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java- bool func that gets an integer parameter and checks duplicates without array

My task is to
write a function that gets an integer as input and
return true if the number has at lest 2 identical digits, false otherwise,
without using special libraries or using arrays
my code, but it only works for consecutive digits
public static boolean check_num(int num) {
int count = 1; //number itself
int tmp = 0;
int tmp1 = 0;
while (num != 0) {
tmp = num % 10;
num /= 10;
tmp1 = num % 10;
while(count>=2) {
if(tmp1==tmp) count++;
}
}
if(count==2 || count>2) return true;
return false;
}
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
System.out.println("enter number");
int num = input.nextInt();
System.out.println(check_num(num));
}
}

To resolve this task we need to track if an appropriate decimal digit has been found in the given number or not. The simplest way could be to use a 10-element boolean array, but as long as the arrays are not allowed, another int number can be used to store the mapping of digits, for example, for a number 1234 the mapping would look like (lower bits last): 11110.
The bit corresponding to a decimal digit is set using right shift and XOR operation ^. If the result of XOR is 0, then the duplicate digit is detected.
An example implementation may look like this (including debug messages):
public static boolean checkNum(int n) {
if (n < 10) return false; // no need to check 1-digit number
int arr = 0;
while (n > 0) {
int digit = n % 10;
int bit = 1 << digit;
arr ^= bit;
if (0 == (arr & bit)) {
System.out.println("Dup found: " + Integer.toBinaryString(arr) + "; digit=" + digit);
return true; // duplicate found
}
n /= 10;
}
System.out.println("No dup found, " + Integer.toBinaryString(arr));
return false;
}
Tests:
for (int x : new int[] {1, 10, 11, 1234, 1020, 12344, 1012345}) {
System.out.printf("checkNum(%d) = %s%n----%n", x, checkNum(x));
}
Output:
checkNum(1) = false
----
No dup found, 11
checkNum(10) = false
----
Dup found: 0; digit=1
checkNum(11) = true
----
No dup found, 11110
checkNum(1234) = false
----
Dup found: 100; digit=0
checkNum(1020) = true
----
Dup found: 0; digit=4
checkNum(12344) = true
----
Dup found: 111101; digit=1
checkNum(1012345) = true
----

Standardly you'd use a Set or an array to record which digits you've seen previously. However, if these are not available you can build upon the approach you're already taking - checking if the two low digits (10s and 1s) are the same. However, you'll need to bring each digit next to every other digit for this to work.
Here's some Java code to illustrate:
static boolean check_num(int n)
{
for(int i = n; i > 10; i /= 10)
for(int j = i; j > 10; j = 10 * (j / 100) + (j % 10))
if(j % 10 == (j / 10) % 10) return true;
return false;
}
Test:
System.out.println(check_num(543521));
System.out.println(check_num(654321));
Output:
true
false

Fibonacci Modified:How to fix this algorithm?

I have this problem in front of me and I can't figure out how to solve it.
It's about the series 0,1,1,2,5,29,866... (Every number besides the first two is the sum of the squares of the previous two numbers (2^2+5^2=29)).
In the first part I had to write an algorithm (Not a native speaker so I don't really know the terminology) that would receive a place in the series and return it's value (6 returned 29)
This is how I wrote it:
public static int mod(int n)
{
if (n==1)
return 0;
if (n==2)
return 1;
else
return (int)(Math.pow(mod(n-1), 2))+(int)(Math.pow(mod(n-2), 2));
}
However, now I need that the algorithm will receive a number and return the total sum up to it in the series (6- 29+5+2+1+1+0=38)
I have no idea how to do this, I am trying but I am really unable to understand recursion so far, even if I wrote something right, how can I check it to be sure? And how generally to reach the right algorithm?
Using any extra parameters is forbidden.
Thanks in advance!

We want:
mod(1) = 0
mod(2) = 0+1
mod(3) = 0+1+1
mod(4) = 0+1+1+2
mod(5) = 0+1+1+2+5
mod(6) = 0+1+1+2+5+29
and we know that each term is defined as something like:
2^2+5^2=29
So to work out mod(7) we need to add the next term in the sequence x to mod(6).
Now we can work out the term using mod:
x = term(5)^2 + term(6)^2
term(5) = mod(5) - mod(4)
term(6) = mod(6) - mod(5)
x = (mod(5)-mod(4))^2 + (mod(6)-mod(5))^2
So we can work out mod(7) by evaluating mod(4),mod(5),mod(6) and combining the results.
Of course, this is going to be incredibly inefficient unless you memoize the function!
Example Python code:
def f(n):
if n<=0:
return 0
if n==1:
return 1
a=f(n-1)
b=f(n-2)
c=f(n-3)
return a+(a-b)**2+(b-c)**2
for n in range(10):
print f(n)
prints:
0
1
2
4
9
38
904
751701
563697636866
317754178345850590849300

How about this? :)
class Main {
public static void main(String[] args) {
final int N = 6; // Your number here.
System.out.println(result(N));
}
private static long result(final int n) {
if (n == 0) {
return 0;
} else {
return element(n) + result(n - 1);
}
}
private static long element(final int n) {
if (n == 1) {
return 0L;
} else if (n == 2) {
return 1L;
} else {
return sqr(element(n - 2)) + sqr(element(n - 1));
}
}
private static long sqr(final long x) {
return x * x;
}
}
Here is the idea that separate function (element) is responsible for finding n-th element in the sequence, and result is responsible for summing them up. Most probably there is a more efficient solution though. However, there is only one parameter.

