LeetCode in Kotlin
2396. Strictly Palindromic Number
Medium
An integer n is strictly palindromic if, for every base b between 2 and n - 2 (inclusive), the string representation of the integer n in base b is palindromic.
Given an integer n, return true if n is strictly palindromic and false otherwise.
A string is palindromic if it reads the same forward and backward.
Example 1:
Input: n = 9
Output: false
Explanation: In base 2: 9 = 1001 (base 2), which is palindromic.
In base 3: 9 = 100 (base 3), which is not palindromic.
Therefore, 9 is not strictly palindromic so we return false.
Note that in bases 4, 5, 6, and 7, n = 9 is also not palindromic.
Example 2:
Input: n = 4
Output: false
Explanation: We only consider base 2: 4 = 100 (base 2), which is not palindromic.
Therefore, we return false.
Constraints:
4 <= n <= 105
Solution
class Solution {
fun isStrictlyPalindromic(n: Int): Boolean {
for (i in 2..n - 2) {
val num = Integer.toString(i)
val s = baseConversion(num, 10, i)
if (!checkPalindrome(s)) {
return false
}
}
return true
}
private fun baseConversion(number: String, sBase: Int, dBase: Int): String {
// Parse the number with source radix
// and return in specified radix(base)
return Integer.toString(number.toInt(sBase), dBase)
}
private fun checkPalindrome(s: String): Boolean {
var start = 0
var end = s.length - 1
while (start <= end) {
if (s[start] != s[end]) {
return false
}
start++
end--
}
return true
}
}