LeetCode in Kotlin

1177. Can Make Palindrome from Substring

Medium

You are given a string s and array queries where queries[i] = [lefti, righti, ki]. We may rearrange the substring s[lefti...righti] for each query and then choose up to ki of them to replace with any lowercase English letter.

If the substring is possible to be a palindrome string after the operations above, the result of the query is true. Otherwise, the result is false.

Return a boolean array answer where answer[i] is the result of the ith query queries[i].

Note that each letter is counted individually for replacement, so if, for example s[lefti...righti] = "aaa", and ki = 2, we can only replace two of the letters. Also, note that no query modifies the initial string s.

Example :

Input: s = “abcda”, queries = [[3,3,0],[1,2,0],[0,3,1],[0,3,2],[0,4,1]]

Output: [true,false,false,true,true]

Explanation:

queries[0]: substring = “d”, is palidrome.

queries[1]: substring = “bc”, is not palidrome.

queries[2]: substring = “abcd”, is not palidrome after replacing only 1 character. q

ueries[3]: substring = “abcd”, could be changed to “abba” which is palidrome. Also this can be changed to “baab” first rearrange it “bacd” then replace “cd” with “ab”.

queries[4]: substring = “abcda”, could be changed to “abcba” which is palidrome.

Example 2:

Input: s = “lyb”, queries = [[0,1,0],[2,2,1]]

Output: [false,true]

Constraints:

• 1 <= s.length, queries.length <= 105
• 0 <= lefti <= righti < s.length
• 0 <= ki <= s.length
• s consists of lowercase English letters.

Solution

class Solution {
    fun canMakePaliQueries(s: String, queries: Array<IntArray>): List<Boolean> {
        return canMakeP(s, queries)
    }

    private fun canMakeP(s: String, qs: Array<IntArray>): List<Boolean> {
        val n = s.length
        val counts = IntArray(n)
        for (i in 0 until n) {
            var m = 0
            if (i > 0) {
                m = counts[i - 1]
            }
            val c = s[i]
            m = m xor (1 shl c.code - 'a'.code)
            counts[i] = m
        }
        val ans: MutableList<Boolean> = ArrayList()
        for (q in qs) {
            ans.add(check(q, counts))
        }
        return ans
    }

    private fun check(q: IntArray, counts: IntArray): Boolean {
        val l = q[0]
        val r = q[1]
        val k = q[2]
        val prev = if (l > 0) counts[l - 1] else 0
        val kk = Integer.bitCount(prev xor counts[r])
        return kk / 2 <= k
    }
}

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