<< problem 112 - Bouncy numbers Counting block combinations I - problem 114 >>

# Problem 113: Non-bouncy numbers

Working from left-to-right if no digit is exceeded by the digit to its left it is called an increasing number; for example, 134468.
Similarly if no digit is exceeded by the digit to its right it is called a decreasing number; for example, 66420.

We shall call a positive integer that is neither increasing nor decreasing a "bouncy" number; for example, 155349.

As n increases, the proportion of bouncy numbers below n increases such that there are only 12951 numbers below one-million
that are not bouncy and only 277032 non-bouncy numbers below 10^10.

How many numbers below a googol (10^100) are not bouncy?

# My Algorithm

My solution is based on a dynamic programming approach:

• solve the problem for one digit
• solve the problem for n+1 digits by using information from n digits
All 9 single digits are not bouncy. Therefore I initialize my array increase and decrease with 1s.
Each of their entries represents how many numbers starting with a digit are not bouncy.
In increase[x][y] you find the count of increasing numbers with x digits, where the front-most digit is y.

In each iteration, increase and decrease are updated:
increase[x][y] is the sum of all increase[x-1][less than or equal to y] (and the other way around for decrease, too).
All not bouncy numbers are either increasing or decreasing. I have to deduct all increasing numbers where the first digit is zero.
On top of that, numbers where all digits are identical are both increasing und decreasing and counted twice, therefore I have to subtract 10.

## Modifications by HackerRank

More than 100 digits are no problem. The result will be a rather large number and has to be printed mod 10^9 + 7.

# Interactive test

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

Number of test cases (1-5):

Input data (separated by spaces or newlines):
Note: Enter the number of digits

This is equivalent to
echo "1 10" | ./113

Output:

Note: the original problem's input 100 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++11 and can be compiled with G++, Clang++, Visual C++. You can download it, too. Or just jump to my GitHub repository.

The code contains #ifdefs to switch between the original problem and the Hackerrank version.
Enable #ifdef ORIGINAL to produce the result for the original problem (default setting for most problems).

       #include <iostream>
#include <vector>

// print result modulo some value
#define ORIGINAL
#ifdef ORIGINAL
const unsigned long long Modulo = 1000000000000000000ULL; // high enough to keep the result unchanged
#else
const unsigned long long Modulo = 1000000007ULL;
#endif

int main()
{
// Googol = 100, but Hackerrank wants more ...
const unsigned int numDigits = 100000;
std::vector<unsigned long long> solutions(numDigits + 1, 0);

// count how many numbers are increasing and/or decreasing
typedef unsigned long long DigitCounter[10]; // some older GCC complain about using this in a vector
std::vector<DigitCounter> increase(numDigits);
std::vector<DigitCounter> decrease(numDigits);

// all one-digit numbers are non-bouncy
unsigned long long sum = 9;
for (auto& x : increase[0])
x = 1;
for (auto& x : decrease[0])
x = 1;

// process digits, beginning from the right side
for (unsigned int i = 1; i < numDigits; i++)
{
// digits 0..9
for (unsigned int current = 0; current <= 9; current++)
{
// add count of all numbers where the next digit is equal or lower
decrease[i][current] = 0;
for (unsigned int smaller = 0; smaller <= current; smaller++)
decrease[i][current] = (decrease[i][current] + decrease[i - 1][smaller]) % Modulo;

// add count of all numbers where the next digit is equal or higher
increase[i][current] = 0;
for (unsigned int bigger = current; bigger <= 9; bigger++)
increase[i][current] = (increase[i][current] + increase[i - 1][bigger]) % Modulo;
}

// compute total sum of increasing and decreasing numbers
for (auto x : increase[i])
sum += x;
for (auto x : decrease[i])
sum += x;

sum -= increase[i][0];

// numbers with identical digits were counted twice
// because they are both increasing and decreasing (e.g. 55555555)
sum -= 10;

// Hackerrank only
sum %= Modulo;
solutions[i] = sum;
}

// lookup results
unsigned int tests;
std::cin >> tests;
while (tests--)
{
unsigned int digits;
std::cin >> digits;
std::cout << solutions[digits - 1] << std::endl; // 0-based array but 1-based input
}

return 0;
}


This solution contains 12 empty lines, 14 comments and 6 preprocessor commands.

# Benchmark

The correct solution to the original Project Euler problem was found in 0.02 seconds on an Intel® Core™ i7-2600K CPU @ 3.40GHz.
Peak memory usage was about 18 MByte.

(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

May 15, 2017 submitted solution

# Hackerrank

My code solves 3 out of 3 test cases (score: 30%)

# Difficulty

Project Euler ranks this problem at 30% (out of 100%).

Hackerrank describes this problem as easy.

Note:
Hackerrank has strict execution time limits (typically 2 seconds for C++ code) and often a much wider input range than the original problem.
In my opinion, Hackerrank's modified problems are usually a lot harder to solve. As a rule thumb: brute-force is rarely an option.

# 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 112 - Bouncy numbers Counting block combinations I - problem 114 >>
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