Problem #4 easy

Largest Palindrome Product

A palindromic number reads the same both ways. The largest palindrome made from the product of two 2-digit numbers is:

\(9009 = 91 \times 99\)

Find the largest palindrome made from the product of two 3-digit numbers.

Solution Approach

This problem requires finding the largest palindromic number that is the product of two 3-digit numbers. The brute force approach checks all possible products in descending order, starting from the largest possible numbers (999 × 999) and working downwards, which ensures we find the largest palindrome first.

cpp javascript ruby go rust

Implementations

View Raw

#include <iostream>
#include <string>
#include <algorithm>

bool is_palindrome(int n) {
    std::string s = std::to_string(n);
    std::string rev = s;
    std::reverse(rev.begin(), rev.end());
    return s == rev;
}

int largest_palindrome_product() {
    int max_palindrome = 0;
    for (int i = 999; i >= 100; --i) {
        for (int j = 999; j >= 100; --j) {
            int product = i * j;
            if (product <= max_palindrome) break;
            if (is_palindrome(product)) {
                max_palindrome = product;
            }
        }
    }
    return max_palindrome;
}

int main() {
    std::cout << largest_palindrome_product() << std::endl;
    return 0;
}

View Raw

function isPalindrome(n) {
    const s = n.toString();
    return s === s.split('').reverse().join('');
}

function largestPalindromeProduct() {
    let maxPalindrome = 0;
    for (let i = 999; i >= 100; i--) {
        for (let j = 999; j >= 100; j--) {
            const product = i * j;
            if (product <= maxPalindrome) break;
            if (isPalindrome(product)) {
                maxPalindrome = product;
            }
        }
    }
    return maxPalindrome;
}

console.log(largestPalindromeProduct());

View Raw

def is_palindrome?(n)
  s = n.to_s
  s == s.reverse
end

def largest_palindrome_product
  max_palindrome = 0
  (100..999).reverse_each do |i|
    (100..999).reverse_each do |j|
      product = i * j
      break if product <= max_palindrome
      max_palindrome = product if is_palindrome?(product)
    end
  end
  max_palindrome
end

puts largest_palindrome_product

View Raw

package main

import (
    "fmt"
    "strconv"
)

func isPalindrome(n int) bool {
    s := strconv.Itoa(n)
    runes := []rune(s)
    for i, j := 0, len(runes)-1; i < j; i, j = i+1, j-1 {
        if runes[i] != runes[j] {
            return false
        }
    }
    return true
}

func largestPalindromeProduct() int {
    maxPalindrome := 0
    for i := 999; i >= 100; i-- {
        for j := 999; j >= 100; j-- {
            product := i * j
            if product <= maxPalindrome {
                break
            }
            if isPalindrome(product) {
                maxPalindrome = product
            }
        }
    }
    return maxPalindrome
}

func main() {
    fmt.Println(largestPalindromeProduct())
}

View Raw

fn is_palindrome(n: u32) -> bool {
    let s = n.to_string();
    let rev: String = s.chars().rev().collect();
    s == rev
}

fn largest_palindrome_product() -> u32 {
    let mut max_palindrome = 0;
    for i in (100..=999).rev() {
        for j in (100..=999).rev() {
            let product = i * j;
            if product <= max_palindrome {
                break;
            }
            if is_palindrome(product) {
                max_palindrome = product;
            }
        }
    }
    max_palindrome
}

fn main() {
    println!("{}", largest_palindrome_product());
}

All Solutions