I have a recursive function below and I was just wondering how can I create an iterative (i.e. loops without recursion) for the same thing. I would really appreciate any help or suggestions thank you!
function countPalindromes(string, count) {
if (string.length <= 1) {
return count;
}
let [ firstLetter ] = string;
let lastLetter = string[string.length - 1];
if (firstLetter === lastLetter) {
let stringWithoutFirstAndLastLetters = string.substring(1, string.length - 1);
return countPalindromes(stringWithoutFirstAndLastLetters, count + 1);
} else {
return count;
}
}
console.log(countPalindromes("level", 0));
console.log(countPalindromes("aya", 0));
Took me a while, but this should work.
function countPalindromes(string, count) {
for (i = 0; i < Math.floor(string.length/2); i++) {
if (string.charAt(i) === string.charAt(string.length - i-1)) {count++};
}
return count;
}
console.log(countPalindromes("level", 0));
console.log(countPalindromes("aya", 0));
You can do something like this:
function countPalindromes(string) {
let count = 0;
for (let i = 0; i < Math.floor(string.length / 2); i++) {
const letterFromTheStart = string[i];
const letterFromTheEnd = string[(string.length - 1) - i];
if (letterFromTheStart !== letterFromTheEnd) {
break;
}
count++;
}
return count;
}
console.log(countPalindromes("level"));
console.log(countPalindromes("aya"));
The important parts are i < Math.floor(string.length / 2) and (string.length - 1) - i.
The first one assures the loop stops before or at the half of the string and the second gets the nth character starting from the string's end.
I didn't get your recursive palindrome function. Just converted the recursive one to iterative one.
function countPalindromes(string, count) {
if (string.length <= 1) {
return count;
}
let i = 0, j = string.length-1;
for (; i < j && string[i] == string[j]; i++, j--);
return i;
}
console.log(countPalindromes("level", 0));
console.log(countPalindromes("aya", 0));
You can reverse the string and compare letters until you hit a mismatch returning the last index that matched plus one.
function isPalindrome(word) {
const drow = word.split('').reverse();
for (i = 0; i < drow.length && drow[i] === word[i]; i++);
return i;
}
console.log(isPalindrome("level"));
I have an issue with a recursive algorithm, that solves the problem of finding the happy numbers.
Here is the code:
function TestingFunction(number){
sumNumberContainer = new Array(0);
CheckIfNumberIsHappy(number);
}
function CheckIfNumberIsHappy(number){
var sumOfTheNumbers = 0;
for (var i = 0; i < number.length; i++) {
sumOfTheNumbers += Math.pow(parseInt(number[i]), 2);
}
console.log(sumOfTheNumbers);
if(sumOfTheNumbers == 1){
return CheckIfNumberIsHappy(sumOfTheNumbers.toString());
//return true;
} else {
sumNumberContainer.push(sumOfTheNumbers);
if(sumNumberContainer.length > 1){
for (var i = 0; i < sumNumberContainer.length - 1; i++) {
for (var j = i + 1; j < sumNumberContainer.length; j++) {
if(sumNumberContainer[i] == sumNumberContainer[j]){
return CheckIfNumberIsHappy(sumOfTheNumbers.toString());
//return false;
}
}
}
}
CheckIfNumberIsHappy(sumOfTheNumbers.toString());
}
}
Algorithm is working ALMOST fine. I've tested it out by calling function with different numbers, and console was displaying correct results. The problem is that I almost can't get any value from the function. There are only few cases in which I can get any value: If the number is build out of ,,0", and ,,1", for example 1000.
Because of that, I figured out, that I have problem with returning any value when the function is calling itself again.
Now I ended up with 2 results:
Returning the
return CheckIfNumberIsHappy(sumOfTheNumbers.toString());
which is giving an infinity looped number. For example when the number is happy, the function is printing in the console number one again and again...
Returning the
//return true
or
//return false
which gives me an undefined value
I'm a little bit in check by this problem, and I'm begging you guys for help.
