So.. I am trying to find the maximum value of a vector and its position in the vector. I am using a for loop, and it's working fine. My problem is, that if the maximum value appears more than once, I want to know all the positions in which it appears in the vector.. So, how could I manage this?
So far, this is the code I am using: (the vector called v has elements that I read from a file but I will not add that part of the code)
std::vector<double>v;
double maxvalue;
int position=0;
maxvalue = v[0];
for (unsigned int i=0; i<v.size(); i++){
if (v[i]> maxvalue){
maxvalue=v[i];
position= i;
}
}
You could modify your approach to keep a vector of indices where the maximum occurred:
#include <cfloat>
#include <iostream>
#include <utility>
#include <vector>
std::pair<double, std::vector<std::size_t>> FindMaxElements(std::vector<double> const& v)
{
std::vector<std::size_t> indices;
double current_max = -DBL_MAX;
for (std::size_t i = 0; i < v.size(); ++i)
{
if (v[i] > current_max)
{
current_max = v[i];
indices.clear();
}
if (v[i] == current_max)
{
indices.push_back(i);
}
}
return std::make_pair(current_max, indices);
}
int main()
{
auto result = FindMaxElements({1, 4, 7, 2, 7, 3});
std::cout << "max: " << result.first << '\n';
std::cout << "indices: ";
for (auto i : result.second)
std::cout << i << ' ';
}
Output
max: 7
indices: 2 4
Here is a two-pass version using the standard library (whereas it might be cleaner without it):
#include <vector>
#include <algorithm>
int main()
{
std::vector<double> v {/* fill it*/ };
std::vector<int> pos;
auto it = std::max_element(std::begin(v), std::end(v));
while (it != std::end(v))
{
pos.push_back(std::distance(std::begin(v), it));
it = std::find(std::next(it), std::end(v), *it);
}
//...
}
The function template below, find_maximums(), returns an std::vector<size_t> that contains the positions where the maximums are in the input vector. Note that it returns an empty vector of indexes if the input vector is empty.
template<typename T>
auto find_maximums(const std::vector<T>& v) {
std::vector<size_t> indexes;
for (auto it_max = std::max_element(v.begin(), v.end()); it_max != v.end();
it_max = std::find(it_max+1, v.end(), *it_max))
{
auto index = std::distance(v.begin(), it_max);
indexes.push_back(index);
}
return indexes;
}
As an example of use:
auto main() -> int {
std::vector<int> v = {11, 7, 3, 11, 0, 7, 1, 11, 11};
auto max_indexes = find_maximums(v);
if (max_indexes.empty())
return 1;
std::cout << "max: " << v[max_indexes.front()] << std::endl;
std::cout << "max at positions: ";
for (auto idx: max_indexes)
std::cout << idx << ' ';
std::cout << '\n';
}
It outputs:
max: 11
max at positions: 0 3 7 8
Passing a couple of iterators and a comparator
template <class It,
class Comp = std::less<typename std::iterator_traits<It>::value_type>>
auto max_elements_indices(It first, It last, Comp cmp = Comp{})
{
// This function returns a vector of indices, so to get the maximum, the caller
// should first check if the returned vector is empty and then use one of
// those indices to retrieve the value.
std::vector<std::size_t> indices;
if (first == last)
return indices;
// Using the first element instead of a sentinel value is easier to generalize
indices.push_back(0);
auto value = *first;
for (auto i = std::next(first); i != last; ++i)
{
// The most common case should be an element below the maximum
if ( cmp(*i, value) )
continue;
else
{
if ( cmp(value, *i) )
{
value = *i;
indices.clear();
}
indices.push_back(std::distance(first, i));
}
}
return indices;
}
It is testable here.
Related
I've searched the Internet and known how to delete an element (with std::erase) and finding duplicates of an element to then delete it (vec.erase(std::unique(vec.begin(), vec.end()),vec.end());). But all methods only delete either an element or its duplicates.
