LeetCode in Kotlin
1771. Maximize Palindrome Length From Subsequences
Hard
You are given two strings, word1 and word2. You want to construct a string in the following manner:
- Choose some non-empty subsequence
subsequence1fromword1. - Choose some non-empty subsequence
subsequence2fromword2. - Concatenate the subsequences:
subsequence1 + subsequence2, to make the string.
Return the length of the longest palindrome that can be constructed in the described manner. If no palindromes can be constructed, return 0.
A subsequence of a string s is a string that can be made by deleting some (possibly none) characters from s without changing the order of the remaining characters.
A palindrome is a string that reads the same forward as well as backward.
Example 1:
Input: word1 = “cacb”, word2 = “cbba”
Output: 5
Explanation: Choose “ab” from word1 and “cba” from word2 to make “abcba”, which is a palindrome.
Example 2:
Input: word1 = “ab”, word2 = “ab”
Output: 3
Explanation: Choose “ab” from word1 and “a” from word2 to make “aba”, which is a palindrome.
Example 3:
Input: word1 = “aa”, word2 = “bb”
Output: 0
Explanation: You cannot construct a palindrome from the described method, so return 0.
Constraints:
1 <= word1.length, word2.length <= 1000word1andword2consist of lowercase English letters.
Solution
class Solution {
fun longestPalindrome(word1: String, word2: String): Int {
val len1 = word1.length
val len2 = word2.length
val len = len1 + len2
val word = word1 + word2
val dp = Array(len) { IntArray(len) }
var max = 0
val arr = word.toCharArray()
for (d in 1..len) {
var i = 0
while (i + d - 1 < len) {
if (arr[i] == arr[i + d - 1]) {
dp[i][i + d - 1] = if (d == 1) 1 else Math.max(dp[i + 1][i + d - 2] + 2, dp[i][i + d - 1])
if (i < len1 && i + d - 1 >= len1) {
max = Math.max(max, dp[i][i + d - 1])
}
} else {
dp[i][i + d - 1] = Math.max(dp[i + 1][i + d - 1], dp[i][i + d - 2])
}
i++
}
}
return max
}
}