Watch 10 video solutions for Minimum Deletions to Make String K-Special, a medium level problem involving Hash Table, String, Greedy. This walkthrough by NeetCodeIO has 13,334 views views. Want to try solving it yourself? Practice on FleetCode or read the detailed text solution.
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.
Example 1:
Input: word = "aabcaba", k = 0
Output: 3
Explanation: 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:
Input: word = "dabdcbdcdcd", k = 2
Output: 2
Explanation: 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:
Input: word = "aaabaaa", k = 2
Output: 1
Explanation: 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.
Constraints:
1 <= word.length <= 1050 <= k <= 105word consists only of lowercase English letters.