Return the sub string [duplicate] - c++

This question already has answers here:
Frequency of ngrams (strings) in tokenized text
(2 answers)
Closed 3 years ago.
How can I use this for x amounts of numbers ? for my approach, its hard coded to 2 substrings. Also is there a better way with less time complexity? There may be a loop hole over here which needs to be fixed about the number of num im passing as I am not using the um parameter at all.

Your current approach has a few problems, including a hard-coded maximum number of ngrams, and the fixed ngram size. In addition, your short variable names and lack of comments do not help explain the code to whoever is reading it.
A simpler solution is to use a map to count the number of times each ngram occurs, and then find the one with the highest count. That would give rougly N.logN time complexity. Alternatively unordered_map would be closer to linear time complexity.
There will of course be an edge case where more than one ngram occurs the same highest count. You would need to decide which of a variety of strategies should be used to resolve that. In my example, I take advantage of intrinsic ordering of std::map to select the ngram with the lowest sort order. If using unordered_map, you'd need a different strategy for resolving contention in a deterministic way.
#include <algorithm>
#include <iostream>
#include <map>
#include <string>
std::string ngram(const std::string &input, int num)
{
if (num <= 0 || num > input.size()) return "";
// Count ngrams of size 'num'
std::map<std::string, int> ngram_count;
for(size_t i = 0; i <= input.size() - num; i++)
{
++ngram_count[input.substr(i, num)];
}
// Select ngram with highest count
std::map<std::string, int>::iterator highest = std::max_element(
ngram_count.begin(), ngram_count.end(),
[](const std::pair<std::string, int>& a, const std::pair<std::string, int>& b)
{
return a.second < b.second;
});
// Return ngram with highest count, otherwise empty string
return highest != ngram_count.end() ? highest->first : "";
}
int main()
{
std::cout << ngram("engineering", 2) << std::endl;
std::cout << ngram("engineering", 3) << std::endl;
return 0;
}

I did it a bit different than paddy, so I thought I would post it. Use std::set. He explains the issue so should get the credit for your answer.
struct test {
test(const std::string& str) :val(str), cnt(0) {}
test(const test& thet) { *this = thet; }
std::string val;
int cnt;
friend bool operator < (const test& a, const test& b) { return a.val < b.val; }
};
using test_set_type = std::set<test>;
const test ngram(std::string A, int num) {
test_set_type set;
for (auto it = A.begin(); it < A.end() - num + 1; ++it)
{
auto found = set.find(std::string(it, it + num));
if (found != set.end())
++const_cast<test&>(*found).cnt;
else
set.insert(std::string(it, it + num));
}
int find = -1;
test_set_type::iterator high = set.begin();
for (auto it = set.begin(); it != set.end(); ++it)
if(it->cnt > find)
++find, high= it;
return *high;
}
int main() {
int num = 2;
std::string word("engineering");
std::cout << ngram(word, num).val << std::endl;
return 0;
}

Related

Sort Integers by The Number of 1 Bits . I used one sort function to sort the vector ? But why sort is not working?

Sort Integers by The Number of 1 Bits
Leetcode : Problem Link
Example Testcase :
Example 1:
Input: arr = [0,1,2,3,4,5,6,7,8]
Output: [0,1,2,4,8,3,5,6,7]
Explantion: [0] is the only integer with 0 bits.
[1,2,4,8] all have 1 bit.
[3,5,6] have 2 bits.
[7] has 3 bits.
The sorted array by bits is [0,1,2,4,8,3,5,6,7]\
Example 2:
Input: arr = [1024,512,256,128,64,32,16,8,4,2,1]
Output: [1,2,4,8,16,32,64,128,256,512,1024]
Explantion: All integers have 1 bit in the binary representation, you should just sort them in ascending order.
My Solution :
class Solution {
public:
unsigned int setBit(unsigned int n){
unsigned int count = 0;
while(n){
count += n & 1;
n >>= 1;
}
return count;
}
vector<int> sortByBits(vector<int>& arr) {
map<int,vector<int>>mp;
for(auto it:arr){
mp[setBit(it)].push_back(it);
}
for(auto it:mp){
vector<int>vec;
vec=it.second;
sort(vec.begin(),vec.end()); //This Sort Function of vector is not working
}
vector<int>ans;
for(auto it:mp){
for(auto ele:it.second){
ans.push_back(ele);
}
}
return ans;
}
};
In my code why sort function is not working ?
