What C++ container should I choose for frequency graph? - c++

I have a little project. I want to make a frequency graph something like this:
a-axis
3| x
2|` x x
1| x x x
0| x x x x
|`_____________`
2 4 6 7 9 b-axis
What container should I use and how do go about implementing that in the simplest way.
The container should also record zero value (frequency).
I find it difficult to correct implement the container. Is there are any sample code I can look at? Or anybody can write a sample for me to study here.
Update:
I was thinking of making std::map<int,std::vector<int>> keyAndMap
How do I access the vector which is inside keyAndMap?
Because I somehow can't do keyAndMap.insert(std::make_pair<int,std::vector<int>>(bAxis,bVec.push_back(bValue)));
The main problem I think is how to generate unique vector for each key in std::map?
const int tries = 21;
std::vector<int>values;
std::map<int,std::vector<int>>keyAndMap;
for (int i = 1; i < tries; i++)
{
int n = i;
int cycle = calculate(n);
/*This next line is an error... why?*/
keyAndMap.insert(std::make_pair<int,std::vector<int>>(cycle,values.push_back(i)));
}
although if i somehow got through that line, my logic here is flawed because the vector is not unique to each key.
Each 'x' in the little pretty graph i drew is gained with each pass in the loop.
So for each b-axis (2,4,6,7, and 9) should be in their own std::vector

What container should I use and how do go about implementing that in the simplest way?
A possible container would be a std::map<key, frequency>, both of which are of type int, with the additional advantage of your values being sorted. Then you could simply read values by something like:
std::map<int, int> data;
int key;
while (std::cin >> key)
{
++data[key]; // increment the frequency of a given key*
}
To access the data you could use iterators, like so:
for (auto it = data.begin; it != data.end(); ++it)
{
// iterator to key
std::cout << it->first <<": "<< it->second <<"\n";
// iterator to frequency
}
Which will print a two-column table of your data.
Second, even simpler approach:
You could use a std::vector<frequency>, where frequency is of type int, and your data would be the indexes of the std::vector, however, you should know the range** of your data so that you can use it as an initial vector size:
i.e. std::vector<int> data(5, 0) could hold values with range 5.
Then if, for example, your data is consisted of the integers: [0, 4], you could have:
std::vector<int> data(5, 0); // five elements with initial value 0
int key;
while(std::cin >> key) // assuming key >= 0 && key < 5
{
data[key] += 1;
}
and then print with:
for (int i= 0; i < data.size(); ++i)
{
std::cout << i <<": "<< data[i] <<"\n";
}
* The default initial frequency value of a distinct key is: 0, so you can simply use the key to access and increment an element.
**range - difference between the lowest and highest values.

Related

C++ How to optimize this algorithm ? (std::map)

