LeetCode in Kotlin
3085. Minimum Deletions to Make String K-Special
Medium
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 <= 105wordconsists only of lowercase English letters.
Solution
import kotlin.math.min
class Solution {
fun minimumDeletions(word: String, k: Int): Int {
val arr = IntArray(26)
for (element in word) {
arr[element.code - 'a'.code]++
}
var min = Int.MAX_VALUE
for (value in arr) {
if (value != 0) {
val u = value + k
var res = 0
for (i in arr) {
if (i < value) {
res += i
} else if (i > u) {
res += (i - u)
}
}
min = min(res, min)
}
}
return min
}
}