I can think of a way of doing this with the constraints in your comments but it's a total hack. You need one method to do two things: find the current value and add previous values. One option is to use negative numbers to flag one of those function:
int f(int n) {
if (n > 0)
return f(-n) + f(n-1);
else if (n > -2)
return 0;
else if (n == -2)
return 1;
else
return f(n+1)*f(n+1)+f(n+2)*f(n+2);
}
The first 8 numbers output (before overflow) are:
0
1
2
4
9
38
904
751701
I don't recommend this solution but it does meet your constraints of being a single recursive method with a single argument.

Here is my proposal.
We know that:
f(n) = 0; n < 2
f(n) = 1; 2 >= n <= 3
f(n) = f(n-1)^2 + f(n-2)^2; n>3
So:
f(0)= 0
f(1)= 0
f(2)= f(1) + f(0) = 1
f(3)= f(2) + f(1) = 1
f(4)= f(3) + f(2) = 2
f(5)= f(4) + f(3) = 5
and so on
According with this behaivor we must implement a recursive function to return:
Total = sum f(n); n= 0:k; where k>0
I read you can use a static method but not use more than one parameter into the function. So, i used a static variable with the static method, just for control the execution of loop:
class Dummy
{
public static void main (String[] args) throws InterruptedException {
int n=10;
for(int i=1; i<=n; i++)
{
System.out.println("--------------------------");
System.out.println("Total for n:" + i +" = " + Dummy.f(i));
}
}
private static int counter = 0;
public static long f(int n)
{
counter++;
if(counter == 1)
{
long total = 0;
while(n>=0)
{
total += f(n);
n--;
}
counter--;
return total;
}
long result = 0;
long n1=0,n2=0;
if(n >= 2 && n <=3)
result++; //Increase 1
else if(n>3)
{
n1 = f(n-1);
n2 = f(n-2);
result = n1*n1 + n2*n2;
}
counter--;
return result;
}
}
the output:
--------------------------
Total for n:1 = 0
--------------------------
Total for n:2 = 1
--------------------------
Total for n:3 = 2
--------------------------
Total for n:4 = 4
--------------------------
Total for n:5 = 9
--------------------------
Total for n:6 = 38
--------------------------
Total for n:7 = 904
--------------------------
Total for n:8 = 751701
--------------------------
Total for n:9 = 563697636866
--------------------------
Total for n:10 = 9011676203564263700
I hope it helps you.
UPDATE: Here is another version without a static method and has the same output:
class Dummy
{
public static void main (String[] args) throws InterruptedException {
Dummy app = new Dummy();
int n=10;
for(int i=1; i<=n; i++)
{
System.out.println("--------------------------");
System.out.println("Total for n:" + i +" = " + app.mod(i));
}
}
private static int counter = 0;
public long mod(int n)
{
Dummy.counter++;
if(counter == 1)
{
long total = 0;
while(n>=0)
{
total += mod(n);
n--;
}
Dummy.counter--;
return total;
}
long result = 0;
long n1=0,n2=0;
if(n >= 2 && n <=3)
result++; //Increase 1
else if(n>3)
{
n1 = mod(n-1);
n2 = mod(n-2);
result = n1*n1 + n2*n2;
}
Dummy.counter--;
return result;
}
}

Non-recursive|Memoized
You should not use recursion since it will not be good in performance.
Use memoization instead.
def FibonacciModified(n):
fib = [0]*n
fib[0],fib[1]=0,1
for idx in range(2,n):
fib[idx] = fib[idx-1]**2 + fib[idx-2]**2
return fib
if __name__ == '__main__':
fib = FibonacciModified(8)
for x in fib:
print x
Output:
0
1
1
2
5
29
866
750797
The above will calculate every number in the series once[not more than that].
While in recursion an element in the series will be calculated multiple times irrespective of the fact that the number was calculated before.
http://www.geeksforgeeks.org/program-for-nth-fibonacci-number/