I would take a step back and reexamine your problem with recursion in mind. The first thing you should think about with recursion is your edge cases — when can you just return a value without recursing. For happy numbers, that's the easy case where the sum of squares === 1 and the harder case where there's a cycle. So test for those and return appropriately. Only after that do you need to recurse. It can then be pretty simple:
function sumSq(num) {
/* simple helper for sums of squares */
return num.toString().split('').reduce((a, c) => c * c + a, 0)
}
function isHappy(n, seen = []) {
/* seen array keeps track of previous values so we can detect cycle */
let ss = sumSq(n)
// two edge cases -- just return
if (ss === 1) return true
if (seen.includes(ss)) return false
// not an edge case, save the value to seen, and recurse.
seen.push(ss)
return isHappy(ss, seen)
}
console.log(isHappy(23))
console.log(isHappy(22))
console.log(isHappy(7839))
Here's a simplified approach to the problem
const digits = x =>
x < 10
? [ x ]
: [ ...digits (x / 10 >> 0), x % 10 ]
const sumSquares = xs =>
xs.reduce ((acc, x) => acc + x * x, 0)
const isHappy = (x, seen = new Set) =>
x === 1
? true
: seen.has (x)
? false
: isHappy ( sumSquares (digits (x))
, seen.add (x)
)
for (let n = 1; n < 100; n = n + 1)
if (isHappy (n))
console.log ("happy", n)
// happy 1
// happy 7
// happy 10
// ...
// happy 97
The program above could be improved by using a technique called memoization
Your code is almost correct. You just forgot to return the result of the recursive call:
function TestingFunction(number){
sumNumberContainer = new Array(0);
if (CheckIfNumberIsHappy(number))
console.log(number);
}
function CheckIfNumberIsHappy(number){
var sumOfTheNumbers = 0;
for (var i = 0; i < number.length; i++) {
sumOfTheNumbers += Math.pow(parseInt(number[i]), 2);
}
if(sumOfTheNumbers == 1){
return true;
} else {
sumNumberContainer.push(sumOfTheNumbers);
if(sumNumberContainer.length > 1){
for (var i = 0; i < sumNumberContainer.length - 1; i++) {
for (var j = i + 1; j < sumNumberContainer.length; j++) {
if(sumNumberContainer[i] == sumNumberContainer[j]){
return false;
}
}
}
}
return CheckIfNumberIsHappy(sumOfTheNumbers.toString());
}
}
for (let i=0; i<100; ++i)
TestingFunction(i.toString()); // 1 7 10 13 ... 91 94 97
I've got the solution, which was given to me in the comments, by the user: Mark_M.
I just had to use my previous
return true / return false
also I had to return the recursive statement in the function, and return the value of the CheckIfTheNumberIsHappy function, which was called in TestingFunction.
The working code:
function TestingFunction(number){
sumNumberContainer = new Array(0);
return CheckIfNumberIsHappy(number);
}
function CheckIfNumberIsHappy(number){
var sumOfTheNumbers = 0;
for (var i = 0; i < number.length; i++) {
sumOfTheNumbers += Math.pow(parseInt(number[i]), 2);
}
console.log(sumOfTheNumbers);
if(sumOfTheNumbers == 1){
return true;
} else {
sumNumberContainer.push(sumOfTheNumbers);
if(sumNumberContainer.length > 1){
for (var i = 0; i < sumNumberContainer.length - 1; i++) {
for (var j = i + 1; j < sumNumberContainer.length; j++) {
if(sumNumberContainer[i] == sumNumberContainer[j]){
return false;
}
}
}
}
return CheckIfNumberIsHappy(sumOfTheNumbers.toString());
}
}
Thanks for the great support :)
Doing an assignment for my JavaScript class that requires me to create a function that adds every other index starting with the first and subtracting all the indexes not previously added and produces the sum. I believe the function below should work but it seems to return undefined.
function questionSix(){
let result = 0;
for(let i = 0; i < arguments.length; i++){
if(i == 0){
result += arguments[i];
}else{
if(i % 2 != 0){
result += arguments[i];
}
if(i % 2 == 0){
result -= arguments[i];
}
}
}
}
Because you aren't returning anything (there is no return statement in your code) :
function questionSix(){
let result = 0;
for(let i = 0; i < arguments.length; i++) {
if(i == 0){
result += arguments[i];
}else{
if(i % 2 != 0){
result += arguments[i];
}
if(i % 2 == 0){
result -= arguments[i];
}
}
}
return result;
}
console.log(questionSix(1,6,5,7,8,9,9,8,4,5));
However, it looks like your code isn't doing exactly what it should, here's a solution to your problem :
function questionSix(){
let result = 0; // the sum
let array = []; // the numbers added
for(let i = 0; i < arguments.length; i++) {
// was the number added ?
if(array.indexOf(i) > -1){ // Yes
result += arguments[i]; // add it to the sum
}else{ // No
result -= arguments[i]; // subtract it from the sum
array.push(arguments[i]); // add it to the array
}
}
return result; // return the sum
}
console.log(questionSix(1,6,5,7,8,9,9,8,4,5));
You haven't included a "return" statement
The clue to this (apart from there being no return statement!) is that you start with a result=0 statement, and yet you are receiving undefined.