The problem is the following: We are given a number 's', s ∈ [0, 10^6], and a number 'n', n ∈ [0, 50000], then n numbers, and we have to find how many number pairs' sum is equal to the 's' number (and we must use either maps or sets to solve it)
Here is the example:
Input:
5 (this is s)
6 (this is n)
1
4
3
6
-1
5
Output:
2
explanation : these are the (1,4) and (6,−1) pairs. (1 +4 = 5 and 6 + (-1) = 5)
Here is my "solution" , I don't even know if it's correct, but it works for the example that we got.
#include <iostream>
#include <map>
#include <iterator>
using namespace std;
int main()
{
cin.tie(0);
ios::sync_with_stdio(false);
int s;
cin >> s;
int n;
cin >> n;
map<int, int> numbers;
int element;
int counter = 0;
for(int i=0; i<n;i++)
{
cin >> element;
numbers.insert(pair<int, int>(element, s-element));
}
for(map<int, int>::iterator it = numbers.begin(); it != numbers.end(); it++)
{
map<int, int>::iterator it2 = it;
while(it2 != numbers.end())
{
if(it->second == it2->first)
{
counter++;
break;
}
it2++;
}
}
cout << counter << "\n";
return 0;
}
Thanks for the answers in advance! I'm still a beginner and I'm learning, sorry.
element, s-element is a good idea but there is no reason to store all the pairs and only then check for duplicates. This removes the O(n^2) loop you have there at the end.
The standard way using hashing would be:
seen=unordered_map<number,count>()
for 1...n:
e = read_int()
if (s-e) in seen:
duplicates+=seen[s-e] # Found new seen[s-e] duplicates.
if e in seen:
seen[e]+=1
else:
seen.insert(e,1)
return duplicates
Here's a brute-force method, using a vector:
int target_s = 0;
int quantity_numbers = 0;
std::cin >> target_s >> quantity_numbers;
std::vector<int> data(quantity_numbers);
for (int i = 0; i < quantity_numbers; ++i)
{
cin >> data[i];
}
int count = 0;
for (int i = 0; i < quantity_numbers; ++i)
{
for (j = 0; j < quantity_numbers; ++j)
{
if (i == j) continue;
int pair_sum = data[i] + data[j];
if (pair_sum == target_s) ++count;
}
}
std::cout << count;
The above code includes the cases where pair <a,b> == s and pair <b,a> == s. Not sure if the requirement only wants pair <a,b> in this case.
As always with this kind of questions, the selection of the appropriate algorithm will improve your solution. Writing some "better" C++ code, will nearly never help. Also, brute forcing is nearly never a solution for such an algorithm.
With the following described approach (which was of course not invented by me), we need just one std::map (or even better, a std::unordered_map) and one for loop. We do not need to store the read values in an additional std::vector or such alike. So, we can come up with low memory condumption and fast computation.
Approach. Any time, after reading a value, we will calculate the delta from the desired sum.
If we look at the required condition that the current value and some previuosly read value, should add up to the desired sum, we can write the following mathematical equations:
currentValue + previouslyReadValue = desiredSum
or
desiredSum - currentValue = previouslyReadValue
or with
delta = desiredSum - currentValue
-->
delta == previouslyReadValue
So, we need to look at the already read values and if they are equal to the delta (Because then they would add up the the desired sum), add their count of occurence the the resulting count of valid pairs.
The already read values and their count of occurence will be stored in a std::unordered_map.
All this will result in a 10 line solution:
#include <iostream>
#include <unordered_map>
int main() {
// Initialize our working variables
int numberOfValues{}, desiredSum{}, currentValue{}, resultingCount{};
// Read basic parameters. Desired sum and overall number of input values.
std::cin >> desiredSum >> numberOfValues;
// Here, we will store all values and their count of occurence
std::unordered_map<int, int> valuesAndCount{};
// Read all values and operate on them
for (int i{}; i < numberOfValues; ++i) {
std::cin >> currentValue; // Read from cin
const int delta{ desiredSum - currentValue }; // Calculate the delta from the desired sum
// Look, if the calculated delta is already in the map. Becuase, if the delta and the
// current value sum up to our desired sum, then we found a valid pair.
if (valuesAndCount.find(delta) != valuesAndCount.end())
// Increase the resulting count, by the number of times that this delta value has already been there
resultingCount += valuesAndCount[delta];
// Nothing special, Just cound the occurence of this value.
valuesAndCount[currentValue]++;
}
return !!(std::cout << resultingCount);
}

Sort a map on the basis of first value of pair

Suppose I have to describe my map as:
map<int, pair<long, int>> mp;
Now I insert elements as:
int y; long x;
pair<long, int> p;
for(int i = 0; i < 5; i++)
{
cin >> x >> y;
p.first = x;
p.second = y;
mp.insert({i, p}); // What is wrong here syntax wise?
}
Further, I want to sort it on the basis of first value of the pair.
You can use a little trick here.
Map in c++ automatically sorts everything by key, so you can do following =>
map <long, (set,vector) < int > > mp; //Create this kind of map
//it will sort elements by first value and depending on your needs, select vector or set
//if you need to sort elements by second value use set
//if you do not care about it use vector
long x;
int y;
for (int i = 0; i < n; i++)
{
cin >> x >> y;
if (mp.find(x) != mp.end()) // if element exist
{
mp[x].push_back(y); // add it to vector
}
else
{
mp[x] = vector < int > (); // if not create new vector
mp[x].push_back(y); // and then push new element
}
}
A std::map in indexed and ordered by its key. Full stop.
I can only imagine 2 possible ways to have it sorted according to its values:
reverse the struct to have the element giving the order as the key (this is #Suspicio's answer)
use what is called a secondary index in database world, that is an auxialiary thing that will be sorted according to your requirements and point to the real data.
Here, I would use either a vector of integers (the keys to your actual map) if you can accept to sort it once before using it (said differently if you map does not change once populated) or a std::multimap if you want to be able to easily add (or remove) items.
multimap<long, int> indices;
for (auto elt : mp) {
indices.insert({ elt.second.first, elt.first });
}
You can now the process your sorted map:
for (auto index : indices) {
auto elt = mp.find(index.second); // *elt will give the elements in order
...
}
You just have to update the indices multimap whenever you add or remove elements to your original mp map.