I want to delete both.
For example, using this vector:
std::vector<int> vec = {2,3,1,5,2,2,5,1};
I want output to be:
{3}
My initial idea was:
void removeDuplicatesandElement(std::vector<int> &vec)
{
std::sort(vec.begin(), vec.end());
int passedNumber = 0; //To tell amount of number not deleted (since not duplicated)
for (int i = 0; i != vec.size(); i = passedNumber) //This is not best practice, but I tried
{
if (vec[i] == vec[i+1])
{
int ctr = 1;
for(int j = i+1; j != vec.size(); j++)
{
if (vec[j] == vec[i]) ctr++;
else break;
}
vec.erase(vec.begin()+i, vec.begin()+i+ctr);
}
else passedNumber++;
}
}
And it worked. But this code is redundant and runs at O(n^2), so I'm trying to find other ways to solve the problem (maybe an STL function that I've never heard of, or just improve the code).
Something like this, perhaps:
void removeDuplicatesandElement(std::vector<int> &vec) {
if (vec.size() <= 1) return;
std::sort(vec.begin(), vec.end());
int cur_val = vec.front() - 1;
auto pred = [&](const int& val) {
if (val == cur_val) return true;
cur_val = val;
// Look ahead to the next element to see if it's a duplicate.
return &val != &vec.back() && (&val)[1] == val;
};
vec.erase(std::remove_if(vec.begin(), vec.end(), pred), vec.end());
}
Demo
This relies heavily on the fact that std::vector is guaranteed to have contiguous storage. It won't work with any other container.
You can do it using STL maps as follows:
#include <iostream>
#include <vector>
#include <unordered_map>
using namespace std;
void retainUniqueElements(vector<int> &A){
unordered_map<int, int> Cnt;
for(auto element:A) Cnt[element]++;
A.clear(); //removes all the elements of A
for(auto i:Cnt){
if(i.second == 1){ // that if the element occurs once
A.push_back(i.first); //then add it in our vector
}
}
}
int main() {
vector<int> vec = {2,3,1,5,2,2,5,1};
retainUniqueElements(vec);
for(auto i:vec){
cout << i << " ";
}
cout << "\n";
return 0;
}
Output:
3
Time Complexity of the above approach: O(n)
Space Complexity of the above approach: O(n)
From what you have searched, we can look in the vector for duplicated values, then use the Erase–remove idiom to clean up the vector.
#include <vector>
#include <algorithm>
#include <iostream>
void removeDuplicatesandElement(std::vector<int> &vec)
{
std::sort(vec.begin(), vec.end());
if (vec.size() < 2)
return;
for (int i = 0; i < vec.size() - 1;)
{
// This is for the case we emptied our vector
if (vec.size() < 2)
return;
// This heavily relies on the fact that this vector is sorted
if (vec[i] == vec[i + 1])
vec.erase(std::remove(vec.begin(), vec.end(), (int)vec[i]), vec.end());
else
i += 1;
}
// Since all duplicates are removed, the remaining elements in the vector are unique, thus the size of the vector
// But we are not returning anything or any reference, so I'm just gonna leave this here
// return vec.size()
}
int main()
{
std::vector<int> vec = {2, 3, 1, 5, 2, 2, 5, 1};
removeDuplicatesandElement(vec);
for (auto i : vec)
{
std::cout << i << " ";
}
std::cout << "\n";
return 0;
}
Output: 3
Time complexity: O(n)
This is a followup of this question. The only difference is the constrain that the two vectors cannot be combined in a struct.
Suppose we have a vector
std::vector<double> v1 = {9.0,5.0,3.0,2.0,1.0};
Now we sort the vector v1. Let v2 be given by
std::vector<std::string> v2 = {"you?","are","how","there","hello"};
How to transform v2 the same way v1 was transformed by sort?
Based on this answer, you can use an array of indices to "sort" the vector of doubles, and just use the resulting index array to index the vector of strings.