Minimum steps to one logic fails for some conditions

The puzzle is to obtain the minimum number of steps it takes to make a number 1. Allowed operations are
1. You can subtract 1 from the number
2. You can divide the number by 2 if it is divisible by 2.
3. You can divide the number by 3 if it is divisible by 3.
At the end, you need to make the number 1 by performing the above operations. I am trying to obtain a solution that gives me the minimum number of the above operations required to make the number 1.
My code (in Java) is as follows.
public int minStepsBottomUp(int n) {
int[] memoArray = new int[n+1];
memoArray[0] = 0;
memoArray[1] = 0;
for(int i=2;i<=n;++i){
int r = 1 + memoArray[i-1];
if(n % 2 == 0) {
r = Math.min(r, 1+memoArray[n/2]);
}
if(n % 3 == 0) {
r = Math.min(r, 1+memoArray[n/3]);
}
memoArray[i] = r;
}
return memoArray[n];
}
But I get some ambiguous results.Example - if the number is 5, I get the minimun number of steps required as 4. Actually it should be 3. Can someone please explain where I have gone wrong?

I suggest reversing the problem: starting from 1 we should reach n by using three kinds of operations:
add 1
multiply by 2
multiply by 3
For instance for 5 we'll have 3 operations (multiply by 3, add 1, add 1):
1 -> 3 -> 4 -> 5
So far so good, now we have standard dynamic programming problem; C# implementation:
private static int Best(int value) {
if (value <= 0)
return -1; // or throw ArgumentOutOfRangeException
else if (value == 1)
return 0;
Dictionary<int, int> best = new Dictionary<int, int>() { {1, 0} };
List<int> agenda = new List<int>() { 1 };
for (int step = 1; ; ++step)
for (int i = agenda.Count - 1; i >= 0; --i) {
int item = agenda[i];
agenda.RemoveAt(i);
int[] next = new int[] { item + 1, item * 2, item * 3 };
foreach (int v in next) {
if (v == value)
return step;
if (!best.ContainsKey(v)) {
best.Add(v, step);
agenda.Add(v);
}
}
}
}
Tests:
// 3
Console.WriteLine(Best(5));
// 3
Console.WriteLine(Best(10));
// 7
Console.WriteLine(Best(100));
// 19
Console.WriteLine(Best(1000000));

Inside your loop, you are using n instead of i.
For instance, n % 2 == 0 should be i % 2 == 0

If the number is 5 you can get 1 by doing:
int x = 5 - 1;
x = x - 1;
x= x / 3;

Comparing integers and creating the smallest integer possible from the digits of the given integers

I need to write the following method: accepts two integer parameters and returns an integer. If either integer is not a 4 digit number than the method should return the smaller integer. Otherwise, the method should return a four digit integer made up of the smallest digit in the thousands place, hundreds place, tens place and ones place. We cannot turn the integers into Strings, or use lists, or arrays.
For example biggestLoser(6712,1234) returns 1212
For example biggestLoser(19,8918) returns 19
Here's how I've started to write it:
public static int biggestLoser(int a, int b){
if(a<9999 || b<9999){
if(a<b)
return a;
else if(b<a)
return b;
}
int at=a/1000;
int ah=a%1000/100;
int an=a%100/10;
int ae=a%10;
int bt=b/1000;
int bh=b%1000/100;
int bn=b%100/10;
int be=a%10;
if(at<bt && ah<bh && an<bn && ae<be)
return at*1000+ah*100+an*10+ae;
else if(at<bt && ah<bh && an<bn && be<ae)
return at*1000+ah*100+an*10+be;
else if(at<bt&& ah<bh && bn<an && ae<be)
else return at*1000+ah*100+bn*10+ae;
However, it looks like I'm going to have to write way too many if statements, is there a shorter way to write the code?

public static int biggestLoser(int a, int b) {
if (a < 1000 || a >= 10000 || b < 1000 || b >= 10000) {
return Math.min(a, b);
} else {
// both a and b are four digits
int result = 0 ;
int multiplier = 1 ;
for (int digit = 0; digit < 4; digit++) {
int nextDigit = Math.min(a % 10, b % 10);
result = result + nextDigit * multiplier ;
multiplier = multiplier * 10 ;
a = a / 10 ;
b = b / 10 ;
}
return result ;
}
}
How does this work? a % 10 is the remainder when a is divided by 10: in other words it is the least significant digit of a (the "ones place").
a = a / 10 performs integer division, so it divides a by 10 and ignores any fraction. So 1234 becomes 123, and on the next iteration 123 becomes 12, etc. In other words, it discards the "ones place".
So the first time through the loop, you look at the "ones" from a and b, find the smallest one, and add it to result. Then you drop the "ones" from both a and b. So what used to be the "tens" are now the "ones". The second time through the loop, you get the smallest "ones" again: but this was originally the smallest "tens". You want to add that to result, but you need to multiply by 10. This is the multiplier: each time through the loop the multiplier is multiplied by 10. So each time, you get the smallest "ones", multiply by the correct thing, add to the result, and then drop the "ones" from a and b.
Just for fun, here's an implementation that needs only one statement (and works if you replace "four digits" with any positive number of digits). You can ask your instructor to explain it ;).
public static final int NUM_DIGITS = 4 ;
public static final int MAX = (int) Math.pow(10, NUM_DIGITS) ;
public static final int MIN = MAX / 10 ;
public static int biggestLoser(int a, int b) {
return (a < MIN || a >= MAX || b < MIN || b >= MAX) ? Math.min(a, b) :
IntStream.iterate(1, multiplier -> multiplier * 10).limit(NUM_DIGITS)
.map(multiplier -> Math.min((a / multiplier) % 10, (b / multiplier) % 10) * multiplier )
.sum();
}

