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Minimum moves to rearrange one string into another

Code Apperentice (229)
I want to rearrange string a into string b in the minimum amount of moves. Both strings are of the same size. Groups of letters can be moved

For example:

Input:
5
3 2 1 4 5
2 3 1 4 5
Output:
3

You can cut the string a, [3], [2], [1, 4, 5], and then rearrange it into [2], [3], [1, 4, 5], which is string b.
Last edited on
Mantorr22 (152)
You can do it like this, and your example needs just one move to accomplish everything.

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#include <iostream>
#include <string>
#include <vector>
#include <sstream>
#include <algorithm>
#include <numeric>
#include <assert.h>

typedef std::vector<std::string>::size_type size_t_t;

size_t_t effectiveFind(std::string val, const std::vector<std::string> &v1, const std::vector<std::string> &v2)
{
	for(std::vector<std::string>::size_type i = 0; i < v1.size(); i++)
	{
		if(v1[i] != v2[i])
		{
			if(v2[i] == val) return static_cast<size_t_t>(i);
		}
	}

	assert(NULL); return 0;
}

std::string addSpace(std::string val) {return (val += ' ');}

std::string rearrange(std::string a, std::string b, size_t_t &numMoves)
{
	std::string val;
	size_t_t position;
	numMoves = 0;

	std::vector<std::string> v1;
	std::vector<std::string> v2;

	std::istringstream iss1(a);
	std::istringstream iss2(b);

	while(iss1 >> val) v1.push_back(val);
	while(iss2 >> val) v2.push_back(val);

	assert(v1.size() && v2.size());
	assert(v1.size() == v2.size());

	std::vector<std::string>::size_type i;
	for(i = 0; i < v1.size(); i++)
	{
		assert(std::count(v1.begin(), v1.end(), v1[i]) == std::count(v2.begin(), v2.end(), v1[i]));
		assert(std::count(v1.begin(), v1.end(), v2[i]) == std::count(v2.begin(), v2.end(), v2[i]));
	}

	for(i = 0; i < v1.size(); i++)
	{
		if(v1[i] != v2[i])
		{
			position = effectiveFind(v1[i], v1, v2);
			std::swap(v2[i], v2[position]); 
			numMoves++;
		}
	}

	std::transform(v2.begin(), v2.end() - 1, v2.begin(), addSpace);

	return std::accumulate(v2.begin(), v2.end(), std::string());
}

int main()
{
	size_t_t numMoves;

	// Emptying these strings mean you have to input them manually
	std::string line1 = "3 2 1 4 5"; // "3 2 1 4 5"
	std::string line2 = "2 3 1 4 5"; // "1 3 2 5 4"

	std::cout << "Enter two strings : " << std::endl;

	if(!line1.size() || !line2.size())
	{
		std::getline(std::cin, line1);
		std::getline(std::cin, line2);
	}
	else
	{
		std::cout << line1 << std::endl;
		std::cout << line2 << std::endl;
	}

	line2 = rearrange(line1, line2, numMoves);

	std::cout << std::endl;
	std::cout << line1 << std::endl;
	std::cout << line2 << std::endl;
	std::cout << numMoves << std::endl;

	return 0;
}


It is easy to understand though, feel free to ask if you have any questions.

Enter two strings :
3 2 1 4 5
2 3 1 4 5

3 2 1 4 5
3 2 1 4 5
1

http://rextester.com/HGY20186
Last edited on
Code Apperentice (229)
How do you also get the number of groups it makes from splitting?
Mantorr22 (152)
How do you also get the number of groups it makes from splitting?

Sorry friend, what do you mean by that?
Code Apperentice (229)
O, I just wasted your time, I reread my question and I poorly worded the question, sorry. The output is three because you have to split the string first then rearranged, as well as minimum number of moves I want the minimum number of groups too. The output should be 3 1
Mantorr22 (152)
The output should be 3 1

I don't understand these numbers. Could you please clarify?
Mantorr22 (152)
Yeah, it is not that much of a challenge though. You just need to add some features so that the program can recognize string groups.