#include <algorithm>
#include <iostream>
#include <string>
#include <numeric>
int main()
{
std::vector<double> v1 = {5.0,9.0,3.0,2.0,1.0};
std::vector<std::string> v2 = {"are", "you?","how","there","hello"};
// Create an array of indices, starting from 0
std::vector<int> index(v1.size());
std::iota(index.begin(), index.end(), 0);
// "Sort" the index array according to the value in the vector of doubles
std::sort(index.begin(), index.end(),
[&](int n1, int n2){ return v1[n1] < v1[n2]; });
// Output results
for (auto i : index )
std::cout << v2[i] << " " << v1[i] << ", index is " << i << "\n";
}
Output:
hello 1, index is 4
there 2, index is 3
how 3, index is 2
are 5, index is 0
you? 9, index is 1
Note:
I changed the original data to illustrate how the index array works.
The abstraction you are missing is the ability to view the vectors as one item. That's the role that a vector of indices is a proxy for in another answer.
I think it is worth mentioning that there are libraries that provide such a concept (often under the name "zip"). For example, with range-v3:
std::vector<double> v1 = {5, 9, 3, 2, 1};
std::vector<std::string> v2 = {"are", "you?", "how", "there", "hello"};
// Sort the vectors
ranges::actions::sort(ranges::views::zip(v1, v2));
// Output results
for (std::size_t i = 0; i < v1.size(); ++i)
std::cout << v2[i] << " " << v1[i] << ", index is " << i << "\n";
A possible solution uses a helper std::vector<int>:
#include <iostream>
#include <vector>
#include <algorithm>
#include <stdexcept>
template<typename T>
void MySort(std::vector<T> t, std::vector<int>& helper)
{
struct StructHelper
{
T t1;
int num;
StructHelper(T t, int i): t1{t}, num{i} {};
bool operator<(const StructHelper& other) const
{ return t1 < other.t1; }
};
std::vector<StructHelper> shVector;
for(int i=0; i<t.size(); ++i)
{
shVector.emplace_back(t[i], i);
}
std::sort(shVector.begin(), shVector.end());
helper = std::vector<int>(t.size());
for(int i=0; i<t.size(); ++i)
{
helper[i] = shVector[i].num;
}
}
template<typename T>
void MySortUsingHelper(std::vector<T>& t1, const std::vector<int>& helper)
{
if(t1.size() != helper.size()) throw std::out_of_range("not same size");
std::vector<T> t2(t1.size());
for(int i=0; i<helper.size(); ++i)
{
t2[i] = t1[helper[i]];
}
t1 = t2;
}
int main() {
std::vector<double> v1 = {9.0,5.0,3.0,2.0,1.0};
std::vector<int> helper;
MySort(v1, helper);
std::vector<std::string> v2 = {"you?","are","how","there","hello"};
MySortUsingHelper(v2, helper);
for(auto elem : v2)
{
std::cout << elem << " ";
}
return 0;
}
You can run the above code online to see the following output:
hello there how are you?
Assume I have a vector with the following elements {1, 1, 2, 3, 3, 4}
I want to write a program with c++ code to remove the unique values and keep only the duplicated once. So the end result will be something like this {1,3}.
So far this is what I've done, but it takes a lot of time,
Is there any way this can be more efficient,
vector <int> g1 = {1,1,2,3,3,4}
vector <int> g2;
for(int i = 0; i < g1.size(); i++)
{
if(count(g1.begin(), g1.end(), g1[i]) > 1)
g2.push_back(g1[i]);
}
v.erase(std::unique(g2.begin(), g2.end()), g2.end());
for(int i = 0; i < g2.size(); i++)
{
cout << g2[i];
}
My approach is to create an <algorithm>-style template, and use an unordered_map to do the counting. This means you only iterate over the input list once, and the time complexity is O(n). It does use O(n) extra memory though, and isn't particularly cache-friendly. Also this does assume that the type in the input is hashable.