maybe it is stupid but try to take advantage of String ( .charAt(int index) )and Integer ( .parseInt( String value ) ) methods , maybe this example help you :
int x=145;
int y=826;
//to know which number have the biggest tens
String a=x+"";
String b=y+"";
if(Integer.parseInt(a.charAt(1)+"")>Integer.parseInt(b.charAt(1)+""))
{
System.out.println("The number which have the biggest tens is "+a);
}
else
{
System.out.println("The number which have the biggest tens is "+b);
}

Using String and StringBuilder
public class Test
{
public static void main(String []args)
{
System.out.println(biggestLooser(6712,1234));
}
public static int biggestLooser(int _a, int _b)
{
String a = String.valueOf(_a);
String b = String.valueOf(_b);
StringBuilder c = new StringBuilder();
if(a.length() < b.length()) return Integer.parseInt(a);
else if(b.length() < a.length()) return Integer.parseInt(b);
else if(a.length() >= 4 && b.length() >= 4)
{
for(int i = 4; i > 0; i--)
{
char ch = '\0';
if(a.charAt(a.length() - i) < b.charAt(b.length() - i))
ch = a.charAt(a.length() - i);
else ch = b.charAt(b.length() - i);
c.append(ch);
}
return Integer.parseInt(c.toString());
}
else return -1;
}
}
//ouput: 1212

Here is the simple answer
public static int biggestLoser(int a, int b) {
if (a < 1000 || b < 1000) {
if (a < b)
return a;
else
return b;
}
int val = 0;
ArrayList<Integer> data1 = new ArrayList<Integer>();
while (a > 0) {
data1.add(a % 10);
a /= 10;
}
Collections.reverse(data1);
ArrayList<Integer> data2 = new ArrayList<Integer>();
while (b > 0) {
data2.add(b % 10);
b /= 10;
}
Collections.reverse(data2);
val = ((data1.get(0) < data2.get(0)) ? data1.get(0) : data2.get(0))
* 1000
+ ((data1.get(1) < data2.get(1)) ? data1.get(1) : data2.get(1))
* 100
+ ((data1.get(2) < data2.get(2)) ? data1.get(2) : data2.get(2))
* 10
+ ((data1.get(3) < data2.get(3)) ? data1.get(3) : data2.get(3));
return val;
}

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

package testing.project;
public class PalindromeThreeDigits {
public static void main(String[] args) {
int value = 0;
for(int i = 100;i <=999;i++)
{
for(int j = i;j <=999;j++)
{
int value1 = i * j;
StringBuilder sb1 = new StringBuilder(""+value1);
String sb2 = ""+value1;
sb1.reverse();
if(sb2.equals(sb1.toString()) && value<value1) {
value = value1;
}
}
}
System.out.println(value);
}
}
This is the code that I wrote in Java... Is there any efficient way other than this.. And can we optimize this code more??

We suppose the largest such palindrome will have six digits rather than five, because 143*777 = 111111 is a palindrome.
As noted elsewhere, a 6-digit base-10 palindrome abccba is a multiple of 11. This is true because a*100001 + b*010010 + c*001100 is equal to 11*a*9091 + 11*b*910 + 11*c*100. So, in our inner loop we can decrease n by steps of 11 if m is not a multiple of 11.
We are trying to find the largest palindrome under a million that is a product of two 3-digit numbers. To find a large result, we try large divisors first:
We step m downwards from 999, by 1's;
Run n down from 999 by 1's (if 11 divides m, or 9% of the time) or from 990 by 11's (if 11 doesn't divide m, or 91% of the time).
We keep track of the largest palindrome found so far in variable q. Suppose q = r·s with r <= s. We usually have m < r <= s. We require m·n > q or n >= q/m. As larger palindromes are found, the range of n gets more restricted, for two reasons: q gets larger, m gets smaller.
The inner loop of attached program executes only 506 times, vs the ~ 810000 times the naive program used.
#include <stdlib.h>
#include <stdio.h>
int main(void) {
enum { A=100000, B=10000, C=1000, c=100, b=10, a=1, T=10 };
int m, n, p, q=111111, r=143, s=777;
int nDel, nLo, nHi, inner=0, n11=(999/11)*11;
for (m=999; m>99; --m) {
nHi = n11; nDel = 11;
if (m%11==0) {
nHi = 999; nDel = 1;
}
nLo = q/m-1;
if (nLo < m) nLo = m-1;
for (n=nHi; n>nLo; n -= nDel) {
++inner;
// Check if p = product is a palindrome
p = m * n;
if (p%T==p/A && (p/B)%T==(p/b)%T && (p/C)%T==(p/c)%T) {
q=p; r=m; s=n;
printf ("%d at %d * %d\n", q, r, s);
break; // We're done with this value of m
}
}
}
printf ("Final result: %d at %d * %d inner=%d\n", q, r, s, inner);
return 0;
}
Note, the program is in C but same techniques will work in Java.