[1024,512,256,128,64,32,16,8,4,2,1]
For the above testcase output is [1024,512,256,128,64,32,16,8,4,2,1] because of sort function is not working. It's correct output is [1,2,4,8,16,32,64,128,256,512,1024]
Note : In the above example testcase every elements of the testcase has only one set-bit(1)
As your iteration in //This sort function ...
refers to mp as the copy of the value inside the map, sort function will not sort the vector inside it, but the copy of it. Which does not affecting the original vector<int> inside the mp. Therefore, no effect occurs. You should refer the vector inside the map as a reference like this:
class Solution {
public:
unsigned int setBit(unsigned int n) {
unsigned int count = 0;
while (n) {
count += n & 1;
n >>= 1;
}
return count;
}
vector<int> sortByBits(vector<int>& arr) {
map<int, vector<int>>mp;
for (auto it : arr) {
mp[setBit(it)].push_back(it);
}
for (auto& it : mp) {
sort(it.second.begin(), it.second.end()); //Now the sort function works
}
vector<int>ans;
for (auto it : mp) {
for (auto ele : it.second) {
ans.push_back(ele);
}
}
return ans;
}
};
Although there is more design problem inside your solution, this will be a solution with minimized modification.
vector<int>vec is a copy of a copy of the one in the map which is then discarded. Try:
for(auto& entry:mp){
vector<int>&vec=entry.second;
sort(vec.begin(),vec.end());
}
Your other for loops should also use references for efficiency but it won't affect the behaviour.
I assume the OP is just learning, so fiddling with various data structures etc. can carry some educational value. Still, only one of the comments pointed out that the starting approach to the problem is wrong, and the whole point of the exercise is to find a custom method of comparing the numbers, by number of bits first, then - by value.
Provided std::sort is allowed (OP uses it), I guess the whole solution comes down to, conceptually, sth likes this (but I haven't verified it against LeetCode):
template <typename T>
struct Comp
{
std::size_t countBits(T number) const
{
size_t count;
while(number) {
count += number & 1;
number>>=1;
}
return count;
}
bool operator()(T lhs, T rhs) const
{
/*
auto lb{countBits(lhs)};
auto rb{countBits(rhs)};
return lb==rb ? lhs < rhs : lb < rb;
* The code above is the manual implementation of the line below
* that utilizes the standard library
*/
return std::tuple{countBits(lhs), lhs} < std::tuple{countBits(rhs), rhs};
}
};
class Solution {
public:
void sortByBits(vector<int>& arr) {
std::sort(begin(arr), end(arr), Comp<int>{});
}
};
Probably it can improved even further, but I'd take it as starting point for analysis.
Here is memory efficient and fast solution. I don't know why you are using map and extra vector. we can solve this questions without any extra memory efficiently. We just have to make a Comparator function which will sort elements according to our own requirements. Please let me know in comments if you require further help in code (or if you find difficult to understand my code). I am using __builtin_popcount() function which will return me number of set bits in a number.
bool sortBits(const int a, const int b){ //Comparator function to sort elements according to number of set bits
int numOfBits1 = __builtin_popcount(a);
int numOfBits2 = __builtin_popcount(b);
if(numOfBits1 == numOfBits2){ //if number of set bits are same, then sorting the elements according to magnitude of element (greater/smaller element)
return a < b;
}
return (numOfBits1 < numOfBits2); //if number of set bits are not same, then sorting the elements according to number of set bits in element
}
class Solution {
public:
vector<int> sortByBits(vector<int>& arr) {
sort(arr.begin(),arr.end(), sortBits);
return arr;
}
};
The problem is already evaluated and the fix is aready explained.
I want to give 2 additional/alternative solution proposals.
In C++17 we have the std::bitset count function. Please see here
And in C++20 we have directly the std::popcount function. Please see here.
(Elderly and grey haired people like me would also find 5 additional most efficient solutions in the Book "Hackers Delight")
Both variants lead to a one statement solution using std::sort with a lambda.
