How to store a sequence of integers in a C++ array? - c++

Assume I have to store 100 integers in array, should I declare an array and store all 100 integers from 1 to 100 or is there any solution that will work better ? and loop them through a for loop or any loop concept in c++
example
int numbers [100] = { 1 ,2 3,......10};
for(){
//manipulate the array elements here
}
or
int numbers = {100};
what if I have more than 100 elements ? i.e. integer number from 1 to 200 or more what is good concept to achieve ?

You can use std::iota to generate a sequence.
For example, like this:
#include <iostream>
#include <numeric>
#include <vector>
int main() {
std::vector<int> vec(100);
std::iota(begin(vec), end(vec), 1);
for (auto i : vec)
std::cout << i << " ";
std::cout << "\n";
}
If you have to use a raw array, you can fill it like this:
int vec[100];
std::iota(vec, vec + 100, 1);

Related

Erasing the first entry of a vector, after the maximum is reached

I have a vector in which i save coordinates.
I perform a series of calculations on each coordinate, thats why i have a limit for the vector size.
Right now i clear the vector, when the limit is reached.
I'm searching for a method, that let's me keep the previous values and only erases the very first value in the vector.
Simplified, something like this (if the maximum size of the vector would be 4).
vector<int> vec;
vec = {1,2,3,4}
vec.push_back(5);
vec = {2,3,4,5}
Is this possible?
As suggested by #paddy, you can use std::deque, it is most performant way to keep N elements if you .push_back(...) new (last) element, and .pop_front() first element.
std::deque gives O(1) complexity for such operations, unlike std::vector which gives O(N) complexity.
Try it online!
#include <deque>
#include <iostream>
int main() {
std::deque<int> d = {1, 2, 3, 4};
for (size_t i = 5; i <= 9; ++i) {
d.push_back(i);
d.pop_front();
// Print
for (auto x: d)
std::cout << x << " ";
std::cout << std::endl;
}
}
Output:
2 3 4 5
3 4 5 6
4 5 6 7
5 6 7 8
6 7 8 9
I think you should properly encapsulate this behaviour in your own vector class, as a std::vector wrapper. You could pass the max capacity as an argument to your constructor. And you could reimplement the methods that may cause "overflow" while just reusing the std::vector ones for the others.
To simplify what you pretend to achieve for the push_back case, using a function and a global variable, you could:
check against a max capacity and,
if that capacity is already reached, rotate your vector contents left by one position; then simply overwrite the last element;
otherwise do a normal push_back.
[Demo]
#include <algorithm> // rotate
#include <iostream> // cout
#include <vector>
const size_t max_capacity{4};
void push_back(std::vector<int>& v, int n)
{
if (v.size() == max_capacity)
{
// Rotate 1 left
std::rotate(std::begin(v), std::begin(v) + 1, std::end(v));
v[v.size() - 1] = n;
}
else
{
v.push_back(n);
}
}
int main()
{
std::vector<int> v{};
for (auto i{1}; i < 9; i++)
{
push_back(v, i);
for (auto&& n : v) { std::cout << n << " "; }
std::cout << "\n";
}
}

How do you generate subarrays of an array with specific number of elements and then store it in another array?

