# Solving Leetcode Interviews in Seconds with AI: Redistribute Characters to Make All Strings Equal


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "1897" 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 (0-indexed). In one operation, pick two distinct indices i and j, where words[i] is a non-empty string, and move any character from words[i] to any position in words[j]. Return true if you can make every string in words equal using any number of operations, and false otherwise.   Example 1:  Input: words = ["abc","aabc","bc"] Output: true Explanation: Move the first 'a' in words[1] to the front of words[2], to make words[1] = "abc" and words[2] = "abc". All the strings are now equal to "abc", so return true.  Example 2:  Input: words = ["ab","a"] Output: false Explanation: It is impossible to make all the strings equal using the operation.    Constraints:  1 <= words.length <= 100 1 <= words[i].length <= 100 words[i] consists of lowercase English letters.  

	# Explanation
	Here's the breakdown of the solution:

*   **Character Count:** The core idea is that we can redistribute characters freely. Therefore, the only thing that matters is the total count of each character across all strings.
*   **Divisibility Check:** If the count of each character is divisible by the number of words, then it's possible to make all strings equal. Otherwise, it's impossible.

*   **Time & Space Complexity:** O(N*M) time, where N is the number of words and M is the maximum length of a word. O(1) space, since we use a fixed-size array (26 elements) for character counts.

	
	# Code
	```python
	def makeEqual(words):
    """
    Checks if it's possible to make all strings in the input array equal
    by moving characters between them.

    Args:
        words: A list of strings.

    Returns:
        True if it's possible to make all strings equal, False otherwise.
    """
    n = len(words)
    counts = [0] * 26  # Character counts for 'a' to 'z'

    for word in words:
        for char in word:
            counts[ord(char) - ord('a')] += 1

    for count in counts:
        if count % n != 0:
            return False

    return True
	```