What is the greatest 6 digit palindrome number?

Mark M. answered • 11/25/17

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The first shall be 250052

from there to 2599952 there are 100.

The next is 2600062

from there to 2699962 there are 100.

The next is 2700072

from there to 2799972 there are 100.

 

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You are not including rangeMin in either of the loops, so you will never test products which involve the lower limit. You want

    for (int p = rangeMax; p >= rangeMin; p--)
        for (int q = rangeMax; q >= rangeMin; q--)

0 in the loop condition:

    for (int p = rangeMax; p >= rangeMin; p--)
        for (int q = rangeMax; q >= rangeMin; q--)

Commutativity

Note that

    for (int p = rangeMax; p >= rangeMin; p--)
        for (int q = rangeMax; q >= rangeMin; q--)

1, so you don't need to test all combinations in the range:

    for (int p = rangeMax; p > rangeMin; p--)
        for (int q = rangeMax; q > rangeMin; q--)

Only the ones where either

    for (int p = rangeMax; p >= rangeMin; p--)
        for (int q = rangeMax; q >= rangeMin; q--)

2 or

    for (int p = rangeMax; p >= rangeMin; p--)
        for (int q = rangeMax; q >= rangeMin; q--)

3, which will reduce your search space by close to 50%! For example, you could change the

    for (int p = rangeMax; p >= rangeMin; p--)
        for (int q = rangeMax; q >= rangeMin; q--)

4 range to start at the current

    for (int p = rangeMax; p >= rangeMin; p--)
        for (int q = rangeMax; q >= rangeMin; q--)

5 value and go down from there:

    for (int p = rangeMax; p >= rangeMin; p--)
        for (int q = p; q >= rangeMin; q--)

Test order: Fastest tests first!

    for (int p = rangeMax; p >= rangeMin; p--)
        for (int q = rangeMax; q >= rangeMin; q--)

6 is an involved function which will take a bit of time. In comparison,

    for (int p = rangeMax; p >= rangeMin; p--)
        for (int q = rangeMax; q >= rangeMin; q--)

7 is blazingly fast. So instead of:

            if (isPal(P))
                if (P > maxP)
                {
                    maxp = p; maxq = q; maxP = P;
                }

how about:

            if (P > maxP)
                if (isPal(P))
                {
                    maxp = p; maxq = q; maxP = P;
                }

Early termination

If

    for (int p = rangeMax; p >= rangeMin; p--)
        for (int q = rangeMax; q >= rangeMin; q--)

8 is ever less than

    for (int p = rangeMax; p >= rangeMin; p--)
        for (int q = rangeMax; q >= rangeMin; q--)

9, then multiplying

    for (int p = rangeMax; p >= rangeMin; p--)
        for (int q = rangeMax; q >= rangeMin; q--)

5 by any smaller value of

    for (int p = rangeMax; p >= rangeMin; p--)
        for (int q = rangeMax; q >= rangeMin; q--)

4 is a waste of time; you can break out of the inner loop, and try the next value of

    for (int p = rangeMax; p >= rangeMin; p--)
        for (int q = rangeMax; q >= rangeMin; q--)

5.

If

    for (int p = rangeMax; p > rangeMin; p--)
        for (int q = rangeMax; q > rangeMin; q--)

3 is ever less than

    for (int p = rangeMax; p >= rangeMin; p--)
        for (int q = rangeMax; q >= rangeMin; q--)

9, and the inner loop only multiplies

    for (int p = rangeMax; p >= rangeMin; p--)
        for (int q = rangeMax; q >= rangeMin; q--)

5 by

    for (int p = rangeMax; p >= rangeMin; p--)
        for (int q = rangeMax; q >= rangeMin; q--)

4 values which a not greater than

    for (int p = rangeMax; p >= rangeMin; p--)
        for (int q = rangeMax; q >= rangeMin; q--)

5, then you can break out of the outer loop, too!

String Manipulation

The following is inefficient, because temporary objects are being created and destroyed during each iteration.

    for (int i = numStr.length() - 1; i >= 0 ; --i)
        rNumStr += numStr.charAt(i);

It is much better to use a

    for (int p = rangeMax; p > rangeMin; p--)
        for (int q = rangeMax; q > rangeMin; q--)

8 to build up strings character by character, because the

    for (int p = rangeMax; p > rangeMin; p--)
        for (int q = rangeMax; q > rangeMin; q--)

8 maintains a mutable buffer for the interm results.

Even better, it includes the

    for (int p = rangeMax; p >= rangeMin; p--)
        for (int q = p; q >= rangeMin; q--)

0 method, which does what you need in one function call.

StringBuilder sb = new StringBuilder(numStr);
String rNumStr = sb.reverse().toString();

Unnecessary Operations

Why convert

    for (int p = rangeMax; p >= rangeMin; p--)
        for (int q = p; q >= rangeMin; q--)

1 to a number using

    for (int p = rangeMax; p >= rangeMin; p--)
        for (int q = p; q >= rangeMin; q--)

2? Isn't the result simply

    for (int p = rangeMax; p >= rangeMin; p--)
        for (int q = p; q >= rangeMin; q--)

3, the value that was passed in to the function?

Why convert

    for (int p = rangeMax; p >= rangeMin; p--)
        for (int q = p; q >= rangeMin; q--)

4 to a number? If

    for (int p = rangeMax; p >= rangeMin; p--)
        for (int q = p; q >= rangeMin; q--)

3 is a palindrome, then aren't

    for (int p = rangeMax; p >= rangeMin; p--)
        for (int q = p; q >= rangeMin; q--)

1 and

    for (int p = rangeMax; p >= rangeMin; p--)
        for (int q = p; q >= rangeMin; q--)

4 equal?

How many 6 digit palindromes are there?

What is the largest palindrome number?

What is the maximum number of 6 letter palindrome?

What is the 6 digit number greatest?