LeetCode in Kotlin
1312. Minimum Insertion Steps to Make a String Palindrome
Hard
Given a string s. In one step you can insert any character at any index of the string.
Return the minimum number of steps to make s palindrome.
A Palindrome String is one that reads the same backward as well as forward.
Example 1:
Input: s = “zzazz”
Output: 0
Explanation: The string “zzazz” is already palindrome we don’t need any insertions.
Example 2:
Input: s = “mbadm”
Output: 2
Explanation: String can be “mbdadbm” or “mdbabdm”.
Example 3:
Input: s = “leetcode”
Output: 5
Explanation: Inserting 5 characters the string becomes “leetcodocteel”.
Constraints:
1 <= s.length <= 500sconsists of lowercase English letters.
Solution
class Solution {
private fun longestPalindrome(a: String, b: String, n: Int): Int {
val dp = Array(n + 1) { IntArray(n + 1) }
for (i in 0 until n + 1) {
for (j in 0 until n + 1) {
if (i == 0 || j == 0) {
dp[i][j] = 0
} else if (a[i - 1] == b[j - 1]) {
dp[i][j] = 1 + dp[i - 1][j - 1]
} else {
dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1])
}
}
}
return dp[n][n]
}
fun minInsertions(s: String): Int {
val n = s.length
if (n < 2) {
return 0
}
val rs = StringBuilder(s).reverse().toString()
val l = longestPalindrome(s, rs, n)
return n - l
}
}