Strictly Palindromic Number LeetCode Solution

Problem – Strictly Palindromic Number LeetCode Solution

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

Strictly Palindromic Number LeetCode Solution in C++

class Solution {
public:
    bool isStrictlyPalindromic(int n) 
    {
        int tmp;                      // store n into a temp val so actual n will not affected
        string base;                  // for storing the value of each base (2 to n-2)
        
        for(int i=2;i<=n-2;i++)
        {
            tmp=n;
            
            while(tmp)                                   // calculating the base i for value tmp 
            {
                base = base + to_string(tmp%i);
                tmp = tmp/i;
            }
            
            int st=0;
            int en=base.size()-1;
            while(st<en)                                // check for palindrom
            {
                if(base[st++]!=base[en--]) return false;
            }
            
            base.clear();                               // clear base each time after iteration
        }
        return true;
    }
};

Strictly Palindromic Number LeetCode Solution in Java

class Solution {
    public boolean isStrictlyPalindromic(int n) {
      
      // toString(int i, int radix) method of Integer class 
      // i: the int value that we want to convert 
      // the radix or base to be used in getting the string representation of int method argument
 
      for(int i=2;i<=n-2;i++){
          if( !checkPalindrome(Integer.toString(n,i)) )   
          return false;
      }
      return true;
    }
    
    public boolean checkPalindrome(String s){
        
        for(int i=0;i<s.length()/2;i++){
            
            if( !(s.charAt(i) == s.charAt(s.length()-i-1)) )
            return false;
        }
        return true;
    }
}

Strictly Palindromic Number LeetCode Solution in Python

class Solution:
    def isStrictlyPalindromic(self, n: int) -> bool:
        for b in range(2, n-1):
            rep = []
            k = n
            while k > 0:
                rep.append(k%b)
                k //= b
            if rep != rep[::-1]: return False
        return True

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