# Solving Leetcode Interviews in Seconds with AI: Longest Common Prefix of K Strings After Removal


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "3485" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> You are given an array of strings words and an integer k. For each index i in the range [0, words.length - 1], find the length of the longest common prefix among any k strings (selected at distinct indices) from the remaining array after removing the ith element. Return an array answer, where answer[i] is the answer for ith element. If removing the ith element leaves the array with fewer than k strings, answer[i] is 0.   Example 1:  Input: words = ["jump","run","run","jump","run"], k = 2 Output: [3,4,4,3,4] Explanation:  Removing index 0 ("jump"):  	 words becomes: ["run", "run", "jump", "run"]. "run" occurs 3 times. Choosing any two gives the longest common prefix "run" (length 3).   Removing index 1 ("run"): 	 words becomes: ["jump", "run", "jump", "run"]. "jump" occurs twice. Choosing these two gives the longest common prefix "jump" (length 4).   Removing index 2 ("run"): 	 words becomes: ["jump", "run", "jump", "run"]. "jump" occurs twice. Choosing these two gives the longest common prefix "jump" (length 4).   Removing index 3 ("jump"): 	 words becomes: ["jump", "run", "run", "run"]. "run" occurs 3 times. Choosing any two gives the longest common prefix "run" (length 3).   Removing index 4 ("run"): 	 words becomes: ["jump", "run", "run", "jump"]. "jump" occurs twice. Choosing these two gives the longest common prefix "jump" (length 4).     Example 2:  Input: words = ["dog","racer","car"], k = 2 Output: [0,0,0] Explanation:  Removing any index results in an answer of 0.     Constraints:  1 <= k <= words.length <= 105 1 <= words[i].length <= 104 words[i] consists of lowercase English letters. The sum of words[i].length is smaller than or equal 105.  

	# Explanation
	Here's the solution:

*   **High-Level Approach:** The solution iterates through the `words` array. For each index `i`, it creates a sub-array excluding the word at index `i`. Then, it iterates through all possible pairs (or more generally, combinations of size *k*) of words within this sub-array to compute the longest common prefix (LCP) of each combination. The maximum LCP length found across all combinations is stored as the result for index `i`. If the sub-array doesn't have at least *k* elements, the result for `i` is 0.

*   **Complexity Analysis:**
    *   **Runtime Complexity:** O(n * (mCk) * l), where n is the length of `words`, m is (n-1), k is the given parameter, and l is the maximum length of the words. In simpler terms, O(n<sup>2</sup> * l) in the worst-case because mCk becomes close to n<sup>k</sup>, however for small constant values of k the problem is in O(n<sup>2</sup> * l)
    *   **Storage Complexity:** O(n), due to the storage of the `answer` array.

	
	# Code
	```python
	def longest_common_prefix(str1, str2):
    """
    Calculates the longest common prefix of two strings.
    """
    min_len = min(len(str1), len(str2))
    i = 0
    while i < min_len and str1[i] == str2[i]:
        i += 1
    return str1[:i]

def longest_common_prefix_k(words, k):
    """
    Calculates the length of the longest common prefix among any k strings in the given list.
    Returns 0 if the list has fewer than k strings.
    """
    n = len(words)
    if n < k:
        return 0

    max_lcp_len = 0
    
    # Iterate through all combinations of k strings.
    import itertools
    for combination in itertools.combinations(words, k):
        # Calculate LCP of the first two strings in the combination.
        lcp = longest_common_prefix(combination[0], combination[1])
        
        # Iterate through the remaining strings and update the LCP.
        for i in range(2, len(combination)):
            lcp = longest_common_prefix(lcp, combination[i])
            
        max_lcp_len = max(max_lcp_len, len(lcp))
    
    return max_lcp_len

def longest_common_prefix_after_removal(words, k):
    """
    Calculates the length of the longest common prefix among any k strings after removing one element.
    """
    n = len(words)
    answer = []
    for i in range(n):
        # Create a new array without the element at index i.
        temp_words = words[:i] + words[i+1:]
        
        # Calculate the length of the longest common prefix.
        lcp_length = longest_common_prefix_k(temp_words, k)
        answer.append(lcp_length)
    
    return answer
	```