Please see:
#include <algorithm>
#include <vector>
#include <iostream>
#include <bitset>
// Solution
class Solution {
public:
std::vector<int> sortByBits(std::vector<int>& arr) {
std::sort(arr.begin(), arr.end(), [](const unsigned int i1, const unsigned int i2)
{ size_t c1{ std::bitset<14>(i1).count() }, c2{ std::bitset<14>(i2).count() }; return c1 == c2 ? i1 < i2 : c1 < c2; });
//{ int c1=std::popcount(i1), c2=std::popcount(i2); return c1 == c2 ? i1 < i2 : c1 < c2; });
return arr;
}
};
// Test
int main() {
std::vector<std::vector<int>> testData{
{0,1,2,3,4,5,6,7,8},
{1024,512,256,128,64,32,16,8,4,2,1}
};
Solution s;
for (std::vector<int>& test : testData) {
for (const int i : s.sortByBits(test)) std::cout << i << ' ';
std::cout << '\n';
}
}

How to erase element from std::vector<string> by its length(erase not working)

i have given a vector `
vector<string> inputArray = { "aba","aa","ad","vcd","aba" };
and i want to return this vector which contains only string with the longest length, in this case i want to return only {"aba","vcd","aba"}, so for now i want to erase elements which length is not equal to the highest `
vector<string> allLongestStrings(vector<string> inputArray) {
int length = inputArray.size();
int longstring = inputArray[0].length();
int count = 0;
vector<string> result;
for (int i = 0; i < length; i++)
{
if (longstring < inputArray[i].length())
{
longstring = inputArray[i].length();
}
count++;
}
for (int = 0; i<count;i++)
{
if (inputArray[i].length() != longstring)
{
inputArray[i].erase(inputArray.begin() + i);
count--;
i--;
}
}
return inputArray;
}
but i get this error no instance of overloaded fucntion "std::basic_string<_Elem,_Traits,_Alloc>::erase[with_Elem=char,_Traits=std::char_traits<char>,_Alloc=std::allocator<char>]" matches the argument list" in inputArray[i].erase(inputArray.begin()+i); this line
what's wrong?
There are other problems, but this specific compiler message is telling you that's not the right way to remove specific character(s) from a string.
However, reading the question in the OP, we see that you wanted to remove a string from a vector. To fix that one specific error, simply change
inputArray[i].erase( /*character position(s) in the string*/ )
to
inputArray.erase( /*some position in the array*/ )
Or you could fix it so it uses an iterator in the string denoted by inputArray[i] to actually delete characters from that string, which of course isn't what you said you wanted to do. The point is, the error message is because you're using the wrong iterator type because you think that you're working with a vector, but you actually told it to work with a string that you got out of the vector.
And then you will compile and have other issues which are well covered in comments already.
The issue with inputArray[i].erase(inputArray.begin() + i); can be fixed as shown in Kenny Ostrom's answer.
I'd like to point out that the OP could make use of the erase-remove idiom or even create a new vector with only the bigger strings instead (the posted code is already copying the source vector).
#include <iostream>
#include <vector>
#include <string>
#include <algorithm>
template <typename InputIt>
auto only_the_longest_of(InputIt first, InputIt last)
{
using value_type = typename std::iterator_traits<InputIt>::value_type;
std::vector<value_type> result;
// find the longest size
auto longest = std::max_element(first, last,
[](value_type const &a, value_type const &b) {
return a.size() < b.size();
});
if ( longest == last )
return result;
// extract only the longest ones, instead of erasing
std::copy_if( first, last, std::back_inserter(result)
, [max_size = longest->size()] (value_type const& v) {
return v.size() >= max_size;
});
return result;
}
template <typename T>
auto erase_the_shortest_from(std::vector<T> &input)
{
// find the longest size
auto longest = std::max_element(input.cbegin(), input.cend(),
[](T const &a, T const &b) {
return a.size() < b.size();
});
if ( longest == input.cend() || longest->size() == 0 )
return input.end();
// implement erase-remove idiom
return input.erase(std::remove_if(
input.begin(), input.end(), [max_size = longest->size()] (T const &v) {
return v.size() < max_size;
}));
}
int main()
{
std::vector<std::string> test = {
"aba", "aa", "ad", "vcd", "aba"
};
// The original vector remain unchanged
auto result = only_the_longest_of(test.cbegin(), test.cend());
for (auto const& str : result)
std::cout << str << '\n';
std::cout << '\n';
// This will change the vector
erase_the_shortest_from(test);
for (auto const& str : test)
std::cout << str << '\n';
}

Sorting one std::vector based on the content of another [duplicate]

This question already has answers here:
How can I sort two vectors in the same way, with criteria that uses only one of the vectors?