What I would do:
Start at 999, working my way backwards to 998, 997, etc
Create the palindrome for my current number.
Determine the prime factorization of this number (not all that expensive if you have a pre-generated list of primes.
Work through this prime factorization list to determine if I can use a combination of the factors to make 2 3 digit numbers.
Some code:
int[] primes = new int[] {2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,
73,79,83,89,97,101,103,107,109,113,,127,131,137,139,149,151,157,163,167,173,
179,181,191,193,197,199,211,223,227,229,233,239,241,251,257,263,269,271,277,281,
283,293,307,311,313,317,331,337,347,349,353,359,367,373,379,383,389,397,401,409,
419,421,431,433,439,443,449,457,461,463,467,479,487,491,499,503,509,521,523,541,
547,557,563,569,571,577,587,593,599,601,607,613,617,619,631,641,643,647,653,659,
661,673,677,683,691,701,709,719,727,733,739,743,751,757,761,769,773,787,797,809,
811,821,823,827,829,839,853,857,859,863,877,881,883,887,907,911,919,929,937,941,
947,953,967,971,977,983,991,997};
for(int i = 999; i >= 100; i--) {
String palstr = String.valueOf(i) + (new StringBuilder().append(i).reverse());
int pal = Integer.parseInt(pal);
int[] factors = new int[20]; // cannot have more than 20 factors
int remainder = pal;
int facpos = 0;
primeloop:
for(int p = 0; p < primes.length; i++) {
while(remainder % p == 0) {
factors[facpos++] = p;
remainder /= p;
if(remainder < p) break primeloop;
}
}
// now to do the combinations here
}

We can translate the task into the language of mathematics.
For a short start, we use characters as digits:
abc * xyz = n
abc is a 3-digit number, and we deconstruct it as 100*a+10*b+c
xyz is a 3-digit number, and we deconstruct it as 100*x+10*y+z
Now we have two mathematical expressions, and can define a,b,c,x,y,z as € of {0..9}.
It is more precise to define a and x as of element from {1..9}, not {0..9}, because 097 isn't really a 3-digit number, is it?
Ok.
If we want to produce a big number, we should try to reach a 9......-Number, and since it shall be palindromic, it has to be of the pattern 9....9. If the last digit is a 9, then from
(100*a + 10*b + c) * (100*x + 10*y + z)
follows that z*c has to lead to a number, ending in digit 9 - all other calculations don't infect the last digit.
So c and z have to be from (1,3,7,9) because (1*9=9, 9*1=9, 3*3=9, 7*7=49).
Now some code (Scala):
val n = (0 to 9)
val m = n.tail // 1 to 9
val niners = Seq (1, 3, 7, 9)
val highs = for (a <- m;
b <- n;
c <- niners;
x <- m;
y <- n;
z <- niners) yield ((100*a + 10*b + c) * (100*x + 10*y + z))
Then I would sort them by size, and starting with the biggest one, test them for being palindromic. So I would omit to test small numbers for being palindromic, because that might not be so cheap.
For aesthetic reasons, I wouldn't take a (toString.reverse == toString) approach, but a recursive divide and modulo solution, but on todays machines, it doesn't make much difference, does it?
// Make a list of digits from a number:
def digitize (z: Int, nums : List[Int] = Nil) : List[Int] =
if (z == 0) nums else digitize (z/10, z%10 :: nums)
/* for 342243, test 3...==...3 and then 4224.
Fails early for 123329 */
def palindromic (nums : List[Int]) : Boolean = nums match {
case Nil => true
case x :: Nil => true
case x :: y :: Nil => x == y
case x :: xs => x == xs.last && palindromic (xs.init) }
def palindrom (z: Int) = palindromic (digitize (z))
For serious performance considerations, I would test it against a toString/reverse/equals approach. Maybe it is worse. It shall fail early, but division and modulo aren't known to be the fastest operations, and I use them to make a List from the Int. It would work for BigInt or Long with few redeclarations, and works nice with Java; could be implemented in Java but look different there.
Okay, putting the things together:
highs.filter (_ > 900000) .sortWith (_ > _) find (palindrom)
res45: Option[Int] = Some(906609)
There where 835 numbers left > 900000, and it returns pretty fast, but I guess even more brute forcing isn't much slower.
Maybe there is a much more clever way to construct the highest palindrom, instead of searching for it.
One problem is: I didn't knew before, that there is a solution > 900000.
A very different approach would be, to produce big palindromes, and deconstruct their factors.

