LeetCode in Kotlin

1745. Palindrome Partitioning IV

Hard

Given a string s, return true if it is possible to split the string s into three non-empty palindromic substrings. Otherwise, return false.

A string is said to be palindrome if it the same string when reversed.

Example 1:

Input: s = “abcbdd”

Output: true

Explanation: “abcbdd” = “a” + “bcb” + “dd”, and all three substrings are palindromes.

Example 2:

Input: s = “bcbddxy”

Output: false

Explanation: s cannot be split into 3 palindromes.

Constraints:

• 3 <= s.length <= 2000
• s consists only of lowercase English letters.

Solution

class Solution {
    fun checkPartitioning(s: String): Boolean {
        val len: Int = s.length
        val ch: CharArray = s.toCharArray()
        val dp = IntArray(len + 1)
        dp[0] = 0x01
        for (i in 0 until len) {
            for (l: Int in intArrayOf(i - 1, i)) {
                var r: Int = i
                var localL = l
                while ((localL >= 0) && (r < len) && (ch[localL] == ch[r])) {
                    dp[r + 1] = dp[r + 1] or (dp[localL] shl 1)
                    localL--
                    r++
                }
            }
        }
        return (dp[len] and 0x08) == 0x08
    }
}

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