Leetcode Largest Palindrome Product problem solution

YASH PAL,

In this Leetcode Largest Palindrome Product problem solution Given an integer n, return the largest palindromic integer that can be represented as the product of two n-digits integers. Since the answer can be very large, return it modulo 1337.

Leetcode Largest Palindrome Product problem solution

Problem solution in Python.

class Solution:
    def largestPalindrome(self, n: int) -> int:
        if n == 1:
            return 9
        min_num = 10 ** (n - 1)
        max_num = 10 ** n - 1       
        
        max_pal = 0
        for i in range(max_num, min_num - 1, -2): 
            if i * i < max_pal:
                break
                
            for j in range(max_num, i - 1, -2):
                product = i * j
                if product % 11 != 0 and product >= max_pal:
                    continue
                if str(product) == str(product)[::-1]:
                    max_pal = product

        return max_pal % 1337

Problem solution in Java.

public int largestPalindrome(int n) {
        if (n == 1) return 9;
        int max = (int) Math.pow(10, n) - 1;
        int max_11 = (max / 11)  * 11;
        long product;
        for (int i=max; i > (int) Math.pow(10, n-1); i--){
            product = Long.parseLong(i + new StringBuilder(String.valueOf(i)).reverse().toString());
            for (long j = max_11; j > (int) Math.pow(10, n-1); j-=11) {
                if ((product/j) / (max+1) == 0){ 
                    if (product%j==0) return (int)(product% 1337);
                }
                else break;
            }
        }
        return -1;
    }

Problem solution in C++.

class Solution {
public:
    int largestPalindrome(int n) {
        long max, min, ans = 0, sum;
        max = static_cast<long>(pow(10, n)) - 1;
        min = max/10 + 1;
        sum = 2 * max;
        while (sum/2 * (sum - sum/2) > ans){
            long i = sum/2, j = sum - sum/2;
            while (j <= max && i >= min){
                long num = i * j;
                if (num > ans && isPalindrome(num)){
                    ans = num;
                    break;
                }
                i--;
                j++;
            }
            sum--;
        }
        return ans % 1337;
    }
    bool isPalindrome(long x){
        long y = 0;
        for (long z = x; z != 0; y = 10 * y + z % 10, z /= 10);
        return x == y;
    }
};

Problem solution in C.

long int creatPalindrome(long int num,int n){
    long int p=num*pow(10,n);
    for(int i=0;i<n;i++){
        p=p+(num/(long int)pow(10,n-i-1))*(long int)pow(10,i);
        num=num%(long int)pow(10,n-i-1);
    }
    return p;
}
int largestPalindrome(int n) {
    if(n==1){return 9;}
    long int p=pow(10,n)-1;
    long int q=p;
    long int temp=pow(10,n-1);
    long int ret=0;
    long int ret1=0;
    while(p>=temp){ 
        ret=creatPalindrome(p,n);
        for(int i=q;i>=temp;i--){
            ret1=ret/i;
            if(ret1>=i){break;}
            if(ret1>=temp&&ret%i==0){return ret%1337;}
        }
        p--;
    }
    return NULL;
}

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