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#include <iostream>
#include <string>
#include <vector>
#include <sstream>
#include <algorithm>
#include <numeric>
#include <assert.h>

typedef std::vector<std::string>::size_type size_t_t;

size_t_t effectiveFind(std::string val, const std::vector<std::string> &v1, const std::vector<std::string> &v2)
{
	for(std::vector<std::string>::size_type i = 0; i < v1.size(); i++)
	{
		if(v1[i] != v2[i])
		{
			if(v2[i] == val) return static_cast<size_t_t>(i);
		}
	}

	assert(NULL); return 0;
}

std::string addSpace(std::string val) {return (val += ' ');}

struct groupRecord
{
	groupRecord(size_t_t _pos, std::string _val) : pos(_pos), val(_val) {}

	size_t_t pos;
	std::string val;
};

void insertProperPosition(std::vector<groupRecord> &target, std::string result, size_t_t position)
{
	std::vector<groupRecord>::iterator it;
	for(it = target.begin(); it != target.end() && it->pos < position; it++);

	if(it == target.end()) target.push_back(groupRecord(position, result));
	else target.insert(it, groupRecord(position, result));
}

std::string rearrange(std::string a, std::string b, size_t_t &numMoves, size_t_t &numGroups)
{
	std::string val;
	size_t_t position;
	numMoves = 0;
	numGroups = 0;

	std::vector<std::string> v1;
	std::vector<std::string> v2;

	std::istringstream iss1(a);
	std::istringstream iss2(b);

	while(iss1 >> val) v1.push_back(val);
	while(iss2 >> val) v2.push_back(val);

	assert(v1.size() && v2.size());
	assert(v1.size() == v2.size());

	std::vector<std::string>::size_type i;
	for(i = 0; i < v1.size(); i++)
	{
		assert(std::count(v1.begin(), v1.end(), v1[i]) == std::count(v2.begin(), v2.end(), v1[i]));
		assert(std::count(v1.begin(), v1.end(), v2[i]) == std::count(v2.begin(), v2.end(), v2[i]));
	}
	
	std::vector<groupRecord> v1_a;
	std::vector<groupRecord> v2_a;

	std::vector<std::string> v1_b;
	std::vector<std::string> v2_b;

	if(std::equal(v1.begin(), v1.end(), v2.begin()))
	{
		numGroups = 1;
		numMoves = 0;

		std::transform(v2.begin(), v2.end() - 1, v2.begin(), addSpace);
		return std::accumulate(v2.begin(), v2.end(), std::string());
	}

	size_t_t group_size, j, k;
	for(group_size = v1.size() - 1; group_size >= 2; group_size--)
	{
		for(i = 0; i <= v1.size() - group_size; i++)
		{
			if(!v1[i].empty())
			{
				for(j = 0; j <= v2.size() - group_size; j++)
				{
					if(!v2[j].empty())
					{
						if(std::equal(v1.begin() + i, v1.begin() + i + group_size, v2.begin() + j))
						{

							std::transform(v1.begin() + i, v1.begin() + i + group_size - 1, v1.begin() + i, addSpace);
							std::transform(v2.begin() + j, v2.begin() + j + group_size - 1, v2.begin() + j, addSpace);

							insertProperPosition(v1_a, std::accumulate(v1.begin() + i, v1.begin() + i + group_size, std::string()), i);
							insertProperPosition(v2_a, std::accumulate(v2.begin() + j, v2.begin() + j + group_size, std::string()), j);

							std::fill(v1.begin() + i, v1.begin() + i + group_size, std::string());
							std::fill(v2.begin() + j, v2.begin() + j + group_size, std::string());
							
							numGroups++;
						}
					}
				}
			}
		}
	}

	for(i = 0; i < v1.size(); i++)
	{
		if(!v1[i].empty())
		{
			insertProperPosition(v1_a, v1[i], i);
		}

		if(!v2[i].empty())
		{
			insertProperPosition(v2_a, v2[i], i);
		}
	}

	std::cout << std::endl;

	for(i = 0; i < v1_a.size(); i++)
	{
		std::cout << v1_a[i].val << " ## ";
	}
	std::cout << std::endl;