#include <algorithm>
#include <iostream>
#include <iterator>
#include <unordered_map>
template <typename InputIt, typename OutputIt>
OutputIt copy_duplicates(
InputIt first,
InputIt last,
OutputIt d_first)
{
std::unordered_map<typename std::iterator_traits<InputIt>::value_type,
std::size_t> seen;
for ( ; first != last; ++first) {
if ( 2 == ++seen[*first] ) {
// only output on the second time of seeing a value
*d_first = *first;
++d_first;
}
}
return d_first;
}
int main()
{
int i[] = {1, 2, 3, 1, 1, 3, 5}; // print 1, 3,
// ^ ^
copy_duplicates(std::begin(i), std::end(i),
std::ostream_iterator<int>(std::cout, ", "));
}
This can output to any kind of iterator. There are special iterators you can use that when written to will insert the value into a container.
Here's a way that's a little more cache friendly than unordered_map answer, but is O(n log n) instead of O(n), though it does not use any extra memory and does no allocations. Additionally, the overall multiplier is probably higher, in spite of it's cache friendliness.
#include <vector>
#include <algorithm>
void only_distinct_duplicates(::std::vector<int> &v)
{
::std::sort(v.begin(), v.end());
auto output = v.begin();
auto test = v.begin();
auto run_start = v.begin();
auto const end = v.end();
for (auto test = v.begin(); test != end; ++test) {
if (*test == *run_start) {
if ((test - run_start) == 1) {
*output = *run_start;
++output;
}
} else {
run_start = test;
}
}
v.erase(output, end);
}
I've tested this, and it works. If you want a generic version that should work on any type that vector can store:
template <typename T>
void only_distinct_duplicates(::std::vector<T> &v)
{
::std::sort(v.begin(), v.end());
auto output = v.begin();
auto test = v.begin();
auto run_start = v.begin();
auto const end = v.end();
for (auto test = v.begin(); test != end; ++test) {
if (*test != *run_start) {
if ((test - run_start) > 1) {
::std::swap(*output, *run_start);
++output;
}
run_start = test;
}
}
if ((end - run_start) > 1) {
::std::swap(*output, *run_start);
++output;
}
v.erase(output, end);
}
Assuming the input vector is not sorted, the following will work and is generalized to support any vector with element type T. It will be more efficient than the other solutions proposed so far.
#include <algorithm>
#include <iostream>
#include <vector>
template<typename T>
void erase_unique_and_duplicates(std::vector<T>& v)
{
auto first{v.begin()};
std::sort(first, v.end());
while (first != v.end()) {
auto last{std::find_if(first, v.end(), [&](int i) { return i != *first; })};
if (last - first > 1) {
first = v.erase(first + 1, last);
}
else {
first = v.erase(first);
}
}
}
int main(int argc, char** argv)
{
std::vector<int> v{1, 2, 3, 4, 5, 2, 3, 4};
erase_unique_and_duplicates(v);
// The following will print '2 3 4'.
for (int i : v) {
std::cout << i << ' ';
}
std::cout << '\n';
return 0;
}
I have 2 improvements for you:
You can change your count to start at g1.begin() + i, everything before was handled by the previous iterations of the loop.
You can change the if to == 2 instead of > 1, so it will add numbers only once, independent of the occurences. If a number is 5 times in the vector, the first 3 will be ignored, the 4th will make it into the new vector and the 5th will be ignored again. So you can remove the erase of the duplicates
Example:
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
int main() {
vector <int> g1 = {1,1,2,3,3,1,4};
vector <int> g2;
for(int i = 0; i < g1.size(); i++)
{
if(count(g1.begin() + i, g1.end(), g1[i]) == 2)
g2.push_back(g1[i]);
}
for(int i = 0; i < g2.size(); i++)
{
cout << g2[i] << " ";
}
cout << endl;
return 0;
}
I'll borrow a principal from Python which is excellent for such operations -
You can use a dictionary where the dictionary-key is the item in the vector and the dictionary-value is the count (start with 1 and increase by one every time you encounter a value that is already in the dictionary).
afterward, create a new vector (or clear the original) with only the dictionary keys that are larger than 1.