How to find a unique number using std::find

Hey here is a trick question asked in class today, I was wondering if there is a way to find a unique number in a array, The usual method is to use two for loops and get the unique number which does not match with all the others I am using std::vectors for my array in C++ and was wondering if find could spot the unique number as I wouldn't know where the unique number is in the array.
Assuming that we know that the vector has at least three
elements (because otherwise, the question doesn't make sense),
just look for an element different from the first. If it
happens to be the second, of course, we have to check the third
to see whether it was the first or the second which is unique,
which means a little extra code, but roughly:
std::vector<int>::const_iterator
findUniqueEntry( std::vector<int>::const_iterator begin,
std::vector<int>::const_iterator end )
{
std::vector<int>::const_iterator result
= std::find_if(
next( begin ), end, []( int value) { return value != *begin );
if ( result == next( begin ) && *result == *next( result ) ) {
-- result;
}
return result;
}
(Not tested, but you get the idea.)
As others have said, sorting is one option. Then your unique value(s) will have a different value on either side.
Here's another option that solves it, using std::find, in O(n^2) time(one iteration of the vector, but each iteration iterates through the whole vector, minus one element.) - sorting not required.
vector<int> findUniques(vector<int> values)
{
vector<int> uniqueValues;
vector<int>::iterator begin = values.begin();
vector<int>::iterator end = values.end();
vector<int>::iterator current;
for(current = begin ; current != end ; current++)
{
int val = *current;
bool foundBefore = false;
bool foundAfter = false;
if (std::find(begin, current, val) != current)
{
foundBefore = true;
}
else if (std::find(current + 1, end, val) != end)
{
foundAfter = true;
}
if(!foundBefore && !foundAfter)
uniqueValues.push_back(val);
}
return uniqueValues;
}
Basically what is happening here, is that I am running ::find on the elements in the vector before my current element, and also running ::find on the elements after my current element. Since my current element already has the value stored in 'val'(ie, it's in the vector once already), if I find it before or after the current value, then it is not a unique value.
This should find all values in the vector that are not unique, regardless of how many unique values there are.
Here's some test code to run it and see:
void printUniques(vector<int> uniques)
{
vector<int>::iterator it;
for(it = uniques.begin() ; it < uniques.end() ; it++)
{
cout << "Unique value: " << *it << endl;
}
}
void WaitForKey()
{
system("pause");
}
int main()
{
vector<int> values;
for(int i = 0 ; i < 10 ; i++)
{
values.push_back(i);
}
/*for(int i = 2 ; i < 10 ; i++)
{
values.push_back(i);
}*/
printUniques(findUniques(values));
WaitForKey();
return -13;
}
As an added bonus:
Here's a version that uses a map, does not use std::find, and gets the job done in O(nlogn) time - n for the for loop, and log(n) for map::find(), which uses a red-black tree.
map<int,bool> mapValues(vector<int> values)
{
map<int, bool> uniques;
for(unsigned int i = 0 ; i < values.size() ; i++)
{
uniques[values[i]] = (uniques.find(values[i]) == uniques.end());
}
return uniques;
}
void printUniques(map<int, bool> uniques)
{
cout << endl;
map<int, bool>::iterator it;
for(it = uniques.begin() ; it != uniques.end() ; it++)
{
if(it->second)
cout << "Unique value: " << it->first << endl;
}
}
And an explanation. Iterate over all elements in the vector<int>. If the current member is not in the map, set its value to true. If it is in the map, set the value to false. Afterwards, all values that have the value true are unique, and all values with false have one or more duplicates.
If you have more than two values (one of which has to be unique), you can do it in O(n) in time and space by iterating a first time through the array and filling a map that has as a key the value, and value the number of occurences of the key.
Then you just have to iterate through the map in order to find a value of 1. That would be a unique number.
This example uses a map to count number occurences. Unique number will be seen only one time:
#include <iostream>
#include <map>
#include <vector>
int main ()
{
std::map<int,int> mymap;
std::map<int,int>::iterator mit;
std::vector<int> v;
std::vector<int> myunique;
v.push_back(10); v.push_back(10);
v.push_back(20); v.push_back(30);
v.push_back(40); v.push_back(30);
std::vector<int>::iterator vit;
// count occurence of all numbers
for(vit=v.begin();vit!=v.end();++vit)
{
int number = *vit;
mit = mymap.find(number);
if( mit == mymap.end() )
{
// there's no record in map for your number yet
mymap[number]=1; // we have seen it for the first time
} else {
mit->second++; // thiw one will not be unique
}
}
// find the unique ones
for(mit=mymap.begin();mit!=mymap.end();++mit)
{
if( mit->second == 1 ) // this was seen only one time
{
myunique.push_back(mit->first);
}
}
// print out unique numbers
for(vit=myunique.begin();vit!=myunique.end();++vit)
std::cout << *vit << std::endl;
return 0;
}
Unique numbers in this example are 20 and 40. There's no need for the list to be ordered for this algorithm.
Do you mean to find a number in a vector which appears only once? The nested loop if the easy solution. I don't think std::find or std::find_if is very useful here. Another option is to sort the vector so that you only need to find two consecutive numbers that are different. It seems overkill, but it is actually O(nlogn) instead of O(n^2) as the nested loop:
void findUnique(const std::vector<int>& v, std::vector<int> &unique)
{
if(v.size() <= 1)
{
unique = v;
return;
}
unique.clear();
vector<int> w = v;
std::sort(w.begin(), w.end());
if(w[0] != w[1]) unique.push_back(w[0]);
for(size_t i = 1; i < w.size(); ++i)
if(w[i-1] != w[i]) unique.push_back(w[i]);
// unique contains the numbers that are not repeated
}
Assuming you are given an array size>=3 which contains one instance of value A, and all other values are B, then you can do this with a single for loop.
int find_odd(int* array, int length) {
// In the first three elements, we are guaranteed to have 2 common ones.
int common=array[0];
if (array[1]!=common && array[2]!=common)
// The second and third elements are the common one, and the one we thought was not.
return common;
// Now search for the oddball.
for (int i=0; i<length; i++)
if (array[i]!=common) return array[i];
}
EDIT:
K what if more than 2 in an array of 5 are different? – super
Ah... that is a different problem. So you have an array of size n, which contains the common element c more than once, and all other elements exactly once. The goal is to find the set of non-common (i.e. unique) elements right?
Then you need to look at Sylvain's answer above. I think he was answering a different question, but it would work for this. At the end, you will have a hash map full of the counts of each value. Loop through the hash map, and every time you see a value of 1, you will know the key is a unique value in the input array.