public class Pin
{
public static boolean isPalin(int num)
{
char[] val = (""+num).toCharArray();
for(int i=0;i<val.length;i++)
{
if(val[i] != val[val.length - i - 1])
{
return false;
}
}
return true;
}
public static void main(String[] args)
{
for(int i=999;i>100;i--)
for(int j=999;j>100;j--)
{
int mul = j*i;
if(isPalin(mul))
{
System.out.printf("%d * %d = %d",i,j,mul);
return;
}
}
}
}

package ex;
public class Main {
public static void main(String[] args) {
int i = 0, j = 0, k = 0, l = 0, m = 0, n = 0, flag = 0;
for (i = 999; i >= 100; i--) {
for (j = i; j >= 100; j--) {
k = i * j;
// System.out.println(k);
m = 0;
n = k;
while (n > 0) {
l = n % 10;
m = m * 10 + l;
n = n / 10;
}
if (m == k) {
System.out.println("pal " + k + " of " + i + " and" + j);
flag = 1;
break;
}
}
if (flag == 1) {
// System.out.println(k);
break;
}
}
}
}

A slightly different approach that can easily calculate the largest palindromic number made from the product of up to two 6-digit numbers.
The first part is to create a generator of palindrome numbers. So there is no need to check if a number is palindromic, the second part is a simple loop.
#include <memory>
#include <iostream>
#include <cmath>
using namespace std;
template <int N>
class PalindromeGenerator {
unique_ptr <int []> m_data;
bool m_hasnext;
public :
PalindromeGenerator():m_data(new int[N])
{
for(auto i=0;i<N;i++)
m_data[i]=9;
m_hasnext=true;
}
bool hasNext() const {return m_hasnext;}
long long int getnext()
{
long long int v=0;
long long int b=1;
for(int i=0;i<N;i++){
v+=m_data[i]*b;
b*=10;
}
for(int i=N-1;i>=0;i--){
v+=m_data[i]*b;
b*=10;
}
auto i=N-1;
while (i>=0)
{
if(m_data[i]>=1) {
m_data[i]--;
return v;
}
else
{
m_data[i]=9;
i--;
}
}
m_hasnext=false;
return v;
}
};
template<int N>
void findmaxPalindrome()
{
PalindromeGenerator<N> gen;
decltype(gen.getnext()) minv=static_cast<decltype(gen.getnext())> (pow(10,N-1));
decltype(gen.getnext()) maxv=static_cast<decltype(gen.getnext())> (pow(10,N)-1);
decltype(gen.getnext()) start=11*(maxv/11);
while(gen.hasNext())
{
auto v=gen.getnext();
for (decltype(gen.getnext()) i=start;i>minv;i-=11)
{
if (v%i==0)
{
auto r=v/i;
if (r>minv && r<maxv ){
cout<<"done:"<<v<<" "<<i<< "," <<r <<endl;
return ;
}
}
}
}
return ;
}
int main(int argc, char* argv[])
{
findmaxPalindrome<6>();
return 0;
}

You can use the fact that 11 is a multiple of the palindrome to cut down on the search space. We can get this since we can assume the palindrome will be 6 digits and >= 111111.
e.g. ( from projecteuler ;) )
P= xyzzyx = 100000x + 10000y + 1000z + 100z + 10y +x
P=100001x+10010y+1100z
P=11(9091x+910y+100z)
Check if i mod 11 != 0, then the j loop can be subtracted by 11 (starting at 990) since at least one of the two must be divisible by 11.

You can try the following which prints
999 * 979 * 989 = 967262769
largest palindrome= 967262769 took 0.015
public static void main(String... args) throws IOException, ParseException {
long start = System.nanoTime();
int largestPalindrome = 0;
for (int i = 999; i > 100; i--) {
LOOP:
for (int j = i; j > 100; j--) {
for (int k = j; k > 100; k++) {
int n = i * j * k;
if (n < largestPalindrome) continue LOOP;
if (isPalindrome(n)) {
System.out.println(i + " * " + j + " * " + k + " = " + n);
largestPalindrome = n;
}
}
}
}
long time = System.nanoTime() - start;
System.out.printf("largest palindrome= %d took %.3f seconds%n", largestPalindrome, time / 1e9);
}
private static boolean isPalindrome(int n) {
if (n >= 100 * 1000 * 1000) {
// 9 digits
return n % 10 == n / (100 * 1000 * 1000)
&& (n / 10 % 10) == (n / (10 * 1000 * 1000) % 10)
&& (n / 100 % 10) == (n / (1000 * 1000) % 10)
&& (n / 1000 % 10) == (n / (100 * 1000) % 10);
} else if (n >= 10 * 1000 * 1000) {
// 8 digits
return n % 10 == n / (10 * 1000 * 1000)
&& (n / 10 % 10) == (n / (1000 * 1000) % 10)
&& (n / 100 % 10) == (n / (100 * 1000) % 10)
&& (n / 1000 % 10) == (n / (10 * 1000) % 10);
} else if (n >= 1000 * 1000) {
// 7 digits
return n % 10 == n / (1000 * 1000)
&& (n / 10 % 10) == (n / (100 * 1000) % 10)
&& (n / 100 % 10) == (n / (10 * 1000) % 10);
} else throw new AssertionError();
}