Look up in google - std::map
Hope this helps.
In general, that task got complexity about O(n*n), that's why it appears slow. Does it have to be a vector? Is that a restriction? Must it be ordered? If not, it better to actually store values as std::map, which eliminates doubles when populated, or as a std::multimap if presence of doubles matters.
I'm supposed to:
Print vector elements sorted without repetition.
Delete the elements that are printed from vector.
Repeat the the previous steps until vector is empty.
But it seems that my code takes more time so, I seek for optimisation. I've tried to do this task with std::vector and std::set.
Here is my approach:
#include <iostream>
#include <algorithm>
#include <vector>
#include <set>
using namespace std;
int main () {
int n;
cin >> n;
vector<int> v(n);
set<int> st;
for (int i = 0; i < n; i++) {
cin >> v[i];
}
while (!v.empty()) {
for (int i = 0; i < v.size(); i++)
st.insert(v[i]);
for (auto x : st) {
cout << x << ' ';
auto it = find(v.begin(), v.end(), x);
if (it != v.end())
v.erase(it);
}
st.clear();
cout << "\n";
}
return 0;
}
For example input is like:
7
1 2 3 3 2 4 3
Output gonna be like this:
1 2 3 4
2 3
3
You might use std::map instead of std::vector/std::set to keep track of numbers:
#include <iostream>
#include <map>
int main () {
map<int, int> m;
int size;
std::cin >> size;
for (int i = 0; i != size; i++) {
int number;
std::cin >> number;
++m[number];
}
while (!m.empty()) {
for (auto it = m.begin(); it != m.end(); /*Empty*/) {
const auto number = it->first;
auto& count = it->second;
std::cout << number << ' ';
if (--count == 0) {
it = m.erase(it);
} else {
++it;
}
}
std::cout << "\n";
}
}
Complexity is now O(n log(n)) instead of O(n²) (with lot of internal allocations).
Due to it overwriting the elements expected to be deleted, std::unique won't be much use for this problem. My solution:
std::sort(v.begin(), v.end());
while (!v.empty())
{
int last = v.front();
std::cout << last << " ";
v.erase(v.begin());
for (auto it = v.begin(); it != v.end(); /* no-op */)
{
if (*it == last)
{
++it;
}
else
{
last = *it;
std::cout << last << " ";
it = v.erase(it);
}
}
std::cout << std::endl;
}
You could probably improve performance further by reversing the sorting of the vector, and then iterating through backwards (since it's cheaper to delete from closer to the back of the vector), but that would complicate the code further, so I'll say "left as an exercise for the reader".
You can use std::map
auto n = 0;
std::cin >> n;
std::map<int, int> mp;
while (--n >= 0) {
auto i = 0;
std::cin >> i;
mp[i] += 1;
}
while (!mp.empty()) {
for (auto& it: mp) {
std::cout << it.first << " ";
it.second--;
}
for (auto it = mp.begin(); it != mp.end(); ++it) {
if (it->second == 0) mp.erase(it);
}
std::cout << "\n";
}
without any erase
auto n = 0;
std::cin >> n;
std::map<int, int> mp;
while (--n >= 0) {
auto i = 0;
std::cin >> i;
mp[i] += 1;
}
auto isDone = false;
while (!isDone) {
isDone = true;
for (auto& it: mp) {
if (it.second > 0) std::cout << it.first << " ";
if (--it.second > 0) isDone = false;
}
std::cout << "\n";
}
Here is a solution using sort and vector. It uses a second vector to hold the unique items and print them.