(9 answers)
Closed 9 months ago.
I have several std::vector, all of the same length. I want to sort one of these vectors, and apply the same transformation to all of the other vectors. Is there a neat way of doing this? (preferably using the STL or Boost)? Some of the vectors hold ints and some of them std::strings.
Pseudo code:
std::vector<int> Index = { 3, 1, 2 };
std::vector<std::string> Values = { "Third", "First", "Second" };
Transformation = sort(Index);
Index is now { 1, 2, 3};
... magic happens as Transformation is applied to Values ...
Values are now { "First", "Second", "Third" };
friol's approach is good when coupled with yours. First, build a vector consisting of the numbers 1…n, along with the elements from the vector dictating the sorting order:
typedef vector<int>::const_iterator myiter;
vector<pair<size_t, myiter> > order(Index.size());
size_t n = 0;
for (myiter it = Index.begin(); it != Index.end(); ++it, ++n)
order[n] = make_pair(n, it);
Now you can sort this array using a custom sorter:
struct ordering {
bool operator ()(pair<size_t, myiter> const& a, pair<size_t, myiter> const& b) {
return *(a.second) < *(b.second);
}
};
sort(order.begin(), order.end(), ordering());
Now you've captured the order of rearrangement inside order (more precisely, in the first component of the items). You can now use this ordering to sort your other vectors. There's probably a very clever in-place variant running in the same time, but until someone else comes up with it, here's one variant that isn't in-place. It uses order as a look-up table for the new index of each element.
template <typename T>
vector<T> sort_from_ref(
vector<T> const& in,
vector<pair<size_t, myiter> > const& reference
) {
vector<T> ret(in.size());
size_t const size = in.size();
for (size_t i = 0; i < size; ++i)
ret[i] = in[reference[i].first];
return ret;
}
typedef std::vector<int> int_vec_t;
typedef std::vector<std::string> str_vec_t;
typedef std::vector<size_t> index_vec_t;
class SequenceGen {
public:
SequenceGen (int start = 0) : current(start) { }
int operator() () { return current++; }
private:
int current;
};
class Comp{
int_vec_t& _v;
public:
Comp(int_vec_t& v) : _v(v) {}
bool operator()(size_t i, size_t j){
return _v[i] < _v[j];
}
};
index_vec_t indices(3);
std::generate(indices.begin(), indices.end(), SequenceGen(0));
//indices are {0, 1, 2}
int_vec_t Index = { 3, 1, 2 };
str_vec_t Values = { "Third", "First", "Second" };
std::sort(indices.begin(), indices.end(), Comp(Index));
//now indices are {1,2,0}
Now you can use the "indices" vector to index into "Values" vector.
Put your values in a Boost Multi-Index container then iterate over to read the values in the order you want. You can even copy them to another vector if you want to.
Only one rough solution comes to my mind: create a vector that is the sum of all other vectors (a vector of structures, like {3,Third,...},{1,First,...}) then sort this vector by the first field, and then split the structures again.
Probably there is a better solution inside Boost or using the standard library.
You can probably define a custom "facade" iterator that does what you need here. It would store iterators to all your vectors or alternatively derive the iterators for all but the first vector from the offset of the first. The tricky part is what that iterator dereferences to: think of something like boost::tuple and make clever use of boost::tie. (If you wanna extend on this idea, you can build these iterator types recursively using templates but you probably never want to write down the type of that - so you either need c++0x auto or a wrapper function for sort that takes ranges)
I think what you really need (but correct me if I'm wrong) is a way to access elements of a container in some order.
Rather than rearranging my original collection, I would borrow a concept from Database design: keep an index, ordered by a certain criterion. This index is an extra indirection that offers great flexibility.
This way it is possible to generate multiple indices according to different members of a class.
using namespace std;
template< typename Iterator, typename Comparator >
struct Index {
vector<Iterator> v;
Index( Iterator from, Iterator end, Comparator& c ){
v.reserve( std::distance(from,end) );
for( ; from != end; ++from ){
v.push_back(from); // no deref!