What I need to do is to create subarrays of an existing array and the subarray should have a given number of elements.
For eg. if I have the array [1,2,3,4] and I need to generate subarrays of it with exactly three elements, I would want the a separate 2-d array to include the following arrays:
[1,2,3]
[1,2,4]
[1,3,4]
[2,3,4]
and so on and so forth.
Thank you in advance.
You can do it by first calculating how many combinations there are by picking 3 elements in a group of 4 and what these combinations look like.
Then you can use these combinations to pick elements from your input and create output like this :
#include <cassert>
#include <algorithm>
#include <iostream>
#include <numeric>
#include <vector>
#include <set>
// calculate numbers of ways in which we can pick number_of_elements_to_pick from
// number_of_elements
auto get_combinations_of_indices_to_pick(std::size_t number_of_elements_to_pick, std::size_t number_of_elements)
{
assert(number_of_elements > number_of_elements_to_pick);
// set which element to pick to true for number_of_elements_to_pick
std::vector<bool> pick_element_n(number_of_elements_to_pick, true);
// and the rest to false (we will not pick out those indices)
pick_element_n.insert(pick_element_n.end(), number_of_elements - number_of_elements_to_pick, 0);
// the total number of combinations is the number of permutations of the bits.
std::vector<std::vector<std::size_t>> combinations;
do
{
std::vector<std::size_t> combination;
for (std::size_t i = 0; i < number_of_elements; ++i)
{
if (pick_element_n[i]) combination.push_back(i);
}
combinations.push_back(combination);
} while (std::prev_permutation(pick_element_n.begin(), pick_element_n.end()));
return combinations;
}
int main()
{
std::vector<int> input{ 1,2,3,4 };
// get combinations of 3 indices to pick from input vector
auto combinations = get_combinations_of_indices_to_pick(3, input.size());
// there will be a vector of vectors as output
std::vector<std::vector<int>> outputs;
outputs.reserve(combinations.size());
// loop over all combinations of indices
for (const auto& combination : combinations)
{
std::vector<int> output;
// foreach combination of indicices to pick
// add input for that index to the output for this combination
for (const auto index : combination)
{
output.push_back(input[index]);
}
// then add this selected subvector to the outputs
outputs.push_back(output);
}
// show the possible output vectors
for (const auto& output : outputs)
{
bool comma = false;
for (const auto& value : output)
{
if (comma) std::cout << ", ";
std::cout << value;
comma = true;
}
std::cout << "\n";
}
}
Note : The number of possible combinations can get very big very quickly. And usually there is a faster way of solving problems then just blindly go over all the possible combinations.

