Remove Element

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.

date this, nums1 has a length of m + n, where the first m elements denote the elements that should be merged, and the last n elements are set to 0 and should be ignored. nums2 has a length of n.

Problem Statement

The problem is to write a function that takes an integer array nums and an integer val as inputs, and removes all occurrences of val from nums in-place, without using any extra memory. The relative order of the remaining elements in the array may be changed, but we need to maintain the order of the elements that are not removed.

The function should return an integer k, which represents the number of elements remaining in the modified array. The first k elements of the nums array should hold the final result, and it does not matter what we leave beyond the first k elements.

In other words, the function should modify the input array nums in-place such that it contains only the elements that are not equal to val, and return the number of such elements.

Examples

Example 1:

Input: nums = [3,2,2,3], val = 3
Output: 2, nums = [2,2,_,_]
Explanation: Your function should return k = 2, with the first two elements of nums being 2.
It does not matter what you leave beyond the returned k (hence they are underscores).

Example 2:

Input: nums = [0,1,2,2,3,0,4,2], val = 2
Output: 5, nums = [0,1,4,0,3,_,_,_]
Explanation: Your function should return k = 5, with the first five elements of nums containing 0, 0, 1, 3, and 4.
Note that the five elements can be returned in any order.
It does not matter what you leave beyond the returned k (hence they are underscores).

Clarifying Questions

1. Can you please confirm that the input array is an integer array?
2. Can you provide an example input and output for this problem?
3. When you say "remove all occurrences of val", do you mean that all instances of val should be removed from the array or only the first instance of val should be removed?
4. Can the input array contain duplicate elements other than the value to be removed?
5. When you say "the relative order of the elements may be changed", can you provide an example of how the order of elements could be changed?
6. Do we need to maintain the original order of the elements in the array, except for the elements we remove?
7. Can we assume that the input array is non-empty, or do we need to handle empty input arrays as well?

Solutions

There are several approaches to solving this problem. Here are a few possible ones:

1. Brute Force

2. Two-pointers approach

3. Using built-in functions

4. Using a stack

5. Using recursion

Brute Force:

We can iterate through the array and remove each occurrence of val by shifting all the subsequent elements one position to the left. This approach has a time complexity of O(n^2) and requires O(1) extra memory.

Algorithm:

1. Initialize a variable count to 0.
2. Iterate through the array from left to right:
   1. If the current element is equal to val, remove it and shift all subsequent elements one position to the left.
   2. Otherwise, increment count.
3. Return count.

      
     
        public int RemoveElement(int[] nums, int val) {
            // Initialize a counter variable to keep track of the number of valid elements.
            int count = 0;
            // Iterate through the array from left to right.
            for (int i = 0; i < nums.Length; i++) {
                // If the current element is not equal to the value to remove,
                // update the element at index count to be the current element,
                // and increment count.
                if (nums[i] != val) {
                    nums[count++] = nums[i];
                }
            }
            // Return the number of valid elements.
            return count;
        }
        
        
      
    

      
     
        def removeElement(nums: List[int], val: int) -> int:
            # Initialize a counter variable to keep track of the number of valid elements.
            count = 0
            # Iterate through the array from left to right.
            i = 0
            while i < len(nums):
                # If the current element is equal to the value to remove,
                # delete it from the list.
                if nums[i] == val:
                    del nums[i]
                else:
                    # Otherwise, update the counter variable and move to the next element.
                    count += 1
                    i += 1
            # Return the number of valid elements.
            return count
    
    
      
    

      
     
    public int removeElement(int[] nums, int val) {
        // Initialize a counter variable to keep track of the number of valid elements.
        int count = 0;
        // Iterate through the array from left to right.
        for (int i = 0; i < nums.length; i++) {
            // If the current element is not equal to the value to remove,
            // update the element at index count to be the current element,
            // and increment count.
            if (nums[i] != val) {
                nums[count++] = nums[i];
            }
        }
        // Return the number of valid elements.
        return count;
    }
        
      
      
    

      
   
    int removeElement(vector<int>& nums, int val) {
        // Initialize a counter variable to keep track of the number of valid elements.
        int count = 0;
        // Iterate through the array from left to right.
        for (int i = 0; i < nums.size(); i++) {
            // If the current element is not equal to the value to remove,
            // update the element at index count to be the current element,
            // and increment count.
            if (nums[i] != val) {
                nums[count++] = nums[i];
            }
        }
        // Return the number of valid elements.
        return count;
    }
    
      
    

Time Complexity:

• The algorithm traverses the entire input array once, performing a constant amount of work at each iteration, except when deleting elements from the list in the Python implementation.
• In the worst case, all elements of the input array are equal to the value to remove, so the algorithm will traverse the entire input array and perform a constant amount of work at each iteration, resulting in a time complexity of O(n).
• In the best case, none of the elements of the input array are equal to the value to remove, so the algorithm will traverse the entire input array and perform a constant amount of work at each iteration, resulting in a time complexity of O(n).
• The average case time complexity is also O(n), assuming a random distribution of the elements in the input array.