You don't have to do a special case for i==0
When i==0, i % 2 will equal 0, so the "inner" if-then-else will do the job adequately, without the if (i==0) segment.
Swap the += and -=
However I wonder whether you have reversed the += and -= ? You want to add the 0th and all even-indexed values, don't you? And subtract the odd ones?
Please review the code below, note the condition of my for loop depends on the step parameter.
Rather than every time the condition is executed it determines which branch to use, I would like to test that once - I had supposed I could create a delegate or of the condition but it doesn't seem to work.
Is it possible in JS to do this?
Code:
function(start, end, step) {
if (step === undefined) {
step = 1;
}
var result = [];
for (; (step < 0) ? start >= end : start <= end; start += step) {
result.push(start);
}
return result;
}
My attempt:
function(start, end, step) {
if (step === undefined) {
step = 1;
}
var condition = (step < 0) ? start >= end : start <= end;
var result = [];
for (; condition; start += step) {
result.push(start);
}
return result;
}
To do this, you need to make condition a function, like below. But even if you do, the condition is still executed at every iteration of the loop.
var condition = (step < 0) ?
function(start){
return start >= end;
} :
function(start){
return start <= end;
};
var result = [];
for (; condition(start); start += step) {
result.push(start);
}
Can anyone tell me what's wrong with the code. Find the largest palindrome made from the product of two 3-digit numbers.
function largestPalindrome(){
for(var i =999; i>100; i--){
for(var j = 999; j>100; j--){
var mul = j*i;
if(isPalin(mul)){
return i * j;
}
}
}
}
function isPalin(i){
return i.toString() == i.toString().split("").reverse().join("");
}
console.log(largestPalindrome());
This answer was close to my question
but still i feel the way i am doing the loop it should return me the largest product.
Yours doesn't work properly since it checks 999*999, then 999*998, then 999*997 until it reaches about 999*583. While it doesn't check 997*995 or something closer to the top
which generates a larger number
function largestPalindrome(){
var arr = [];
for(var i =999; i>100; i--){
for(var j = 999; j>100; j--){
var mul = j*i;
if(isPalin(mul)){
arr.push(j * i);
}
}
}
return Math.max.apply(Math, arr);
}
function isPalin(i){
return i.toString() == i.toString().split("").reverse().join("");
}
console.log(largestPalindrome());
Here is another approach, store all palindrome generated by 3 numbers in an array, then use Math.max on the array to get the largest palindrome
I think if you apply maths to the problem you can decrease the guesswork really significantly.
I will write the three digit numbers as 1000 - a and 1000 - b which means the palindrome is 1 000 000 - 1000(a+b) + ab.
First, let's find solutions where ab < 1000. Then the three leftmost digits are 1000 - (a+b) and the three rightmost digits are ab.
Then I will say this is a palindrome with digits x,y,z:
100x+10y+z=ab
100z+10y+x=1000-a-b
thus
99x-99z = ab+a+b-1000
x-z = 1/99(ab+a+b-10)-10
So then (ab+a+b-10) is divisible by 99 and we also know that x and z being digits the left side is between -9 and 0 (the whole shebang is symmetrical so we can presume x <= z) so then 1/99(ab+a+b-10) is between 1 and 9. We can rewrite ab+a+b-10 as ab+a+b+1-11=99p so (a+1)(b+1)=99p+11=11*(9p+1) where p runs between 1 and 9. That's really easy:
for ($p = 1; $p <= 9; $p++) {
$n = 9 * $p + 1;
// This could be vastly optimized further.
for ($j = 1; $j <= $n; $j++) {
if ($n % $j === 0) {
$a = 1001 - $n / $j;
$b = 1001 - 11 * $j;
$test = $a * $b;
if (strrev($test) === (string) $test) {
print "$a $b " . $a * $b . "\n";
}
}
}
}
Now this prints only one solution which is the correct one.