Finding Duplicates in an array using a Set in C++

I am currently practicing for coding interviews and am working on a function that takes in an array and the size of that array and prints out which numbers in it are duplicates. I have gotten this to work using the two for loop method but want an optimized solution using sets. Snippet of the code I have is below,
#include <iostream>
#include <set>
using namespace std;
void FindDuplicate(int integers[], int n){
set<int>setInt;
for(int i = 0; i < n; i++){
//if this num is not in the set then it is not a duplicate
if(setInt.find(integers[i]) != setInt.end()){
setInt.insert({integers[i]});
}
else
cout << integers[i] << " is a duplicate";
}
}
int main() {
int integers [] = {1,2,2,3,3};
int n = sizeof(integers)/sizeof(integers[0]);
FindDuplicate(integers, n);
}
Any helpful advice is appreciated, thanks
I think your comparison is not needed, insert do it for you:
https://en.cppreference.com/w/cpp/container/set/insert
Returns a pair consisting of an iterator to the inserted element (or
to the element that prevented the insertion) and a bool value set to
true if the insertion took place.
Just insert element and check what insert function returns (false on second element of pair in case of duplication) :)
my solution proposal is :
count the frequencies of each element (algo for frequencies are explained here frequency
display elements with frequency more than 1 (it is a duplicate)
In each operation, you do not use imbricated loops.
#include <iostream>
#include <unordered_map>
using namespace std;
void FindDuplicate(int integers[], int n)
{
unordered_map<int, int> mp;
// Traverse through array elements and
// count frequencies
for (int i = 0; i < n; i++)
{
mp[integers[i]]++;
}
cout << "The repeating elements are : " << endl;
for (int i = 0; i < n; i++) {
if (mp[integers[i]] > 1)
{
cout << integers[i] << endl;
mp[integers[i]] = -1;
}
}
}
int main()
{
int integers [] = {1,1,0,0,2,2,3,3,3,6,7,7,8};
int n = sizeof(integers)/sizeof(integers[0]);
FindDuplicate(integers, n);
}
This is my feedback:
#include <iostream>
#include <vector>
#include <set>
// dont' do this, in big projects it's not done (nameclash issues)
// using namespace std;
// pass vector by const reference you're not supposed to change the input
// the reference will prevent data from being copied.
// naming is important, do you want to find one duplicate or more...
// renamed to FindDuplicates because you want them all
void FindDuplicates(const std::vector<int>& input)
{
std::set<int> my_set;
// don't use index based for loops if you don't have to
// range based for loops are more safe
// const auto is more refactorable then const int
for (const auto value : input)
{
//if (!my_set.contains(value)) C++ 20 syntax
if (my_set.find(value) == my_set.end())
{
my_set.insert(value);
}
else
{
std::cout << "Array has a duplicate value : " << value << "\n";
}
}
}
int main()
{
// int integers[] = { 1,2,2,3,3 }; avoid "C" style arrays they're a **** to pass around safely
// int n = sizeof(integers) / sizeof(integers[0]); std::vector (or std::array) have size() methods
std::vector input{ 1,2,2,3,3 };
FindDuplicates(input);
}
You do not need to use a set.
To find the duplicates:
Sort array with numbers
Iterate over the array (start with second element) and copy elements where previous element equals
current element into a new vector "duplicates"
(Optional) use unique on the "duplicates" if you like to know which number is a duplicate and do not care if it is 2, 3 or 4 times in the numbers array
Example Implementation:
#include <algorithm>
#include <iostream>
#include <vector>
void
printVector (std::vector<int> const &numbers)
{
for (auto const &number : numbers)
{
std::cout << number << ' ';
}
std::cout << std::endl;
}
int
main ()
{
auto numbers = std::vector<int>{ 1, 2, 2, 42, 42, 42, 3, 3, 42, 42, 1, 2, 3, 4, 5, 6, 7, 7 };
std::sort (numbers.begin (), numbers.end ());
auto duplicates = std::vector<int>{};
std::for_each (numbers.begin () + 1, numbers.end (), [prevElement = numbers.begin (), &duplicates] (int currentElement) mutable {
if (currentElement == *prevElement)
{
duplicates.push_back (currentElement);
}
prevElement++;
});
duplicates.erase (std::unique (duplicates.begin (), duplicates.end ()), duplicates.end ());
printVector (duplicates);
}
edit:
If you have no problem with using more memory and more calculations but like it more expressive:
Sort numbers
Create a new array with unique numbers "uniqueNumbers"
Use "set_difference" to calculate (numbers-uniqueNumbers) which leads to an new array with all the duplicates
(Optional) use unique on the "duplicates" if you like to know which number is a duplicate and do not care if it is 2, 3 or 4 times in the numbers array
Example Implementation:
#include <algorithm>
#include <iostream>
#include <vector>
void
printVector (std::vector<int> const &numbers)
{
for (auto const &number : numbers)
{
std::cout << number << ' ';
}
std::cout << std::endl;
}
int
main ()
{
auto numbers = std::vector<int>{ 2, 2, 42, 42, 42, 3, 3, 42, 42, 1, 2, 3, 4, 5, 6, 7, 7 };
std::sort (numbers.begin (), numbers.end ());
auto uniqueNumbers = std::vector<int>{};
std::unique_copy (numbers.begin (), numbers.end (), std::back_inserter (uniqueNumbers));
auto duplicates = std::vector<int>{};
std::set_difference (numbers.begin (), numbers.end (), uniqueNumbers.begin (), uniqueNumbers.end (), std::back_inserter (duplicates));
std::cout << "duplicate elements: ";
printVector (duplicates);
std::cout << "unique duplicate elements: ";
printVector ({ duplicates.begin (), std::unique (duplicates.begin (), duplicates.end ()) });
}
here's a quick solution use an array of size N (try a big number)
and whenever a number is added into the other array on the large array add 1 to the position like:
array_of_repeated[user_input]++;
so if the program asks how many times (for example) number 234 was repeated?
std::cout<<array_of_repeated[requested_number]<<std::endl;
so in this way you wont spend time looking for a number inside the other list