Given a vector with integers from 0 to n, but not all included, how do I efficiently get the non-included integers?

Given a vector with integers from 0 to n, but not all included, how do I efficiently get the non-included integers?
For example if I have a vector with 1 2 3 5, I need to get the vector that contains 0 4.
But I need to do it very efficiently.
Since the vector is already sorted, this becomes trivial:
vector<int> v = {1,2,3,5};
vector<int> ret;
v.push_back(n+1); // this is to enforce a limit using less branches in the loop
for(int i = 0, j = 0; i <= n; ++i){
int present = v[j++];
while(i < present){
ret.push_back(i++);
}
}
return ret;
Additionally, if it wasn't sorted, you could either sort it and apply the above algorithm, or, if you know the range of n, and you can afford the extra memory, you could instead create an array of boolean (or a bitset) and mark the index corresponding to every element you encounter (e.g. bitset[v[j++]] = true;), subsequently iterating from 0 to n and inserting into your vector every element whose bitset position has not been marked.
Basically the idea presented here is that we know the number of missing items beforehand if we can assume sorted input without duplicate values.
Then it is possible to pre-allocate enough space to hold the missing values beforehand (no later dynamic allocation required). Then we can also exploit the possible shortcut when all missing values were found.
If the input vector is not sorted or contains duplicate values, a wrapper function can be used that establishes this precondition.
#include <iostream>
#include <set>
#include <vector>
inline std::vector<int> find_missing(std::vector<int> const & input) {
// assuming non-empty, sorted input, no duplicates
// number of items missing
int n_missing = input.back() - input.size() + 1;
// pre-allocate enough memory for missing values
std::vector<int> result(n_missing);
// iterate input vector with shortcut if all missing values were found
auto input_it = input.begin();
auto result_it = result.begin();
for (int i = 0; result_it != result.end() && input_it != input.end(); ++i) {
if (i < *input_it) (*result_it++) = i;
else ++input_it;
}
return result;
}
// use this if the input vector is not sorted/unique
inline std::vector<int> find_missing_unordered(std::vector<int> const & input) {
std::set<int> values(input.begin(), input.end());
return find_missing(std::vector<int>(values.begin(), values.end()));
}
int main() {
std::vector<int> input = {1,2,3,5,5,5,7};
std::vector<int> result = find_missing_unordered(input);
for (int i : result)
std::cout << i << " ";
std::cout << "\n";
}
The output is:
$ g++ test.cc -std=c++11 && ./a.out
0 4 6