}
sort( v.begin(), v.end(), c );
}
};
template< typename Iterator, typename Comparator >
Index<Iterator,Comparator> index ( Iterator from, Iterator end, Comparator& c ){
return Index<Iterator,Comparator>(from,end,c);
}
struct mytype {
string name;
double number;
};
template< typename Iter >
struct NameLess : public binary_function<Iter, Iter, bool> {
bool operator()( const Iter& t1, const Iter& t2 ) const { return t1->name < t2->name; }
};
template< typename Iter >
struct NumLess : public binary_function<Iter, Iter, bool> {
bool operator()( const Iter& t1, const Iter& t2 ) const { return t1->number < t2->number; }
};
void indices() {
mytype v[] = { { "me" , 0.0 }
, { "you" , 1.0 }
, { "them" , -1.0 }
};
mytype* vend = v + _countof(v);
Index<mytype*, NameLess<mytype*> > byname( v, vend, NameLess<mytype*>() );
Index<mytype*, NumLess <mytype*> > bynum ( v, vend, NumLess <mytype*>() );
assert( byname.v[0] == v+0 );
assert( byname.v[1] == v+2 );
assert( byname.v[2] == v+1 );
assert( bynum.v[0] == v+2 );
assert( bynum.v[1] == v+0 );
assert( bynum.v[2] == v+1 );
}
A slightly more compact variant of xtofl's answer for if you are just looking to iterate through all your vectors based on the of a single keys vector. Create a permutation vector and use this to index into your other vectors.
#include <boost/iterator/counting_iterator.hpp>
#include <vector>
#include <algorithm>
std::vector<double> keys = ...
std::vector<double> values = ...
std::vector<size_t> indices(boost::counting_iterator<size_t>(0u), boost::counting_iterator<size_t>(keys.size()));
std::sort(begin(indices), end(indices), [&](size_t lhs, size_t rhs) {
return keys[lhs] < keys[rhs];
});
// Now to iterate through the values array.
for (size_t i: indices)
{
std::cout << values[i] << std::endl;
}
ltjax's answer is a great approach - which is actually implemented in boost's zip_iterator http://www.boost.org/doc/libs/1_43_0/libs/iterator/doc/zip_iterator.html
It packages together into a tuple whatever iterators you provide it.
You can then create your own comparison function for a sort based on any combination of iterator values in your tuple. For this question, it would just be the first iterator in your tuple.
A nice feature of this approach is that it allows you to keep the memory of each individual vector contiguous (if you're using vectors and that's what you want). You also don't need to store a separate index vector of ints.
This would have been an addendum to Konrad's answer as it an approach for a in-place variant of applying the sort order to a vector. Anyhow since the edit won't go through I will put it here
Here is a in-place variant with a slightly higher time complexity that is due to a primitive operation of checking a boolean. The additional space complexity is of a vector which can be a space efficient compiler dependent implementation. The complexity of a vector can be eliminated if the given order itself can be modified.
Here is a in-place variant with a slightly higher time complexity that is due to a primitive operation of checking a boolean. The additional space complexity is of a vector which can be a space efficient compiler dependent implementation. The complexity of a vector can be eliminated if the given order itself can be modified. This is a example of what the algorithm is doing.
If the order is 3 0 4 1 2, the movement of the elements as indicated by the position indices would be 3--->0; 0--->1; 1--->3; 2--->4; 4--->2.
template<typename T>
struct applyOrderinPlace
{
void operator()(const vector<size_t>& order, vector<T>& vectoOrder)
{
vector<bool> indicator(order.size(),0);
size_t start = 0, cur = 0, next = order[cur];
size_t indx = 0;
T tmp;
while(indx < order.size())
{
//find unprocessed index
if(indicator[indx])
{
++indx;
continue;
}
start = indx;
cur = start;
next = order[cur];
tmp = vectoOrder[start];
while(next != start)
{
vectoOrder[cur] = vectoOrder[next];
indicator[cur] = true;
cur = next;
next = order[next];
}
vectoOrder[cur] = tmp;
indicator[cur] = true;
}
}
};
Here is a relatively simple implementation using index mapping between the ordered and unordered names that will be used to match the ages to the ordered names:
void ordered_pairs()
{
std::vector<std::string> names;
std::vector<int> ages;
// read input and populate the vectors
populate(names, ages);
// print input
print(names, ages);
// sort pairs
std::vector<std::string> sortedNames(names);
std::sort(sortedNames.begin(), sortedNames.end());