How to count duplicate entries of a vector in C++

I'm using Armadillo to do linear algebra calculation in C++.
For example, there is a
vector a = (1,1,2,2,0,2,1,0)
I wish return a matrix
(0, 2) //means 0 shows 2 times in the vector
(1, 3) //1 shows 3 times
(2, 3) //2 shows 3 times
Is there any function can fulfill such job?
As mentioned in comments you could use a std::map to collect the results. Then you can convert to a matrix as you see fit. You could skip the map step and use a matrix directly if it's already pre-initialised with the rows you're after.
As for a function to do this, you can use std::for_each from <algorithm> along with a lambda expression, although it seems overkill when a loop would be fine.
#include <algorithm>
#include <iostream>
#include <vector>
#include <map>
using namespace std;
int main()
{
vector<int> v{1,1,2,2,0,2,1,0};
map<int,int> dup;
for_each( v.begin(), v.end(), [&dup]( int val ){ dup[val]++; } );
for( auto p : dup ) {
cout << p.first << ' ' << p.second << endl;
}
return 0;
}
Here's another solution, using only Armadillo functions, and a C++11 compiler:
vec a = {1,1,2,2,0,2,1,0}; // vec holds elements of type 'double'
vec b = unique(a);
uvec c = hist(a,b); // uvec holds unsigned integers
mat X(b.n_rows, 2);
X.col(0) = b;
X.col(1) = conv_to<vec>::from(c);
X.print("X:");
Explanation:
vec b = unique(a) creates a vector containing the unique elements of a, sorted in ascending order
uvec c = hist(a,b) creates a histogram of counts of elements in a, using b as the bin centers
conv_to<vec>::from(c) converts c (vector with unsigned integers) to the same vector type as a

