Two strings are considered **close** if you can attain one from the other using the following operations:

*   Operation 1: Swap any two **existing** characters.
    *   For example, `abcde -> aecdb`
*   Operation 2: Transform **every** occurrence of one **existing** character into another **existing** character, and do the same with the other character.
    *   For example, `aacabb -> bbcbaa` (all `a`'s turn into `b`'s, and all `b`'s turn into `a`'s)

You can use the operations on either string as many times as necessary.

Given two strings, `word1` and `word2`, return `true` _if_ `word1` _and_ `word2` _are **close**, and_ `false` _otherwise._

**Example 1:**

**Input:** word1 =  "abc ", word2 =  "bca "
**Output:** true
**Explanation:** You can attain word2 from word1 in 2 operations.
Apply Operation 1:  "abc " ->  "acb "
Apply Operation 1:  "acb " ->  "bca "

**Example 2:**

**Input:** word1 =  "a ", word2 =  "aa "
**Output:** false
**Explanation:** It is impossible to attain word2 from word1, or vice versa, in any number of operations.

**Example 3:**

**Input:** word1 =  "cabbba ", word2 =  "abbccc "
**Output:** true
**Explanation:** You can attain word2 from word1 in 3 operations.
Apply Operation 1:  "cabbba " ->  "caabbb "
`Apply Operation 2:  "`caabbb " ->  "baaccc "
Apply Operation 2:  "baaccc " ->  "abbccc "

**Constraints:**

*   `1 <= word1.length, word2.length <= 105`
*   `word1` and `word2` contain only lowercase English letters.


def get_winner(arr, k):
    winner = arr[0]
    consecutive_wins = 0

    for i in range(1, len(arr)):
        if arr[i] > winner:
            winner = arr[i]
            consecutive_wins = 0

        consecutive_wins += 1
        if consecutive_wins == k:
            break

    return winner