Now we know 906609 is a solution so then is there a solution where ab > 1000 and 1000(a+b) - ab < 93391 ? There is not :)
As explained in #VisioN's comment:
995*583 = 580085 is a palindrome.
993*913 = 906609 is also a (larger) palindrome.
Your code checks 995*583 before 993*913 and exits at the first palindrome found, so it doesn't return the largest palindrome.
Solution: get the largest palindromes starting from 999*999 = 998001 downwards and check if they can be written as xyz*abc.
Or simply use the accepted solution from the question you linked :). Your solution, but instead of returning when you find the first palindrome, check if it is larger than the largest one already found, in which case you need to replace it. You can stop as soon as the largest palindrome is larger than i*999.
A bit more optimized version with comments included. Notice, there is no need of fast return, just store the max and optimize the cycles to not recalculate j*i if i*j has already been checked.
function largestPalindrome() {
var max = 0;
// not using i >= 100 since 100*100 is not palindrome! :)
for (var i = 999; i > 100; i--) {
// because i * j === j * i, no need of both i and j
// to count down from 999
for (var j = i; j > 100; j--) {
var mul = j * i;
if (isPalin(mul) && mul > max) {
max = i * j;
}
}
}
return max;
}
function isPalin(i) {
// adding empty string to i instead using of .toString
// avoids unnecessary wrapping in String object on the left side
i = '' + i;
// don't rely on ==, use === instead
return i === i.split("").reverse().join("");
}
console.log(largestPalindrome());
Suggesting a solution using underscore.js. First, find all palindromes and then loop through them starting from the largest one and return the one which has two 3-digit prime factors.
function isPalindrome(num) {
var str = num.toString();
return str.split('').reverse().join('') === str;
}
function palindromes() {
var max = 999 * 999;
var min = 100 * 100;
return _.select(_.range(max, min, -1), isPalindrome);
}
palindromes().find(function (x) {
if (_.find(_.range(999, 100, -1), function (y) {
return (x % y === 0 && y != x / y && x / y < 1000) ? true : false;
})) return true;
})
#define MAX(a, b) ((a) > (b) ? (a) : (b))
int largestPalindrome()
{
int ret = 0;
for (int i = 999; i > 100; --i)
{
int jLimit = MAX(ret / i, 100);
for (int j = i; j > jLimit; --j)
{
int ans = i * j;
if (isPalin(ans))
{
ret = MAX(ans, ret);
}
}
}
return ret;
}
Reasons explained above.
We can recompute the range of j when we find a palindrome product.This should be faster.
The above solution will work perfectly fine but we will have issue ONLY when we try to find-out what are those 2 numbers (i = 913 and j = 993)
I will just modify the solution proposed by Azder
int max = 0;
int no1 = 0;
int no2 = 0;
// not using i >= 100 since 100*100 is not palindrome! :)
for (var i = 999; i > 100; i--) {
// because i * j === j * i, no need of both i and j
// to count down from 999
for (var j = i; j > 100; j--) {
var mul = j * i;
if (isPalin(mul)) {
if ((i+j) > max) {
max = i+j;
no1 = i; no2 = j;
}
}
}
}
//Now we can get the 2 numbers (no1=993 and no2=913)
return (no1*no2);
This is how I did it. I used the old fashioned way to check for a palindrome. It appears to run faster on my computer but I may be wrong. Pushing to an array, as in the above post, was definitely very slow on my computer. Noticeable lag at the console. I would recommend just checking to see if your product is greater than your current max, if it is, store that instead of pushing everything to an array. Please feel free to correct me if I'm wrong. Much appreciated.
//should find the largest palindrome made from the product of two 3 digit numbers
var largestPalindrome = function() {
var max = 0,
product = 0;
for (var num1 = 999; num1 >= 100; num1--) {
for (var num2 = 999; num2 >= 100; num2--) {
product = num1 * num2;
product > max && isPalindrome(product.toString()) ? max = product : 0;
}
}
return max;
};
//check to see if product is a palindrome
var isPalindrome = function(product) {
var palindromeCheck = true;
for (var i = 0; i < product.length / 2; i++) {
if (product[i] != product[product.length - i - 1])
palindromeCheck = false;
}
return palindromeCheck;
//return product === product.split("").reverse().join("");
};
I think you can go for code given at this link
http://www.mathblog.dk/project-euler-problem-4/
As this save your CPU cycle from multiplication, which is quite costly operation.