Finding Frequency of numbers in a given group of numbers

Suppose we have a vector/array in C++ and we wish to count which of these N elements has maximum repetitive occurrences and output the highest count. Which algorithm is best suited for this job.
example:
int a = { 2, 456, 34, 3456, 2, 435, 2, 456, 2}
the output is 4 because 2 occurs 4 times. That is the maximum number of times 2 occurs.
Sort the array and then do a quick pass to count each number. The algorithm has O(N*logN) complexity.
Alternatively, create a hash table, using the number as the key. Store in the hashtable a counter for each element you've keyed. You'll be able to count all elements in one pass; however, the complexity of the algorithm now depends on the complexity of your hasing function.
Optimized for space:
Quicksort (for example) then iterate over the items, keeping track of largest count only.
At best O(N log N).
Optimized for speed:
Iterate over all elements, keeping track of the separate counts.
This algorithm will always be O(n).
If you have the RAM and your values are not too large, use counting sort.
A possible C++ implementation that makes use of STL could be:
#include <iostream>
#include <algorithm>
#include <map>
// functor
struct maxoccur
{
int _M_val;
int _M_rep;
maxoccur()
: _M_val(0),
_M_rep(0)
{}
void operator()(const std::pair<int,int> &e)
{
std::cout << "pair: " << e.first << " " << e.second << std::endl;
if ( _M_rep < e.second ) {
_M_val = e.first;
_M_rep = e.second;
}
}
};
int
main(int argc, char *argv[])
{
int a[] = {2,456,34,3456,2,435,2,456,2};
std::map<int,int> m;
// load the map
for(unsigned int i=0; i< sizeof(a)/sizeof(a[0]); i++)
m [a[i]]++;
// find the max occurence...
maxoccur ret = std::for_each(m.begin(), m.end(), maxoccur());
std::cout << "value:" << ret._M_val << " max repetition:" << ret._M_rep << std::endl;
return 0;
}
a bit of pseudo-code:
//split string into array firts
strsplit(numbers) //PHP function name to split a string into it's components
i=0
while( i < count(array))
{
if(isset(list[array[i]]))
{
list[array[i]]['count'] = list + 1
}
else
{
list[i]['count'] = 1
list[i]['number']
}
i=i+1
}
usort(list) //usort is a php function that sorts an array by its value not its key, Im assuming that you have something in c++ that does this
print list[0]['number'] //Should contain the most used number
The hash algorithm (build count[i] = #occurrences(i) in basically linear time) is very practical, but is theoretically not strictly O(n) because there could be hash collisions during the process.
An interesting special case of this question is the majority algorithm, where you want to find an element which is present in at least n/2 of the array entries, if any such element exists.
Here is a quick explanation, and a more detailed explanation of how to do this in linear time, without any sort of hash trickiness.
If the range of elements is large compared with the number of elements, I would, as others have said, just sort and scan. This is time n*log n and no additional space (maybe log n additional).
THe problem with the counting sort is that, if the range of values is large, it can take more time to initialize the count array than to sort.
Here's my complete, tested, version, using a std::tr1::unordered_map.
I make this approximately O(n). Firstly it iterates through the n input values to insert/update the counts in the unordered_map, then it does a partial_sort_copy which is O(n). 2*O(n) ~= O(n).
#include <unordered_map>
#include <vector>
#include <algorithm>
#include <iostream>
namespace {
// Only used in most_frequent but can't be a local class because of the member template
struct second_greater {
// Need to compare two (slightly) different types of pairs
template <typename PairA, typename PairB>
bool operator() (const PairA& a, const PairB& b) const
{ return a.second > b.second; }
};
}
template <typename Iter>
std::pair<typename std::iterator_traits<Iter>::value_type, unsigned int>
most_frequent(Iter begin, Iter end)
{
typedef typename std::iterator_traits<Iter>::value_type value_type;
typedef std::pair<value_type, unsigned int> result_type;
std::tr1::unordered_map<value_type, unsigned int> counts;
for(; begin != end; ++begin)
// This is safe because new entries in the map are defined to be initialized to 0 for
// built-in numeric types - no need to initialize them first
++ counts[*begin];
// Only need the top one at this point (could easily expand to top-n)
std::vector<result_type> top(1);
std::partial_sort_copy(counts.begin(), counts.end(),
top.begin(), top.end(), second_greater());
return top.front();
}
int main(int argc, char* argv[])
{
int a[] = { 2, 456, 34, 3456, 2, 435, 2, 456, 2 };
std::pair<int, unsigned int> m = most_frequent(a, a + (sizeof(a) / sizeof(a[0])));
std::cout << "most common = " << m.first << " (" << m.second << " instances)" << std::endl;
assert(m.first == 2);
assert(m.second == 4);
return 0;
}
It wil be in O(n)............ but the thing is the large no. of array can take another array with same size............
for(i=0;i
mar=count[o];
index=o;
for(i=0;i
then the output will be......... the element index is occured for max no. of times in this array........
here a[] is the data array where we need to search the max occurance of certain no. in an array.......
count[] having the count of each element..........
Note : we alrdy knw the range of datas will be in array..
say for eg. the datas in that array ranges from 1 to 100....... then have the count array of 100 elements to keep track, if its occured increament the indexed value by one........
Now, in the year 2022 we have
namespace aliases
more modern containers like std::unordered_map
CTAD (Class Template Argument Deduction)
range based for loops
using statment
the std::ranges library
more modern algorithms
projections
structured bindings
With that we can now write:
#include <iostream>
#include <vector>
#include <unordered_map>
#include <algorithm>
namespace rng = std::ranges;
int main() {
// Demo data
std::vector data{ 2, 456, 34, 3456, 2, 435, 2, 456, 2 };
// Count values
using Counter = std::unordered_map<decltype (data)::value_type, std::size_t> ;
Counter counter{}; for (const auto& d : data) counter[d]++;
// Get max
const auto& [value, count] = *rng::max_element(counter, {}, &Counter::value_type::second);
// Show output
std::cout << '\n' << value << " found " << count << " times\n";
}