You are given a string word and an integer k.
We consider word to be k-special if |freq(word[i]) - freq(word[j])| <= k for all indices i and j in the string.
Here, freq(x) denotes the frequency of the character x in word, and |y| denotes the absolute value of y.
Return the minimum number of characters you need to delete to make word k-special.
\n
Example 1:
\n\nInput: word = "aabcaba", k = 0
\n\nOutput: 3
\n\nExplanation: We can make word 0-special by deleting 2 occurrences of "a" and 1 occurrence of "c". Therefore, word becomes equal to "baba" where freq('a') == freq('b') == 2.
Example 2:
\n\nInput: word = "dabdcbdcdcd", k = 2
\n\nOutput: 2
\n\nExplanation: We can make word 2-special by deleting 1 occurrence of "a" and 1 occurrence of "d". Therefore, word becomes equal to "bdcbdcdcd" where freq('b') == 2, freq('c') == 3, and freq('d') == 4.
Example 3:
\n\nInput: word = "aaabaaa", k = 2
\n\nOutput: 1
\n\nExplanation: We can make word 2-special by deleting 1 occurrence of "b". Therefore, word becomes equal to "aaaaaa" where each letter's frequency is now uniformly 6.
\n
Constraints:
\n\n1 <= word.length <= 1050 <= k <= 105word consists only of lowercase English letters.给你一个字符串 word 和一个整数 k。
如果 |freq(word[i]) - freq(word[j])| <= k 对于字符串中所有下标 i 和 j 都成立，则认为 word 是 k 特殊字符串。
此处，freq(x) 表示字符 x 在 word 中的出现频率，而 |y| 表示 y 的绝对值。
返回使 word 成为 k 特殊字符串 需要删除的字符的最小数量。
\n\n
示例 1：
\n\n输入：word = \"aabcaba\", k = 0
\n\n输出：3
\n\n解释：可以删除 2 个 \"a\" 和 1 个 \"c\" 使 word 成为 0 特殊字符串。word 变为 \"baba\"，此时 freq('a') == freq('b') == 2。
示例 2：
\n\n输入：word = \"dabdcbdcdcd\", k = 2
\n\n输出：2
\n\n解释：可以删除 1 个 \"a\" 和 1 个 \"d\" 使 word 成为 2 特殊字符串。word 变为 \"bdcbdcdcd\"，此时 freq('b') == 2，freq('c') == 3，freq('d') == 4。
示例 3：
\n\n输入：word = \"aaabaaa\", k = 2
\n\n输出：1
\n\n解释：可以删除 1 个 \"b\" 使 word 成为 2特殊字符串。因此，word 变为 \"aaaaaa\"，此时每个字母的频率都是 6。
\n\n
提示：
\n\n1 <= word.length <= 1050 <= k <= 105word 仅由小写英文字母组成。x be the character with the smallest frequency in the answer. It can be shown that out of the selected characters, the optimal solution will never delete an occurrence of character x to obtain the answer.",
"We will fix a character c and assume that it will be the character with the smallest frequency in the answer. Suppose its frequency is x.",
"Then, for every other character, we will count the number of occurrences that will be deleted. Suppose that the current character has y occurrences. \\u7248\\u672c\\uff1a \\u7f16\\u8bd1\\u65f6\\uff0c\\u5c06\\u4f1a\\u91c7\\u7528 \\u4e3a\\u4e86\\u4f7f\\u7528\\u65b9\\u4fbf\\uff0c\\u5927\\u90e8\\u5206\\u6807\\u51c6\\u5e93\\u7684\\u5934\\u6587\\u4ef6\\u5df2\\u7ecf\\u88ab\\u81ea\\u52a8\\u5bfc\\u5165\\u3002<\\/p>\"],\"java\":[\"Java\",\" \\u7248\\u672c\\uff1a \\u4e3a\\u4e86\\u65b9\\u4fbf\\u8d77\\u89c1\\uff0c\\u5927\\u90e8\\u5206\\u6807\\u51c6\\u5e93\\u7684\\u5934\\u6587\\u4ef6\\u5df2\\u88ab\\u5bfc\\u5165\\u3002<\\/p>\\r\\n\\r\\n \\u5305\\u542b Pair \\u7c7b: https:\\/\\/docs.oracle.com\\/javase\\/8\\/javafx\\/api\\/javafx\\/util\\/Pair.html <\\/p>\"],\"python\":[\"Python\",\" \\u7248\\u672c\\uff1a \\u4e3a\\u4e86\\u65b9\\u4fbf\\u8d77\\u89c1\\uff0c\\u5927\\u90e8\\u5206\\u5e38\\u7528\\u5e93\\u5df2\\u7ecf\\u88ab\\u81ea\\u52a8 \\u5bfc\\u5165\\uff0c\\u5982\\uff1aarray<\\/a>, bisect<\\/a>, 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