#include <iostream>
#include <vector>
#include <algorithm>
#include <iterator>
int main()
{
std::vector<int> v{1,2,3,3,2,4,3};
std::sort(v.begin(), v.end());
std::vector<int>::iterator vit;
while(!v.empty()){
std::vector<int> printer;
std::vector<int>::iterator pit;
vit = v.begin();
while (vit != v.end()){
pit = find(printer.begin(), printer.end(), *vit);
if (pit == printer.end()){
printer.push_back(*vit);
vit = v.erase(vit);
} else {
++vit;
}
}
std::copy(printer.begin(), printer.end(), std::ostream_iterator<int>(std::cout, " "));
std::cout << '\n';
}
}
Output:
1 2 3 4
2 3
3
It's not clear (at least to me) exactly what you're talking about when you mention "efficiency". Some people use it to refer solely to computational complexity. Others think primarily in terms of programmer's time, while still others think of overall execution speed, regardless of whether that's obtained via changes in computational complexity, or (for one example) improved locality of reference leading to better cache utilization.
So, with that warning, I'm not sure whether this really improves what you care about or not, but it's how I think I'd do the job anyway:
#include <iostream>
#include <vector>
#include <algorithm>
#include <iterator>
// preconditions: input range is sorted
template <class BidiIt>
BidiIt partition_unique(BidiIt begin, BidiIt end) {
auto pivot = end;
for (auto pos = begin; pos != pivot; ++pos) {
auto mid = std::next(pos);
for ( ; mid < pivot && *mid == *pos; ++mid, --pivot)
;
std::rotate(std::next(pos), mid, end);
}
return pivot;
}
template <class It>
void show(It b, It e, std::ostream &os) {
while (b != e) {
os << *b << ' ';
++b;
}
os << '\n';
}
int main() {
std::vector<int> input{ 1, 2, 3, 3, 2, 4, 3 };
std::sort(input.begin(), input.end());
auto begin = input.begin();
auto pos = begin;
while ((pos = partition_unique(begin, input.end())) != input.end()) {
show(begin, pos, std::cout);
begin = pos;
}
show(begin, input.end(), std::cout);
}
I'm not really sure it's possible to improve the computational complexity much over what this does (but it might be--I haven't thought about it enough to be sure one way or the other). Compared to some versions I see posted already, there's a decent chance this will improve overall speed (e.g., since it just moves things around inside the same vector, it's likely to get better locality than those that copy data from one vector to another.
The code is in java but the idea remains the same.
At first, I sort the array. Now, the idea is to create buckets.
This means that each line of sorted elements is like a bucket. So, find the count of each element. Now, put that element into each bucket, count number of times. If it so happens that bucket size is less, create a new bucket and add the current element to it.
In the end, print all buckets.
Time Complexity is O(nlog(n)) for sorting and O(n) for the buckets since you have to visit each and every element to print it. So, it's O(nlog(n)) + O(n) = O(nlog(n)) asymptotically.
Code:
import java.util.*;
public class GFG {
public static void main(String[] args){
int[] arr1 = {1,2,3,3,2,4,3};
int[] arr2 = {45,98,65,32,65,74865};
int[] arr3 = {100,100,100,100,100};
int[] arr4 = {100,200,300,400,500};
printSeries(compute(arr1,arr1.length));
printSeries(compute(arr2,arr2.length));
printSeries(compute(arr3,arr3.length));
printSeries(compute(arr4,arr4.length));
}
private static void printSeries(List<List<Integer>> res){
int size = res.size();
for(int i=0;i<size;++i){
System.out.println(res.get(i).toString());
}
}
private static List<List<Integer>> compute(int[] arr,int N){
List<List<Integer>> buckets = new ArrayList<List<Integer>>();
Arrays.sort(arr);
int bucket_size = 0;
for(int i=0;i<N;++i){
int last_index = i;
if(bucket_size > 0){
buckets.get(0).add(arr[i]);
}else{
buckets.add(newBucket(arr[i]));
bucket_size++;
}
for(int j=i+1;j<N;++j){
if(arr[i] != arr[j]) break;
if(j-i < bucket_size){
buckets.get(j-i).add(arr[i]);
}else{
buckets.add(newBucket(arr[i]));
bucket_size++;
}
last_index = j;
}
i = last_index;
}
return buckets;
}
private static List<Integer> newBucket(int value){
List<Integer> new_bucket = new ArrayList<>();
new_bucket.add(value);
return new_bucket;
}
}
OUTPUT
[1, 2, 3, 4]
[2, 3]
[3]
[32, 45, 65, 98, 74865]
[65]
[100]
[100]
[100]
[100]
[100]
[100, 200, 300, 400, 500]
This is what i came up with:
http://coliru.stacked-crooked.com/a/b3f06693a74193e5
The key idea:
sort vector
print by iterating. just print a value if it differs from last printed
remove unique elements. i have done this with what i called inverse_unique. the std library comes with an algorithm called unique, which will remove all duplicates. i inverted this so that it will just keep all dublicates.
