Getting distance between two elements in a 2D map - not quite correct -solved by Brian

Question

I am working on the 2D Map challenge, so careful spoiler alert...
Anyways, somehow my solution is flawed but I do not know where the mistake is. Any help or hints would be appreciated :)

Here is the code:

#include<iostream>
#include<string>
#include<vector>
#include<sstream>
#include<algorithm>

int main()
{
std::string str = "PXXXX,XXXXX,XXXXP,XXXXX,XXXXX";
//std::string str;
//std::getline(std::cin, str);
std::vector<std::string> vec;
std::stringstream ss(str);
std::string strTemp;
std::vector<int> nums;
std::vector<int> block;
std::vector<int> option;
int result = 0;

//first tokenize strings and create vector of strings
while(std::getline(ss, strTemp, ','))
{
vec.push_back(strTemp);
}

//find index of 'P' in each string
for(auto &val : vec)
{
std::cout << val << std::endl;
std::size_t found = val.find('P');
if (found != std::string::npos)
{
nums.push_back(found);
}
}

//find instance of 'P' in string
for(size_t i = 0; i < vec.size(); i++)
{
if (std::find(vec[i].begin(), vec[i].end(), 'P') != vec[i].end())
{
block.push_back(i);
}
}

//if two Ps are found in the same string...
if(block.size()<2)
{
for(unsigned int i = 0; i< vec[block[0]].size(); i++)
{
if(vec[block[0]][i] == 'P')
{
option.push_back(i);
}
}
result = option[1] - option[0];
std::cout << result;
}

//subtract position of indices and position in blocks and add result
else
{
result = (nums[1]-nums[0]) + (block[1]-block[0]);
std::cout << result;
}
}

Answer 1

The_Fox I found the problem. Yes, abs() is the solution.

Consider what happens when the 2nd P is in a lower row and to the left of the first P.

Answer 2

The_Fox the distance between two points is the length of their hypotenuse: sqrt(sqr(x2-x1)+sqr(y2-y1))

Conveniently, there is a standard library function for that.

#include <cmath>

result = hypot(x2-x1, y2-y1);

Answer 3

The_Fox now I understand this is a Code Coach task. I will take another look.

Answer 4

You are welcome. The_Fox, I wish the same for you!

Answer 5

Brian I guess I looked too long so I did not see it ;) Thanks a lot!! Happy New Year to you and your loved ones. Until next time.

Answer 6

Brian Thank you. Not sure if that will help me in my specific case but I will try! Happy holidays :)

Answer 7

@Brian The hypotenuse does not help me here because I need to 'walk' the distance through the different x-fields (if that makes any sense).
e.g.

XPXX
XXXX
XXXP

Distance = 4 -> 2 down, 2 to the right (cannot walk diagonally).

Answer 8

Exactly… it might have something to do with one of the values. Not sure. Someone mentioned using absolute numbers only…