Space Complexity:

• The algorithm performs the operation in-place, modifying the input array directly without using any extra space.
• The space used by the algorithm is therefore constant and does not depend on the size of the input array.
• The space complexity of the Brute force approach is O(1).

Two-pointers:

We can use two pointers, one to track the current position in the original array, and another to track the current position in the modified array. We iterate through the array, and whenever we encounter an element that is not equal to val, we copy it to the modified array using the second pointer. After we have processed all the elements, we return the index of the second pointer, which represents the number of elements remaining in the modified array. This approach has a time complexity of O(n) and requires O(1) extra memory.

Algorithm:

1. Initialize two pointers, i and j, to 0.
2. While i is less than the length of the array:
   1. If nums[i] is not equal to val, set nums[j] to nums[i] and increment j.
   2. Increment i.
3. Return j.

      
        public int RemoveElement(int[] nums, int val) {
            int i = 0; // i is the slow-runner pointer
            for (int j = 0; j < nums.Length; j++) { // j is the fast-runner pointer
                if (nums[j] != val) { // if the current element is not equal to val
                    nums[i++] = nums[j]; // move the element to the i-th position and increment i
                }
            }
            return i; // return i as the number of elements in the modified array that are not equal to val
        }
        
      
      
    

      
     
        def removeElement(nums: List[int], val: int) -> int:
            i = 0 # i is the slow-runner pointer
            for j in range(len(nums)): # j is the fast-runner pointer
                if nums[j] != val: # if the current element is not equal to val
                    nums[i] = nums[j] # move the element to the i-th position
                    i += 1 # increment i
            return i # return i as the number of elements in the modified array that are not equal to val
    
    
      
    

      
        public int removeElement(int[] nums, int val) {
            int i = 0; // i is the slow-runner pointer
            for (int j = 0; j < nums.length; j++) { // j is the fast-runner pointer
                if (nums[j] != val) { // if the current element is not equal to val
                    nums[i++] = nums[j]; // move the element to the i-th position and increment i
                }
            }
            return i; // return i as the number of elements in the modified array that are not equal to val
        }
        
      
      
    

      
 
int removeElement(vector<int>& nums, int val) {
    int i = 0; // i is the slow-runner pointer
    for (int j = 0; j < nums.size(); j++) { // j is the fast-runner pointer
        if (nums[j] != val) { // if the current element is not equal to val
            nums[i++] = nums[j]; // move the element to the i-th position and increment i
        }
    }
    return i; // return i as the number of elements in the modified array that are not equal to val
}

      
    

Time complexity: O(n)

Space complexity: O(1)

The time complexity is O(n) because the algorithm performs a single pass through the input array, and the number of iterations is proportional to the length of the array, which is n. In each iteration, the algorithm performs constant-time operations, such as checking the value of the current element and copying it to the appropriate position in the array. Therefore, the overall time complexity is linear in the length of the array.

The space complexity is O(1) because the algorithm modifies the input array in place, without using any additional space. The only extra space used by the algorithm is the integer variable i, which is used to keep track of the current position in the modified array, but this variable does not depend on the size of the input array, so the space complexity is constant.

Using built-in functions:

Many programming languages provide built-in functions to remove elements from arrays. For example, in Python, we can use the remove() method of the list class to remove all occurrences of val from the list. This approach is easy to implement, but it may not meet the requirement of not using extra memory, depending on how the language implements the built-in functions.

Algorithm:

1. Call the built-in function to remove all occurrences of val from the array.
2. Return the length of the modified array.

      
        public int RemoveElement(int[] nums, int val) {
            // Use Array.FindAll to filter out the values that equal to val
            nums = Array.FindAll(nums, num => num != val);
            // Get the length of the resulting array
            int length = nums.Length;
            return length;
        }
        
        
      
    

      
        def removeElement(nums: List[int], val: int) -> int:
            # Use list comprehension to filter out the values that equal to val
            nums[:] = [num for num in nums if num != val]
            # Get the length of the resulting list
            length = len(nums)
            return length
    
    
      
    

      
        public int removeElement(int[] nums, int val) {
            // Use Arrays.stream to filter out the values that equal to val
            nums = Arrays.stream(nums).filter(num -> num != val).toArray();
            // Get the length of the resulting array
            int length = nums.length;
            return length;
        }
        
      
      
    

      
  
    int removeElement(vector<int>& nums, int val) {
        // Use std::remove to move the elements that equal to val to the end of the vector
        nums.erase(std::remove(nums.begin(), nums.end(), val), nums.end());
        // Get the length of the resulting vector
        int length = nums.size();
        return length;
    }
    
    
      
    

Note that the above code snippets modify the input array or vector in place to remove the values that equal to val. They use built-in functions such as Array.FindAll in C#, list comprehension in Python, Arrays.stream and toArray in Java, and std::remove and vector::erase in C++, to filter out the elements that satisfy a certain condition. The resulting array or vector has the same order as the input array, except that the values that equal to val have been removed. The length of the resulting array or vector is returned as the output. Since the time complexity of the built-in functions is usually implementation-dependent, we cannot give a precise analysis of the time complexity. However, it is usually reasonable to assume that the time complexity is proportional to the size of the input array or vector. The space complexity is O(1) because the built-in functions do not use any extra space, apart from the input and output arrays or vectors.