std::vector<int> indexMap;
for(unsigned int i = 0; i < sortedNames.size(); ++i)
{
for (unsigned int j = 0; j < names.size(); ++j)
{
if (sortedNames[i] == names[j])
{
indexMap.push_back(j);
break;
}
}
}
// use the index mapping to match the ages to the names
std::vector<int> sortedAges;
for(size_t i = 0; i < indexMap.size(); ++i)
{
sortedAges.push_back(ages[indexMap[i]]);
}
std::cout << "Ordered pairs:\n";
print(sortedNames, sortedAges);
}
For the sake of completeness, here are the functions populate() and print():
void populate(std::vector<std::string>& n, std::vector<int>& a)
{
std::string prompt("Type name and age, separated by white space; 'q' to exit.\n>>");
std::string sentinel = "q";
while (true)
{
// read input
std::cout << prompt;
std::string input;
getline(std::cin, input);
// exit input loop
if (input == sentinel)
{
break;
}
std::stringstream ss(input);
// extract input
std::string name;
int age;
if (ss >> name >> age)
{
n.push_back(name);
a.push_back(age);
}
else
{
std::cout <<"Wrong input format!\n";
}
}
}
and:
void print(const std::vector<std::string>& n, const std::vector<int>& a)
{
if (n.size() != a.size())
{
std::cerr <<"Different number of names and ages!\n";
return;
}
for (unsigned int i = 0; i < n.size(); ++i)
{
std::cout <<'(' << n[i] << ", " << a[i] << ')' << "\n";
}
}
And finally, main() becomes:
#include <iostream>
#include <sstream>
#include <string>
#include <vector>
#include <algorithm>
void ordered_pairs();
void populate(std::vector<std::string>&, std::vector<int>&);
void print(const std::vector<std::string>&, const std::vector<int>&);
//=======================================================================
int main()
{
std::cout << "\t\tSimple name - age sorting.\n";
ordered_pairs();
}
//=======================================================================
// Function Definitions...
**// C++ program to demonstrate sorting in vector
// of pair according to 2nd element of pair
#include <iostream>
#include<string>
#include<vector>
#include <algorithm>
using namespace std;
// Driver function to sort the vector elements
// by second element of pairs
bool sortbysec(const pair<char,char> &a,
const pair<int,int> &b)
{
return (a.second < b.second);
}
int main()
{
// declaring vector of pairs
vector< pair <char, int> > vect;
// Initialising 1st and 2nd element of pairs
// with array values
//int arr[] = {10, 20, 5, 40 };
//int arr1[] = {30, 60, 20, 50};
char arr[] = { ' a', 'b', 'c' };
int arr1[] = { 4, 7, 1 };
int n = sizeof(arr)/sizeof(arr[0]);
// Entering values in vector of pairs
for (int i=0; i<n; i++)
vect.push_back( make_pair(arr[i],arr1[i]) );
// Printing the original vector(before sort())
cout << "The vector before sort operation is:\n" ;
for (int i=0; i<n; i++)
{
// "first" and "second" are used to access
// 1st and 2nd element of pair respectively
cout << vect[i].first << " "
<< vect[i].second << endl;
}
// Using sort() function to sort by 2nd element
// of pair
sort(vect.begin(), vect.end(), sortbysec);
// Printing the sorted vector(after using sort())
cout << "The vector after sort operation is:\n" ;
for (int i=0; i<n; i++)
{
// "first" and "second" are used to access
// 1st and 2nd element of pair respectively
cout << vect[i].first << " "
<< vect[i].second << endl;
}
getchar();
return 0;`enter code here`
}**
with C++11 lambdas and the STL algorithms based on answers from Konrad Rudolph and Gabriele D'Antona:
template< typename T, typename U >
std::vector<T> sortVecAByVecB( std::vector<T> & a, std::vector<U> & b ){
// zip the two vectors (A,B)
std::vector<std::pair<T,U>> zipped(a.size());
for( size_t i = 0; i < a.size(); i++ ) zipped[i] = std::make_pair( a[i], b[i] );
// sort according to B
std::sort(zipped.begin(), zipped.end(), []( auto & lop, auto & rop ) { return lop.second < rop.second; });
// extract sorted A
std::vector<T> sorted;
std::transform(zipped.begin(), zipped.end(), std::back_inserter(sorted), []( auto & pair ){ return pair.first; });
return sorted;
}
So many asked this question and nobody came up with a satisfactory answer. Here is a std::sort helper that enables to sort two vectors simultaneously, taking into account the values of only one vector. This solution is based on a custom RadomIt (random iterator), and operates directly on the original vector data, without temporary copies, structure rearrangement or additional indices:
C++, Sort One Vector Based On Another One

count of elements in a set (not total count)

I have a set of strings in c++.
