:

• Initialize an integer k to 0 to act as a pointer while traversing the array and maintain the size of the array with size.
• Use a while loop to iterate through the array until k is less than size.
• Inside the loop, check if the current element at index k equals the target value.
  • If it does, replace the current element with the last element in the active region of the array and decrement size.
  • If it does not, increment k to continue checking the next element.
• The loop exits when all elements have been checked, resulting in size representing the new length of the array after all instances of the target have been removed.
• Return size, which now indicates how many elements remain in the modified array without the target values.

This approach ensures the array is modified in place, reducing the need for additional memory, and it efficiently handles scenarios where the target value appears multiple times or not at all.

int eraseElement(int* arr, int length, int value) {
    int index = 0;
    int newLength = length;
    while (index < newLength) {
        if (arr[index] == value) {
            arr[index] = arr[newLength - 1];
            newLength--;
        } else {
            index++;
        }
    }
    return newLength;
}

This solution presents a C language function eraseElement that takes three parameters: a pointer to an integer array arr, the length of the array length, and the integer value to be removed from the array. The function iterates through the given array, and each occurrence of the specified value is removed by replacing it with the last element in the array, which effectively decreases the array's length each time a matching value is found. The iteration continues until all elements of the array have been checked. Once finished, the function returns the new length of the modified array.

Here's what happens throughout the code execution:

• Initialize index to 0 to start checking the array from the beginning.
• Set newLength equal to length as the initial modified length of the array.
• Use a while loop to iterate over the array elements until index is less than newLength.
• In each iteration, check if the current element equals value.
  • If it does, move the last element in the array to the current index and reduce newLength by one.
  • If it does not match, increment index to continue with the next element.
• The loop continues until each element has been inspected or moved if needed.
• After completing the iterations, the function returns the new length of the array which denotes the count of remaining elements after the specified value has been removed.

This method allows the array to be modified in-place without using additional memory (aside from loop control variables), providing an efficient way to remove elements.

var filterArray = function (elements, value) {
    let index = 0;
    let length = elements.length;
    while (index < length) {
        if (elements[index] == value) {
            elements[index] = elements[length - 1];
            length--;
        } else {
            index++;
        }
    }
    return length;
};

This solution defines a JavaScript function filterArray designed to remove specified elements from an array. The function takes two arguments: elements (the array from which to remove elements) and value (the value of the elements to remove). Here's how the function operates:

• Initialize index to 0 to track the current position in the array and length to the total number of elements in the array.
• Use a while loop to iterate through the array until index is less than length.
• Within the loop, check if the current element (at index) is equal to value. If true:
  • Replace the element at the current index with the element at the last valid index (length - 1).
  • Decrement length to effectively reduce the array size by one.
• If the current element does not match value, increment index to move to the next element.
• Finally, after the loop completes, the function returns length, representing the new size of the array with the specified elements removed.

This approach modifies the array in place without using extra space for another array, which makes it efficient in terms of space complexity. Use this function to clean arrays quickly, removing all occurrences of a specific value while maintaining the array's remaining elements' order up to the new length.

class Solution:
    def delete_value(self, elements: List[int], target: int) -> int:
        index = 0
        size = len(elements)
        while index < size:
            if elements[index] == target:
                elements[index] = elements[size - 1]
                size -= 1
            else:
                index += 1
        return size

The provided Python solution demonstrates a method to remove all occurrences of a specified element (target) from an array (elements) in-place and returns the new length of the array after removal. The delete_value function uses a two-pointer approach:

• Initialize an index to track the current position in the array and size to keep track of the effective length of the array.
• Iterate through the array with the index pointer. If the element at index matches target, replace it with the last element in the effective part of the list and reduce size. This avoids shifting all elements and achieves the removal in O(1) time complexity for each replacement.
• If the element does not match, simply move index forward.
• At the completion of the loop, size indicates the new length of the array, which does not contain the target element.

This approach ensures efficient removal of the specified element with minimal space usage since the operations modify the array in place. The return value gives the number of remaining elements.