Well even in this you can make some more to make to make it more like, you can modify its while loop a bit
while (!found) {
firstHalf--;
palin = makePalindrome(firstHalf);
for (int i = 999; i > 99; i--) {
if ((palin / i) > 999 || i*i < palin) {
break;
}
if ((palin % i == 0)) {
found = true;
factors[0] = palin / i;
factors[1] = i;
break;
}
}
}
So here instead of moving from i=999 : 100, we can write it as i=sqrt(palin):100, as you can find factorial of number within its square root. Refer link How to find Number is prime number or not!
And also you can change if(condition) to if(!(palin%i)) as comparing with zero is usually not considered a good practice also comparing takes more CPU cycle compared to your simple negating bits.
instead of creating an Array or ArrayList to store all palindromes, I just created another variable max and stored highest valued palindrome in it.
My code is in Java, but you can understand the logic from it.
Here is my code to better explain what I said (read comments):
package euler;
import java.util.ArrayList; import java.util.Collections;
public class Problem4 {
public static void main (String[] args)
{
int product=0;
int max=0;
for(int i=999;i>100;i--)
{
for (int j=i;j>100;j--)
{
product=i*j;
if(isPalindrome(product))
{
//Just store maximum value of product.
//Following if loop is required in your code,in place of return i*j;
if(product>max)
{ max=product; }
}
}
}
System.out.println(max);
}
//might be inefficient to create StringBuilder and again String to compare.
public static boolean isPalindrome(int product)
{
boolean isPalindrome=false;
StringBuilder temp = new StringBuilder(Integer.toString(product)).reverse();
if(temp.toString().equals(Integer.toString(product)))
{
isPalindrome=true;
}
return isPalindrome;
}
}
What you are doing is returning and breaking out of the loop as soon as you get the first palindrome. Which in your case is not the maximum value palindrome.
Instead use an if condition and keep a track of maximum values and let the loop continue till end.
I have added the if condition that lets the loop running and registers the value.
Got the correct answer from this code.
PS. Thanks Xan for your input. I guess I could've explained it better first time.
I have seen a lot of posts for this question, this is the solution that i have come up with:
Smallest number that is multiple of two 3 digits number is 10000(100*100)
Largest number that is multiple of two 3 digits number is 998001(999*999)
Our palindrome lies between these two number, write a program to loop through these number and whenever you get a palindrome check whether its perfectly divisible by a 3 digit number and quotient is also a 3 digit number.
Below is my program in C#, the last number that it prints is our required answer, enjoy.
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text.RegularExpressions;
using System.Collections;
namespace E
{
public class Program
{
public static void Main(string[] args)
{
//Your code goes here
for(int i=10000;i<=998001;i++)
{
string s1 = i.ToString();
char[] array = s1.ToCharArray();
Array.Reverse(array);
string s2 = new String(array);
if(s1==s2)
{
for(int j=100;j<=999;j++)
{
if(i%j==0 && i/j <= 999)
{
System.Console.WriteLine(i);
continue;
}
}
}
}
System.Console.WriteLine("done");
}
}
}
I believe this should be optimal
#include <functional>
#include <algorithm>
#include <iostream>
using namespace std;
template <typename T>
bool IsPalindrome(const T num) {
T reverse = 0;
T n = num;
while (n > 0) {
reverse = (reverse * 10) + n % 10;
n /= 10;
}
return reverse == num;
}
template <typename T = long long int>
T LongestPalindromeFromProductOfNDigitNums(int n) {
T result = 0, val = 0, max_n_digit_num = std::pow(10, n)-1,
least_n_digit_num = std::pow(10, n-1);
int n_checks = 0;
for (T i = max_n_digit_num; i >= least_n_digit_num; --i) {
if ((i*i) < result) {//found the highest palindrome
break;
}
