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.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.",
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