Using recursion:

We can implement the removal of elements recursively by removing the first occurrence of val in the array and then calling the function recursively on the remaining part of the array. This approach has a time complexity of O(n^2) in the worst case and requires O(n) extra memory due to the recursive call stack.

Algorithm:

1. If the array is empty, return 0.
2. If the first element of the array is equal to val, remove it and return the result of recursively calling the function on the remaining part of the array.
3. Otherwise, increment the result of recursively calling the function on the remaining part of the array by 1 and return it.

      
     
        public int RemoveElement(int[] nums, int val) {
            // Call the helper function to remove val recursively
            int length = RemoveElement(nums, 0, val);
            return length;
        }
        
        private int RemoveElement(int[] nums, int index, int val) {
            // Base case: return 0 if the index exceeds the array length
            if (index == nums.Length) {
                return 0;
            }
            // Recursive case 1: if the current element equals to val, 
            // return the length of the remaining array after removing val recursively
            if (nums[index] == val) {
                int remainingLength = RemoveElement(nums, index + 1, val);
                return remainingLength;
            }
            // Recursive case 2: if the current element is not equal to val,
            // move the element to the front of the array by swapping it with the element at the next available position,
            // and return the length of the remaining array after removing val recursively
            else {
                int remainingLength = RemoveElement(nums, index + 1, val);
                nums[index - remainingLength] = nums[index];
                return remainingLength + 1;
            }
        }
        
      
        
      
    

      
        def removeElement(nums: List[int], val: int) -> int:
            # Call the helper function to remove val recursively
            length = removeElementHelper(nums, 0, val)
            return length
    
        def removeElementHelper(nums: List[int], index: int, val: int) -> int:
            # Base case: return 0 if the index exceeds the list length
            if index == len(nums):
                return 0
            # Recursive case 1: if the current element equals to val, 
            # return the length of the remaining list after removing val recursively
            if nums[index] == val:
                remainingLength = removeElementHelper(nums, index + 1, val)
                return remainingLength
            # Recursive case 2: if the current element is not equal to val,
            # move the element to the front of the list by swapping it with the element at the next available position,
            # and return the length of the remaining list after removing val recursively
            else:
                remainingLength = removeElementHelper(nums, index + 1, val)
                nums[index - remainingLength] = nums[index]
                return remainingLength + 1
        
    
      
    

      
     
        public int removeElement(int[] nums, int val) {
            // Call the helper function to remove val recursively
            int length = removeElementHelper(nums, 0, val);
            return length;
        }
        
        private int removeElementHelper(int[] nums, int index, int val) {
            // Base case: return 0 if the index exceeds the array length
            if (index == nums.length) {
                return 0;
            }
            // Recursive case 1: if the current element equals to val, 
            // return the length of the remaining array after removing val recursively
            if (nums[index] == val) {
                int remainingLength = removeElementHelper(nums, index + 1, val);
                return remainingLength;
            }
            // Recursive case 2: if the current element is not equal to val,
            // move the element to the front of the array by swapping it with the element at the next available position,
            // and return the length of the remaining array after removing val recursively
            else {
                int remainingLength = removeElementHelper(nums, index + 1, val);
                nums[index - remainingLength] = nums[index];
                return remainingLength + 1;
            }
        }
        
      
    

      
     
        // function to remove all occurrences of val in nums using recursion
        int removeElementRecursive(vector<int>& nums, int val) {
            // base case: if the array is empty, return 0
            if (nums.empty()) {
                return 0;
            }
            
            // recursive case: remove the element recursively from the subarray nums[1:] 
            int k = removeElementRecursive(vector<int>(nums.begin()+1, nums.end()), val);
            
            // if the first element is equal to val, remove it by copying the rest of the array
            if (nums[0] == val) {
                for (int i = 0; i < k; i++) {
                    nums[i] = nums[i+1];
                }
            }
            
            // return k if the first element is equal to val, else return k+1
            return (nums[0] == val) ? k : k+1;
        }
       
      
    

Time complexity: O(n^2)

Space complexity: O(n)

The time complexity of the Using recursion approach is O(n^2) in the worst case, when all the elements of the array are equal to val. This is because each recursive call removes one element from the array, and there can be up to n recursive calls in the worst case. The space complexity is O(n) in the worst case, due to the recursion stack.