so we have no memory allocation at all. i cant see how one could make the algorithm more efficient. we are just doing the bare minimum and its exactly done the way a human thinks about.
i tested it with several combinations. hope its bug free ;-P
code:
#include <iostream>
#include <algorithm>
#include <vector>
template<class ForwardIt>
ForwardIt inverse_unique(ForwardIt first, ForwardIt last)
{
if (first == last)
return last;
auto one_ahead = first+1;
auto dst = first;
while(one_ahead != last)
{
if(*first == *one_ahead)
{
*dst = std::move(*first);
++dst;
}
++first;
++one_ahead;
}
return dst;
}
void print_unique(std::vector<int> const& v)
{
if(v.empty()) return;
// print first
std::cout << v[0] << ' ';
auto last_printed = v.cbegin();
// print others
for(auto it = std::next(std::cbegin(v)); it != std::cend(v); ++it)
{
if(*it != *last_printed)
{
std::cout << *it << ' ';
last_printed = it;
}
}
std::cout << "\n";
}
void remove_uniques(std::vector<int> & v)
{
auto new_end = inverse_unique(std::begin(v), std::end(v));
v.erase(new_end, v.end());
}
int main ()
{
std::vector<int> v = {1, 2, 3, 3, 2, 4, 3};
std::sort(std::begin(v), std::end(v));
while (!v.empty())
{
print_unique(v);
remove_uniques(v);
}
return 0;
}
Edit: updated inverse_unique function. should be easy to understand now.
Half baked at http://coliru.stacked-crooked.com/a/c45df1591d967075
Slightly modified counting sort.
#include <iostream>
#include <vector>
#include <algorithm>
#include <iterator>
#include <map>
int main() {
std::vector<int> v{1,2,3,3,2,4,3};
std::map<int, int> map;
for (auto x : v)
++map[x];
while(map.size()) {
for(auto pair = map.begin(); pair != map.end(); ) {
std::cout << pair->first << ' ';
if (!--pair->second)
pair = map.erase(pair);
else
++pair;
}
std::cout << "\n";
}
return 0;
}
In my program I have a set: ( std::set<int> myset; ) and after some treatment I finally take one iterator which I have decelerated like this ( set<int>::iterator findnumb; ). Well let say I have to do this 5 times in MYSET, 5 times define differences iterators (it1, it2 ,....it5) and not only this but after that to sort them by the content that they show. My idea it was to create a loop and in every time put the specific iterator in a set and sort but I don't know how I declare this.
I create an simple example that take for user 5 number and finds the nearest bigger from our set and print it. But in every loop the old iterator replace from the new one and in my real program I want to store this iterator int a set. Well based on this can you modify this example and store the iterators in a set sort by the content that they show.