i am inserting into that set as :
m.insert("1-2-35-2");
m.insert("1-2-36-1");
m.insert("1-2-37-2");
m.insert("1-2-38-1");
m.insert("1-2-39-2");
m.insert("2-2-40-1");
m.insert("2-2-41-2");
m.insert("2-2-42-1");
m.insert("1-2-43-2");
m.insert("1-2-44-1");
m.insert("1-2-45-2");
m.insert("1-2-46-1");
m.insert("1-2-47-2");
i want to calculate the count of all the strings inside the set which start with "2-"(count =3) and also which start with "1-"(count=10).
is there any way to do it.
I tried with lower_bound and upper_bound but its giving me some errors.
errors are coming for the statement:
int i=it_upper-it_lower;
I am using solaris SPARC OS.
i just tested this program
#include <iostream>
#include <iterator>
#include <list>
using namespace std;
int main () {
list<int> mylist;
for (int i=0; i<10; i++) mylist.push_back (i*10);
list<int>::iterator first = mylist.begin();
list<int>::iterator last = mylist.end();
cout << "The distance is: " << distance(first,last) << endl;
return 0;
}
it gives me compilation error:
line 13: Error: Could not find a match for std::distance<std::ForwardIterator, std::Distance>(std::list<int, std::allocator<int>>::iterator, std::list<int, std::allocator<int>>::iterator).
1 Error(s) detected.
Sorry. Wrong answer
Update:
count_if is an algorithm to count elements based on function. Try like in this example:
bool struct key_part: public std::unary_function< std::string, bool >
{
std::string _part;
key_part(const std::string part):_part(part){}
bool operator()(std::string &s)
{
return s.find(_part)!=std::string::npos;
}
}
std::count_if( m.begin(), m.end(), key_part("1-") );
It will count all elements that contains "1-" as part of key
If you have a modern compiler that supports lambdas, you could use those as the predicate to count_if:
auto if_s_1 = [](const std::string &s) { return s.find("1-") == 0; }
auto if_s_2 = [](const std::string &s) { return s.find("2-") == 0; }
int count1 = std::count_if(m.begin(), m.end(), if_s_1);
int count2 = std::count_if(m.begin(), m.end(), if_s_2);

Sorting a set<string> on the basis of length

My question is related to this.
I wanted to perform a sort() operation over the set with the help of a lambda expression as a predicate.
My code is
#include <set>
#include <string>
#include <iostream>
#include <algorithm>
int main() {
using namespace std;
string s = "abc";
set<string> results;
do {
for (int n = 1; n <= s.size(); ++n) {
results.insert(s.substr(0, n));
}
} while (next_permutation(s.begin(), s.end()));
sort (results.begin(),results.end());[](string a, string b)->bool{
size_t alength = a.length();
size_t blength = b.length();
return (alength < blength);
});
for (set<string>::const_iterator x = results.begin(); x != results.end(); ++x) {
cout << *x << '\n';
}
return 0;
}
But the numbers and types of errors were so complex that I couldn't understand how to fix them. Can someone tell me whats wrong with this code.
Edit: Note that Steve Townsend's solution is actually the one you're searching for, as he inlines as a C++0x Lambda what I write as C++03 code below.
Another solution would be to customize the std::set ordering function:
The std::set is already ordered...
The std::set has its own ordering, and you are not supposed to change it once it is constructed. So, the following code:
int main(int argc, char* argv[])
{
std::set<std::string> aSet ;
aSet.insert("aaaaa") ;
aSet.insert("bbbbb") ;
aSet.insert("ccccccc") ;
aSet.insert("ddddddd") ;
aSet.insert("e") ;
aSet.insert("f") ;
outputSet(aSet) ;
return 0 ;
}
will output the following result:
- aaaaa
- bbbbb
- ccccccc
- ddddddd
- e
- f
... But you can customize its ordering function
Now, if you want, you can customize your set by using your own comparison function:
struct MyStringLengthCompare
{
bool operator () (const std::string & p_lhs, const std::string & p_rhs)
{
const size_t lhsLength = p_lhs.length() ;
const size_t rhsLength = p_rhs.length() ;
if(lhsLength == rhsLength)
{
return (p_lhs < p_rhs) ; // when two strings have the same
// length, defaults to the normal
// string comparison
}
return (lhsLength < rhsLength) ; // compares with the length
}
} ;
In this comparison functor, I did handle the case "same length but different content means different strings", because I believe (perhaps wrongly) that the behaviour in the original program is an error. To have the behaviour coded in the original program, please remove the if block from the code.