for (T j = i; j >= least_n_digit_num; --j) {
val = i*j;
++n_checks;
if (val < result) // any product i*j for the value of 'j' after this will be less than result
break;
if (IsPalindrome(val)) {
if (val > result)
result = val;
break; // whenever a palindrome is found break since we only need highest one
}
}
}
std::cout << " Total num of checks = " << n_checks << std::endl;
return result;
}
int main() {
int n = 3;
std::cout << " LongestPalindromeFromProductOfNDigitNums for n = "
<< n << " is " << LongestPalindromeFromProductOfNDigitNums(n) << std::endl;
n = 4;
std::cout << " LongestPalindromeFromProductOfNDigitNums for n = "
<< n << " is " << LongestPalindromeFromProductOfNDigitNums(n) << std::endl;
return 0;
}
http://ideone.com/WoNSJP
Swift 3:
// my approach is to make 6-digit palindrome first and then
// check if I can divide it by 3-digit number
// (you can see some visual listing at the end of the code)
// execution time on my laptop is around: 2.75409698486328 sec
import Foundation
func maxPalindrom() -> Int {
var result = 999999
var i = 9
var j = 9
var k = 9
while true {
while true {
while true {
print("in K loop: \(result) k = \(k)")
if isDivisible(number: result) {
return result
}
if k <= 0 {
k = 9
result += 9900
break
}
result -= 1100
k -= 1
}
print("in J loop: \(result)")
if isDivisible(number: result) {
return result
}
if j < 0 {
j = 9
result += 90090
break
}
result -= 10010
j -= 1
}
print("in I loop: \(result)")
if isDivisible(number: result) {
return result
}
if i < 0 {
break
}
result -= 100001
i -= 1
}
if result == 100001 {
return -1
}
return -1
}
func isDivisible(number: Int) -> Bool {
var i = 999
while true {
if number % i == 0 && number / i < 999 {
return true
}
if i < 500 {
return false
}
i -= 1
}
}
let start = NSDate()
print(maxPalindrom()) // 906609
let end = NSDate()
print("executio time: \(end.timeIntervalSince(start as Date)) sec") // ~ execution time: 2.75409698486328 sec
//in K loop: 999999 k = 9
//in K loop: 998899 k = 8
//in K loop: 997799 k = 7
//in K loop: 996699 k = 6
//in K loop: 995599 k = 5
//in K loop: 994499 k = 4
//in K loop: 993399 k = 3
//in K loop: 992299 k = 2
//in K loop: 991199 k = 1
//in K loop: 990099 k = 0
//in J loop: 999999
//in K loop: 989989 k = 9
//in K loop: 988889 k = 8
//in K loop: 987789 k = 7
//in K loop: 986689 k = 6
//in K loop: 985589 k = 5
//in K loop: 984489 k = 4
//in K loop: 983389 k = 3
.....
Most of the answers here are correct. If you want to save going through 900*900 loops, you can just loop through all palindromes between 10000 and 998001 and find if they are divisible by 3 digit number.
static void largestpalindromeproduct(){
int a=999,b=999,c=a*b,loopcounter=0;
while(c>10000){
loopcounter++;
c--;
if(isPalindrome(c))
if(isDivisible(c))
break;
}
System.out.println(" largest : " + c+ "\nloops:"+ loopcounter);
}
static boolean isDivisible(int n){
int a=999;
while(a>=100){
if(n%a==0){
if(secondDividerIs3Digit(n,a))
return true;
}
a--;
}
return false;
}
static boolean secondDividerIs3Digit(int n, int a){
Integer b=n/a;
if(b.toString().length()==3)
return true;
return false;
}
static boolean isPalindrome(int n){
Integer i=new Integer(n);
String p=i.toString();
StringBuffer s=new StringBuffer(i.toString());
s.reverse();
if(p.equals(s.toString()))
return true;
return false;
}
As a very simple solution, this one works
public class LargestPallendrome {
public static void main(String[] args) {
int a = 999;
int b = 999;
long max = 0;
while (a > 100) {
long num = a * b;
if (checkPallendrome(num)) {
if (num > max)
max = num;
}
if (b >= 100)
b--;
else {
a--;
b = 999;
}
}
System.out.println(max);
}
public static boolean checkPallendrome(long num) {
String a = num + "";
String b = new StringBuffer(num + "").reverse().toString();
if (a.equals(b))
return true;
return false;
}
}
Another Simple Solution in JavaScript
function reverseNumber(n)
{
n = n + "";
return n.split("").reverse().join("");
}
function palindrom(){