#include <iostream>
#include <set>
#include <vector>
using namespace std;
int main()
{
std::set<int> myset;
for (int i=1; i<=20; ++i)
{myset.insert(i*10);}
vector<int> vect;
int numb;
for (int i=0; i<5; ++i)
{
cout << "give a number" <<endl;
cin >> numb;
vect.push_back(numb);
}
set<int>::iterator findnumb;
for (vector<int>::iterator it = vect.begin(); it != vect.end(); ++it)
{
findnumb=myset.find(*it);
while ( (*findnumb)!=(*it))
{ ++(*it); findnumb=myset.find(*it);}
cout << *findnumb<<" ";
}
cout << endl;
return 0;
}
I am not sure if I understood your question correctly.
struct dereferenced_less {
template<typename I>
bool operator()(const I& lh, const I& rh) const { return *lh < *rh; }
};
int main()
{
using namespace std;
set<int> myset = { 10, 20, 30, 40, 50, 60, 70, 80, 90, 100 };
vector<int> vect = { 17, 32, 4, 16, 28, 120 };
set<set<int>::const_iterator, dereferenced_less> findnumbs;
for (int i : vect)
{
auto findnumb = myset.upper_bound(i);
if (findnumb != end(myset))
findnumbs.insert(findnumb);
}
// findnumbs: {10, 20, 30, 40}
}
If you only want the iterators ordered it might be better to sort the input vector and process the set skipping dupklicate elements.
vector<set<int>::const_iterator> result;
sort(begin(vect), end(vect));
for (int i : vect)
{
auto find = myset.upper_bound(i);
if (find != end(myset) && (result.empty() || *result.back() != *find))
result.push_back(find);
}
Don't use sets where you can use vectors.
and this
for (vector<int>::iterator it = vect.begin(); it != vect.end(); ++it)
is NOT how you traverse a vector!!! Iterators are made to be used in template algorithms, an on containers that cannot be randomly accessed. This is how you use vectors
for (size_t i = 0; i < vect.size(); ++ i)
vec[i] // the i-th element
Or with the new syntax
for (int x : vect)
x // the i-th element
Also find will allways find the element in the set, when not it will return the end iterator, that's after the very last element
so this
findnumb=myset.find(*it);
while ( (*findnumb)!=(*it))
{ ++(*it); findnumb=myset.find(*it);}
is useless and wrong. To test if find found the element use
if (findnumb != myset.end())
Of course if all you want to know is whether or not the element is in the set, it's better to use count and forget about the iterators at all.
if (myset.count(x) > 0)
And don't store iterators if you only want to read the value not write it. Just store the value itself
with the help of hansmaad I found out what I need (baste on my code) thanks a lot.!!!!!!!!!!! Also I have a last question, how I use FIND in set findnumbs!!!! like set::const_iterator, dereferenced_less>::iterator ithelp; ithelp=findnumbs.find(20)
#
include <iostream>
#include <set>
#include <vector>
#include <algorithm>
struct dereferenced_less {
template<typename I>
bool operator()(const I& lh, const I& rh) const { return *lh < *rh; }
};
int main()
{
using namespace std;
set<int> myset;
for (int i=1; i<=10; ++i)
{myset.insert(i*10);}
int vc[] = { 17, 32, 4, 16, 28 };
vector<int> vect ;
vect.assign (vc, vc+5);
set<set<int>::const_iterator, dereferenced_less> findnumbs;
set<int>::iterator findnumb;
for (std::vector<int>::iterator i = vect.begin(); i != vect.end(); ++i)
{ cout << *i << endl; }
for (std::vector<int>::iterator i = vect.begin(); i != vect.end(); ++i)
{
findnumb = myset.find(*i);
while ( (*findnumb)!=(*i))
{ ++(*i); findnumb=myset.find(*i);}
findnumbs.insert(findnumb);
cout << *findnumb<<" ";
}
for (set<set<int>::const_iterator, dereferenced_less>::iterator it=findnumbs.begin(); it!=findnumbs.end(); ++it)
std::cout << ' ' << *(*it);
std::cout << '\n';
return 0;
}