And now, you construct the set:
int main(int argc, char* argv[])
{
std::set<std::string, MyStringLengthCompare> aSet ;
aSet.insert("aaaaa") ;
aSet.insert("bbbbb") ;
aSet.insert("ccccccc") ;
aSet.insert("ddddddd") ;
aSet.insert("e") ;
aSet.insert("f") ;
outputSet(aSet) ;
return 0 ;
}
The set will now use the functor MyStringLengthCompare to order its items, and thus, this code will output:
- e
- f
- aaaaa
- bbbbb
- ccccccc
- ddddddd
But beware of the ordering mistake!
When you create your own ordering function, it must follow the following rule:
return true if (lhs < rhs) is true, return false otherwise
If for some reason your ordering function does not respect it, you'll have a broken set on your hands.
std::sort rearranges the elements of the sequence you give it. The arrangement of the sequence in the set is fixed, so the only iterator you can have is a const iterator.
You'll need to copy results into a vector or deque (or such) first.
vector sortable_results( results.begin(), results.end() );
You can customize the ordering of the elements in the set by providing a custom predicate to determine ordering of added elements relative to extant members. set is defined as
template <
class Key,
class Traits=less<Key>,
class Allocator=allocator<Key>
>
class set
where Traits is
The type that provides a function
object that can compare two element
values as sort keys to determine their
relative order in the set. This
argument is optional, and the binary
predicate less is the default
value.
There is background on how to use lambda expression as a template parameter here.
In your case this translates to:
auto comp = [](const string& a, const string& b) -> bool
{ return a.length() < b.length(); };
auto results = std::set <string, decltype(comp)> (comp);
Note that this will result in set elements with the same string length being treated as duplicates which is not what you want, as far as I can understand the desired outcome.
sort requires random access iterators which set doesn't provide (It is a bidirectional iterator). If you change the code to use vector it compiles fine.
You cannot sort a set. It's always ordered on keys (which are elements themselves).
To be more specific, std::sort requires random access iterators. The iterators provided by std::set are not random.
Since I wrote the original code you're using, perhaps I can expand on it... :)
struct cmp_by_length {
template<class T>
bool operator()(T const &a, T const &b) {
return a.length() < b.length() or (a.length() == b.length() and a < b);
}
};
This compares by length first, then by value. Modify the set definition:
set<string, cmp_by_length> results;
And you're good to go:
int main() {
using namespace std;
string s = "abc";
typedef set<string, cmp_by_length> Results; // convenience for below
Results results;
do {
for (int n = 1; n <= s.size(); ++n) {
results.insert(s.substr(0, n));
}
} while (next_permutation(s.begin(), s.end()));
// would need to add cmp_by_length below, if I hadn't changed to the typedef
// i.e. set<string, cmp_by_length>::const_iterator
// but, once you start using nested types on a template, a typedef is smart
for (Results::const_iterator x = results.begin(); x != results.end(); ++x) {
cout << *x << '\n';
}
// of course, I'd rather write... ;)
//for (auto const &x : results) {
// cout << x << '\n';
//}
return 0;
}
std::set is most useful to maintain a sorted and mutating list. It faster and smaller to use a vector when the set itself wont change much once it's been built.
#include <vector>
#include <string>
#include <iostream>
#include <algorithm>
int main() {
using namespace std;
string s = "abc";
vector<string> results;
do {
for (size_t n = 1; n <= s.size(); ++n) {
results.push_back(s.substr(0, n));
}
} while (next_permutation(s.begin(), s.end()));
//make it unique
sort( results.begin(), results.end() );
auto end_sorted = unique( results.begin(), results.end() );
results.erase( end_sorted, results.end() );
//sort by length
sort (results.begin(),results.end());
[](string lhs, string rhs)->bool
{ return lhs.length() < rhs.length(); } );
for ( const auto& result: results ) {
cout << result << '\n';
}
}
I used the classic, sort/unique/erase combo to make the results set unique.I also cleaned up your code to be a little bit more c++0x-y.