var r= 1 , y =1;
var largest = 0;
while(r <= 1000){
var num1 = r;
var num2 = 0;
while(num1 <= 1000 && num2 <= num1){
product = num1 * num2;
if (product == reverseNumber(product)){
console.log(`${num1} x ${num2} = ${product}`);
if(product > largest){
largest = product;
}
}
num1 = num1 + 1;
num2= num2 + 1;
}
r++;
}
console.log(``)
console.log(`The largest is ${largest}`);
}
console.log(palindrom());
public static void main(String[] args) {
int tempAns = 0;
int max = 999;
for (int i = 100; i <= max; i++) {
for (int j = max; j >= i; j--) {
if (findPalindrome(i * j) && (i * j) > tempAns) {
System.out.println("Palindrome: " + j + " * " + i + " = " + j * i);
tempAns = i * j;
}
}
}
}
private static boolean findPalindrome(int n) {
String nString = String.valueOf(n);
int j = 0;
int stringLength = nString.length() - 1;
for (int i = stringLength; i >= 0; i--) {
if (nString.charAt(j) == nString.charAt(i)) {
if (i == 0) {
return true;
}
j++;
} else if (nString.charAt(j) != nString.charAt(i)) {
return false;
}
}
return false;
}
This is better because its using O(N) time complexity to find all the palindrome (As calculating palindrome of a six digit no is constant) and O(N2) nearly to find the actual palindrome that too worst case the moment its finding its first no we don't have to do any more calculation and here we are actually using the worst case on possible palindromic no. So I think its better
package ProjectEuler;
import java.util.ArrayList;
import java.util.Arrays;
public class Largest_Palindrome_Product {
public static void main(String[] args) {
int count=0;
for(int i=10000;i<998002;i++) {
int x=i,c=0;
while(x!=0) {
c=c*10+x%10;
x/=10;
}
if(c==i) {
count++;
}
}
int a[]=new int[count],count1=0;
for(int i=10000;i<998002;i++) {
int x=i,c=0;
while(x!=0) {
c=c*10+x%10;
x/=10;
}
if(c==i) {
a[count1]=i;
count1++;
}
}
Arrays.sort(a);
tp:for(int i=count-1;i>=0;i--)
{
for(int j=999;j>100;j--)
if(a[i]%j==0&&a[i]/j<=999) {
System.out.println(a[i]+" "+j+" "+a[i]/j);
break tp;
}
}
}
}
This is how I did it in Javascript. Simple & easy!
let num1 = 999;
let num2 = 999;
let arr = [];
function check(x, y)
{
if(String(x*y) == String(x*y).split("").reverse().join(""))
{
return true;
}
return false;
}
for(let i=0; i<999999; i++)
{
if(check(num1, num2))
{
arr.push(num1*num2);
num1--;
num2 = num1+1;
}
num2--;
}
console.log(arr.sort((x, y) => y-x)[0]);
I check it some times with random.randint. In python 3.7.1, you should run it with CMD and after 20 sec you will get the right answer.
import random
x,y,z,a,b=100,100,' ','',0
while 100<=x<=999 and 100<=y<=999:
a=x*y
x=random.randint(900,999)
y=random.randint(900,999)
print(x,' x ',y,'=')
z=len(str(a))
if z==6:
if str(a)[0] == str(a)[5]:
if str(a)[1] == str(a)[4]:
if str(a)[2] == str(a)[3]:
print(a,'yes')
exit(a)
else:
pass
#906609
Readable option:
function maxPalindrome(num) {
let maxPalindrome = 1;
for (let i = num; i > 0; i--) {
for (let j = num; j > 0; j--) {
const product = i * j;
if (
product.toString() === product.toString().split("").reverse().join("")
&& product > maxPalindrome
) {
maxPalindrome = product;
}
}
}
return maxPalindrome;
}
console.log(maxPalindrome(999));
This is how I have done with C#:
public static void maxPali() {
int max = 0;
for (int i = 99; i >= 10; i--) {
for (int j = 99; j >= 10; j--) {
if (i*j == reverse(i*j))
max = max >= (i*j) ? max : (i*j);
}
}
Console.WriteLine(max);
}
public static int reverse(int num) {
int rev = 0;
while (num > 0) {
int rem = num % 10;
rev = (rev * 10) + rem;
num /= 10;
}
return rev;
}
JavaScript solution:
(function main() {
let start = 100,
stop = 999,
step = 1;
let arr = Array(Math.ceil((stop + step - start) /
step)).fill(start).map((x, y) => x + y * step);
let max = 0;
arr.slice(0).reverse().map(function(i) {
arr.slice(0).reverse().map(function(j) {
if (i*j == (i*j).toString().split('').reverse().join(''))
if (max < (i*j))
max = i*j;
});
});
console.log(max); }());