i did this my way , but m not sure if this is the most efficient way of doing this .
package problems;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
public class P_4 {
/**
* #param args
* #throws IOException
*/
static int[] arry = new int[6];
static int[] arry2 = new int[6];
public static boolean chk()
{
for(int a=0;a<arry.length;a++)
if(arry[a]!=arry2[a])
return false;
return true;
}
public static void main(String[] args) throws IOException {
// TODO Auto-generated method stub
InputStreamReader ir = new InputStreamReader(System.in);
BufferedReader br = new BufferedReader(ir);
int temp,z,i;
for(int x=999;x>100;x--)
for(int y=999;y>100;y--)
{
i=0;
z=x*y;
while(z>0)
{
temp=z%10;
z=z/10;
arry[i]=temp;
i++;
}
for(int k = arry.length;k>0;k--)
arry2[arry.length- k]=arry[k-1];
if(chk())
{
System.out.print("pelindrome = ");
for(int l=0;l<arry2.length;l++)
System.out.print(arry2[l]);
System.out.println(x);
System.out.println(y);
}
}
}
}

This is code in C, a little bit long, but gets the job done.:)
#include <stdio.h>
#include <stdlib.h>
/*
A palindromic number reads the same both ways. The largest palindrome made from the product of two
2-digit numbers is 9009 = 91 99.
Find the largest palindrome made from the product of two 3-digit numbers.*/
int palndr(int b)
{
int *x,*y,i=0,j=0,br=0;
int n;
n=b;
while(b!=0)
{
br++;
b/=10;
}
x=(int *)malloc(br*sizeof(int));
y=(int *)malloc(br*sizeof(int));
int br1=br;
while(n!=0)
{
x[i++]=y[--br]=n%10;
n/=10;
}
int ind = 1;
for(i=0;i<br1;i++)
if(x[i]!=y[i])
ind=0;
free(x);
free(y);
return ind;
}
int main()
{
int i,cek,cekmax=1;
int j;
for(i=100;i<=999;i++)
{
for(j=i;j<=999;j++)
{
cek=i*j;
if(palndr(cek))
{
if(pp>cekmax)
cekmax=cek;
}
}
}
printf("The largest palindrome is: %d\n\a",cekmax);
}

You can actually do it with Python, it's easy just take a look:
actualProduct = 0
highestPalindrome = 0
# Setting the numbers. In case it's two digit 10 and 99, in case is three digit 100 and 999, etc.
num1 = 100
num2 = 999
def isPalindrome(number):
number = str(number)
reversed = number[::-1]
if number==reversed:
return True
else:
return False
a = 0
b = 0
for i in range(num1,num2+1):
for j in range(num1,num2+1):
actualProduct = i * j
if (isPalindrome(actualProduct) and (highestPalindrome < actualProduct)):
highestPalindrome = actualProduct
a = i
b = j
print "Largest palindrome made from the product of two %d-digit numbers is [ %d ] made of %d * %d" % (len(str(num1)), highestPalindrome, a, b)

Since we are not cycling down both iterators (num1 and num2) at the same time, the first palindrome number we find will be the largest. We don’t need to test to see if the palindrome we found is the largest. This significantly reduces the time it takes to calculate.
package testing.project;
public class PalindromeThreeDigits {
public static void main(String[] args) {
int limit = 99;
int max = 999;
int num1 = max, num2, prod;
while(num1 > limit)
{
num2 = num1;
while(num2 > limit)
{
total = num1 * num2;
StringBuilder sb1 = new StringBuilder(""+prod);
String sb2 = ""+prod;
sb1.reverse();
if( sb2.equals(sb1.toString()) ) { //optimized here
//print and exit
}
num2--;
}
num1--;
}
}//end of main
}//end of class PalindromeThreeDigits