	for(i = 0; i < v2_a.size(); i++)
	{
		std::cout << v2_a[i].val << " ## ";
	}
	std::cout << std::endl;

	numGroups += v1.size() - std::count(v1.begin(), v1.end(), std::string());

	assert(v1_a.size() && v2_a.size());
	assert(v1_a.size() == v2_a.size());

	for(i = 0; i < v1_a.size(); i++)
	{
		v1_b.push_back(v1_a[i].val); 
		v2_b.push_back(v2_a[i].val); 
	}

	for(i = 0; i < v1_b.size(); i++) 
	{
		if(v1_b[i] != v2_b[i])
		{
			position = effectiveFind(v1_b[i], v1_b, v2_b);
			std::swap(v2_b[i], v2_b[position]); 
			numMoves++;
		}
	}

	std::transform(v2_b.begin(), v2_b.end() - 1, v2_b.begin(), addSpace);

	return std::accumulate(v2_b.begin(), v2_b.end(), std::string());

}

int main()
{
	size_t_t numMoves;
	size_t_t numGroups;

	// Emptying these strings mean you have to input them manually
	std::string line1 = "3 2 1 4 5"; // "3 2 1 4 5"
	std::string line2 = "2 3 1 4 5"; // "2 3 1 4 5"

	std::cout << "Enter two strings : " << std::endl;

	if(!line1.size() || !line2.size())
	{
		std::getline(std::cin, line1);
		std::getline(std::cin, line2);
	}
	else
	{
		std::cout << line1 << std::endl;
		std::cout << line2 << std::endl;
	}

	line2 = rearrange(line1, line2, numMoves, numGroups);

	std::cout << std::endl;
	std::cout << line1 << std::endl;
	std::cout << line2 << std::endl;
	std::cout << numGroups << ' ' << numMoves << std::endl;

	return 0;
}


Enter two strings :
3 2 1 4 5
2 3 1 4 5

3 ## 2 ## 1 4 5 ##
2 ## 3 ## 1 4 5 ##

3 2 1 4 5
3 2 1 4 5
3 1

http://rextester.com/PETP40773
Code Apperentice (229)
Last thing, is there a way to just calculate the groups without actually rearranging the strings? I became intrigued after I made that initial mistake in my wording of the question
Mantorr22 (152)
Last thing, is there a way to just calculate the groups without
actually rearranging the strings? I became intrigued after I made that
initial mistake in my wording of the question

I already gave the code you ask for, it is up to you how to implement it.
If you copy that challenge from somewhere, please provide the link.

jonnin (11427)
code aside, I think the answer is always 1 move per group. Getting there might be tricky, but I can't see any place where it would take more than 1.

Mantorr22 (152)
Code aside, I think the answer is always 1 move per group. Getting there might be tricky, but I can't see any place where it would take more than 1.

How about this? It takes two moves.

3 2 1 4 5
5 3 4 2 1

3 ## 2 1 ## 4 ## 5 ##
5 ## 3 ## 4 ## 2 1 ##

3 2 1 4 5
3 2 1 4 5
4 2
fasravo (7)
Hey, new to the forum.
Can I ask how people do it? I mean how they did everything, I don't understand what they are doing with the code.

By the way, I tested such a long string and it still performs marvelously.

"Enter two strings :
1 2 0 ax 3 5 100 0 C C++ 4 1 1 0 ax ax ax ax ax hello >
C C++ 1 2 0 ax ax ax 1 3 4 hello ax ax 5 0 ax 0 1 > 100

...

1 2 0 ax 3 5 100 0 C C++ 4 1 1 0 ax ax ax ax ax hello >
1 2 0 ax 3 5 100 0 C C++ 4 1 1 0 ax ax ax ax ax hello >
14 11"
jonnin (11427)
one move per substring, I meant, as in should be able concat all the substrings at the end and its done. Figuring out where each substring goes and how to slice it up front is the work, and getting the least # of substrings.


Last edited on
Code Apperentice (229)
Thanks for the help everryone
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