LeetCode 2396: Strictly Palindromic Number
LeetCode 2396 Solution Explanation
Explanation:
To solve this problem, we need to check if a given number n is strictly palindromic. A number is strictly palindromic if, for every base b between 2 and n - 2 inclusive, the string representation of n in base b is palindromic.
We can iterate over the bases from 2 to n - 2 and convert the number n to each base to check if it is a palindrome in that base. If we find a base where n is not a palindrome, we can immediately return false. If we successfully iterate over all bases without finding any base where n is not a palindrome, we return true.
Algorithm:
- Iterate over bases from 2 to
n - 2. - Convert number
nto the current base. - Check if the converted number is a palindrome.
- If any base conversion is not a palindrome, return false. Otherwise, return true.
Time Complexity:
The time complexity of this algorithm is O(log2(n) * n) since we iterate over each base from 2 to n - 2 and convert the number n to each base.
Space Complexity: The space complexity is O(log2(n)) to store the string representation of the number in different bases.
LeetCode 2396 Solutions in Java, C++, Python
class Solution {
public boolean isStrictlyPalindromic(int n) {
for (int base = 2; base <= n - 2; base++) {
String numInBase = Integer.toString(n, base);
if (!isPalindrome(numInBase)) {
return false;
}
}
return true;
}
private boolean isPalindrome(String s) {
int left = 0, right = s.length() - 1;
while (left < right) {
if (s.charAt(left) != s.charAt(right)) {
return false;
}
left++;
right--;
}
return true;
}
}Interactive Code Editor for LeetCode 2396
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