<< problem 156 - Counting Digits Digital root sums of factorisations - problem 159 >>

# Problem 158: Exploring strings for which only one character comes lexicographically after its neighbour to the left

Taking three different letters from the 26 letters of the alphabet, character strings of length three can be formed.
Examples are 'abc', 'hat' and 'zyx'.

When we study these three examples we see that for 'abc' two characters come lexicographically after its neighbour to the left.
For 'hat' there is exactly one character that comes lexicographically after its neighbour to the left.
For 'zyx' there are zero characters that come lexicographically after its neighbour to the left.

In all there are 10400 strings of length 3 for which exactly one character comes lexicographically after its neighbour to the left.

We now consider strings of n <= 26 different characters from the alphabet.
For every n, p(n) is the number of strings of length n for which exactly one character comes lexicographically after its neighbour to the left.

What is the maximum value of p(n)?

# My Algorithm

Each valid string s consists of two sub-strings: s = left + right
left and right must be strictly monotonically descending (each letter/character is smaller than its left neighbor)
and there is a "break" between left and right such that the last character of left is bigger than the first character of right.

My function count(n, alphabet) returns the number of words with n characters out of an alphabet of size alphabet (which is 26) that match all conditions.
It considers the simplified case where only the first characters are chosen. That mean that for n=3 only "a", "b" and "c" are taken from the alphabet.
The "break" can be at any position from 2 to n-1 and therefore a simple loop analyzes each possible break:

• there are \binom{n}{i} ways to build left using only i out of the first n characters (see choose(n, k), taken from problem 116)
• any of the first n characters which are not part of left must be part of right
• if the highest i characters are all part of left then there is no break between left and right → we must not count that combination
To convert the simplified case to the general case, I multiply the result by the number of ways to choose n characters from the whole alphabet.

## Alternative Approaches

I tried a Dynamic Programming approach, too, but had a bug somewhere. The next day I came up with the must simpler version you see below.

## Note

The for-loop can be transformed to a closed formula → speeding up the program.
But the program already finishes all calculations in less than 0.01 seconds, therefore I don't bother with finding the correct formula.

# Interactive test

You can submit your own input to my program and it will be instantly processed at my server:

Input data (separated by spaces or newlines):
Note: Enter the size of the alphabet and the maximum length of words

This is equivalent to
echo "26 3" | ./158

Output:

Note: the original problem's input 26 26 cannot be entered
because just copying results is a soft skill reserved for idiots.

(this interactive test is still under development, computations will be aborted after one second)

# My code

… was written in C++ and can be compiled with G++, Clang++, Visual C++. You can download it, too. Or just jump to my GitHub repository.

       #include <iostream>

// number of ways to choose n elements from k available
// code taken from problem 116
unsigned long long choose(unsigned long long n, unsigned long long k)
{
// n! / (n-k)!k!
unsigned long long result = 1;
// reduce overflow by dividing as soon as possible to keep numbers small
for (unsigned long long invK = 1; invK <= k; invK++)
{
result *= n;
result /= invK;
n--;
}
return result;
}

// count number words with n characters
unsigned long long count(unsigned int n, unsigned int alphabet)
{
// invalid parameters: must not use each letter of the alphabet more than once
if (n > alphabet)
return 0;

// count how many word with n characters use the characters 1..n
unsigned long long result = 0;
// there are n places where the "break" between s1 and s2 can occur
// count all possible characters chosen for s1 and s2
for (unsigned int i = 1; i < n; i++)
result += choose(n, i) - 1; // minus 1 because there is always one combination with no break between s1 and s2

// general case: use characters 1..Alphabet instead of 1..n
return result * choose(alphabet, n);
}

int main()
{
// bonus feature: user-defined alphabet size and maximum word length
unsigned int alphabet = 26;
unsigned int size     =  3;
std::cin >> alphabet >> size;

// all "words" with 2..size characters
unsigned long long best = 0;
for (unsigned int i = 2; i <= size; i++)
{
unsigned long long current = count(i, alphabet);
// more than before ?
if (best < current)
best = current;
}

std::cout << best << std::endl;
return 0;
}


This solution contains 7 empty lines, 13 comments and 1 preprocessor command.

# Benchmark

The correct solution to the original Project Euler problem was found in less than 0.01 seconds on an Intel® Core™ i7-2600K CPU @ 3.40GHz.
(compiled for x86_64 / Linux, GCC flags: -O3 -march=native -fno-exceptions -fno-rtti -std=gnu++11 -DORIGINAL)

See here for a comparison of all solutions.

Note: interactive tests run on a weaker (=slower) computer. Some interactive tests are compiled without -DORIGINAL.

# Changelog

June 13, 2017 submitted solution

# Difficulty

Project Euler ranks this problem at 55% (out of 100%).

# Heatmap

Please click on a problem's number to open my solution to that problem:

 green solutions solve the original Project Euler problem and have a perfect score of 100% at Hackerrank, too yellow solutions score less than 100% at Hackerrank (but still solve the original problem easily) gray problems are already solved but I haven't published my solution yet blue solutions are relevant for Project Euler only: there wasn't a Hackerrank version of it (at the time I solved it) or it differed too much orange problems are solved but exceed the time limit of one minute or the memory limit of 256 MByte red problems are not solved yet but I wrote a simulation to approximate the result or verified at least the given example - usually I sketched a few ideas, too black problems are solved but access to the solution is blocked for a few days until the next problem is published [new] the flashing problem is the one I solved most recently

I stopped working on Project Euler problems around the time they released 617.
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The 310 solved problems (that's level 12) had an average difficulty of 32.6% at Project Euler and
I scored 13526 points (out of 15700 possible points, top rank was 17 out of ≈60000 in August 2017) at Hackerrank's Project Euler+.

My username at Project Euler is stephanbrumme while it's stbrumme at Hackerrank.

Look at my progress and performance pages to get more details.

 << problem 156 - Counting Digits Digital root sums of factorisations - problem 159 >>
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