I tried the solution by Tobin joy and vickyhacks and both of them produce the result 580085 which is wrong here is my solution, though very clumsy:
import java.util.*;
class ProjEu4
{
public static void main(String [] args) throws Exception
{
int n=997;
ArrayList<Integer> al=new ArrayList<Integer>();
outerloop:
while(n>100){
int k=reverse(n);
int fin=n*1000+k;
al=findfactors(fin);
if(al.size()>=2)
{
for(int i=0;i<al.size();i++)
{
if(al.contains(fin/al.get(i))){
System.out.println(fin+" factors are:"+al.get(i)+","+fin/al.get(i));
break outerloop;}
}
}
n--;
}
}
private static ArrayList<Integer> findfactors(int fin)
{
ArrayList<Integer> al=new ArrayList<Integer>();
for(int i=100;i<=999;i++)
{
if(fin%i==0)
al.add(i);
}
return al;
}
private static int reverse(int number)
{
int reverse = 0;
while(number != 0){
reverse = (reverse*10)+(number%10);
number = number/10;
}
return reverse;
}
}

Most probably it is replication of one of the other solution but it looks simple owing to pythonified code ,even it is a bit brute-force.
def largest_palindrome():
largest_palindrome = 0;
for i in reversed(range(1,1000,1)):
for j in reversed(range(1, i+1, 1)):
num = i*j
if check_palindrome(str(num)) and num > largest_palindrome :
largest_palindrome = num
print "largest palindrome ", largest_palindrome
def check_palindrome(term):
rev_term = term[::-1]
return rev_term == term

What about : in python
>>> for i in range((999*999),(100*100), -1):
... if str(i) == str(i)[::-1]:
... print i
... break
...
997799
>>>

I believe there is a simpler approach: Examine palindromes descending from the largest product of two three digit numbers, selecting the first palindrome with two three digit factors.
Here is the Ruby code:
require './palindrome_range'
require './prime'
def get_3_digit_factors(n)
prime_factors = Prime.factors(n)
rf = [prime_factors.pop]
rf << prime_factors.shift while rf.inject(:*) < 100 || prime_factors.inject(:*) > 999
lf = prime_factors.inject(:*)
rf = rf.inject(:*)
lf < 100 || lf > 999 || rf < 100 || rf > 999 ? [] : [lf, rf]
end
def has_3_digit_factors(n)
return !get_3_digit_factors(n).empty?
end
pr = PalindromeRange.new(0, 999 * 999)
n = pr.downto.find {|n| has_3_digit_factors(n)}
puts "Found #{n} - Factors #{get_3_digit_factors(n).inspect}, #{Prime.factors(n).inspect}"
prime.rb:
class Prime
class<<self
# Collect all prime factors
# -- Primes greater than 3 follow the form of (6n +/- 1)
# Being of the form 6n +/- 1 does not mean it is prime, but all primes have that form
# See http://primes.utm.edu/notes/faq/six.html
# -- The algorithm works because, while it will attempt non-prime values (e.g., (6 *4) + 1 == 25),
# they will fail since the earlier repeated division (e.g., by 5) means the non-prime will fail.
# Put another way, after repeatedly dividing by a known prime, the remainder is itself a prime
# factor or a multiple of a prime factor not yet tried (e.g., greater than 5).
def factors(n)
square_root = Math.sqrt(n).ceil
factors = []
while n % 2 == 0
factors << 2
n /= 2
end
while n % 3 == 0
factors << 3
n /= 3
end
i = 6
while i < square_root
[(i - 1), (i + 1)].each do |f|
while n % f == 0
factors << f
n /= f
end
end
i += 6
end
factors << n unless n == 1
factors
end
end
end
palindrome_range.rb:
class PalindromeRange
FIXNUM_MAX = (2**(0.size * 8 -2) -1)
def initialize(min = 0, max = FIXNUM_MAX)
#min = min
#max = max
end
def downto
return enum_for(:downto) unless block_given?
n = #max
while n >= #min
yield n if is_palindrome(n)
n -= 1
end
nil
end
def each
return upto
end
def upto
return enum_for(:downto) unless block_given?
n = #min
while n <= #max
yield n if is_palindrome(n)
n += 1
end
nil
end
private
def is_palindrome(n)
s = n.to_s
i = 0
j = s.length - 1
while i <= j
break if s[i] != s[j]
i += 1
j -= 1
end
i > j
end
end

public class ProjectEuler4 {
public static void main(String[] args) {
int x = 999; // largest 3-digit number
int largestProduct = 0;
for(int y=x; y>99; y--){
int product = x*y;
if(isPalindormic(x*y)){
if(product>largestProduct){
largestProduct = product;
System.out.println("3-digit numbers product palindormic number : " + x + " * " + y + " : " + product);
}
}
if(y==100 || product < largestProduct){y=x;x--;}
}
}
public static boolean isPalindormic(int n){
int palindormic = n;
int reverse = 0;
while(n>9){
reverse = (reverse*10) + n%10;
n=n/10;
}
reverse = (reverse*10) + n;
return (reverse == palindormic);
}
}