# Solving Leetcode Interviews in Seconds with AI: Make Number of Distinct Characters Equal


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "2531" 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 two 0-indexed strings word1 and word2. A move consists of choosing two indices i and j such that 0 <= i < word1.length and 0 <= j < word2.length and swapping word1[i] with word2[j]. Return true if it is possible to get the number of distinct characters in word1 and word2 to be equal with exactly one move. Return false otherwise.   Example 1:  Input: word1 = "ac", word2 = "b" Output: false Explanation: Any pair of swaps would yield two distinct characters in the first string, and one in the second string.  Example 2:  Input: word1 = "abcc", word2 = "aab" Output: true Explanation: We swap index 2 of the first string with index 0 of the second string. The resulting strings are word1 = "abac" and word2 = "cab", which both have 3 distinct characters.  Example 3:  Input: word1 = "abcde", word2 = "fghij" Output: true Explanation: Both resulting strings will have 5 distinct characters, regardless of which indices we swap.    Constraints:  1 <= word1.length, word2.length <= 105 word1 and word2 consist of only lowercase English letters.  

	# Explanation
	Here's a breakdown of the approach and the Python code:

*   **High-Level Approach:**
    *   Calculate the initial distinct character counts in both strings.
    *   Iterate through all possible swap combinations (one character from `word1` with one from `word2`). After each swap, recalculate distinct character counts and check for equality. If equal, return `True`.
    *   If no swap results in equal distinct counts after trying all combinations, return `False`.

*   **Complexity Analysis:**
    *   Runtime: O(n*m), where n and m are the lengths of word1 and word2, respectively.
    *   Storage: O(1) - Constant extra space.

	
	# Code
	```python
	def equal_distinct(word1: str, word2: str) -> bool:
    """
    Checks if it is possible to get the number of distinct characters in word1 and word2 to be equal with exactly one move.
    """

    def count_distinct(s: str) -> int:
        """Helper function to count distinct characters in a string."""
        return len(set(s))

    n = len(word1)
    m = len(word2)

    for i in range(n):
        for j in range(m):
            # Create temporary copies to simulate the swap
            temp_word1 = list(word1)
            temp_word2 = list(word2)

            # Perform the swap
            temp_word1[i], temp_word2[j] = temp_word2[j], temp_word1[i]

            # Calculate the number of distinct characters
            distinct_word1 = count_distinct("".join(temp_word1))
            distinct_word2 = count_distinct("".join(temp_word2))

            # Check if the number of distinct characters are equal
            if distinct_word1 == distinct_word2:
                return True

    return False
	```