Find the biggest 3 numbers in a vector

I'm trying to make a function to get the 3 biggest numbers in a vector. For example:
Numbers: 1 6 2 5 3 7 4
Result: 5 6 7
I figured I could sort them DESC, get the 3 numbers at the beggining, and after that resort them ASC, but that would be a waste of memory allocation and execution time. I know there is a simpler solution, but I can't figure it out. And another problem is, what if I have only two numbers...
BTW: I use as compiler BorlandC++ 3.1 (I know, very old, but that's what I'll use at the exam..)
Thanks guys.
LE: If anyone wants to know more about what I'm trying to accomplish, you can check the code:
#include<fstream.h>
#include<conio.h>
int v[1000], n;
ifstream f("bac.in");
void citire();
void afisare_a();
int ultima_cifra(int nr);
void sortare(int asc);
void main() {
clrscr();
citire();
sortare(2);
afisare_a();
getch();
}
void citire() {
f>>n;
for(int i = 0; i < n; i++)
f>>v[i];
f.close();
}
void afisare_a() {
for(int i = 0;i < n; i++)
if(ultima_cifra(v[i]) == 5)
cout<<v[i]<<" ";
}
int ultima_cifra(int nr) {
return nr - 10 * ( nr / 10 );
}
void sortare(int asc) {
int aux, s;
if(asc == 1)
do {
s = 0;
for(int i = 0; i < n-1; i++)
if(v[i] > v[i+1]) {
aux = v[i];
v[i] = v[i+1];
v[i+1] = aux;
s = 1;
}
} while( s == 1);
else
do {
s = 0;
for(int i = 0; i < n-1; i++)
if(v[i] < v[i+1]) {
aux = v[i];
v[i] = v[i+1];
v[i+1] = v[i];
s = 1;
}
} while(s == 1);
}
Citire = Read
Afisare = Display
Ultima Cifra = Last digit of number
Sortare = Bubble Sort
If you were using a modern compiler, you could use std::nth_element to find the top three. As is, you'll have to scan through the array keeping track of the three largest elements seen so far at any given time, and when you get to the end, those will be your answer.
For three elements that's a trivial thing to manage. If you had to do the N largest (or smallest) elements when N might be considerably larger, then you'd almost certainly want to use Hoare's select algorithm, just like std::nth_element does.
You could do this without needing to sort at all, it's doable in O(n) time with linear search and 3 variables keeping your 3 largest numbers (or indexes of your largest numbers if this vector won't change).
Why not just step through it once and keep track of the 3 highest digits encountered?
EDIT: The range for the input is important in how you want to keep track of the 3 highest digits.
Use std::partial_sort to descending sort the first c elements that you care about. It will run in linear time for a given number of desired elements (n log c) time.
If you can't use std::nth_element write your own selection function.
You can read about them here: http://en.wikipedia.org/wiki/Selection_algorithm#Selecting_k_smallest_or_largest_elements
Sort them normally and then iterate from the back using rbegin(), for as many as you wish to extract (no further than rend() of course).
sort will happen in place whether ASC or DESC by the way, so memory is not an issue since your container element is an int, thus has no encapsulated memory of its own to manage.
Yes sorting is good. A especially for long or variable length lists.
Why are you sorting it twice, though? The second sort might actually be very inefficient (depends on the algorithm in use). A reverse would be quicker, but why even do that? If you want them in ascending order at the end, then sort them into ascending order first ( and fetch the numbers from the end)
I think you have the choice between scanning the vector for the three largest elements or sorting it (either using sort in a vector or by copying it into an implicitly sorted container like a set).
If you can control the array filling maybe you could add the numbers ordered and then choose the first 3 (ie), otherwise you can use a binary tree to perform the search or just use a linear search as birryree says...
Thank #nevets1219 for pointing out that the code below only deals with positive numbers.
I haven't tested this code enough, but it's a start:
#include <iostream>
#include <vector>
int main()
{
std::vector<int> nums;
nums.push_back(1);
nums.push_back(6);
nums.push_back(2);
nums.push_back(5);
nums.push_back(3);
nums.push_back(7);
nums.push_back(4);
int first = 0;
int second = 0;
int third = 0;
for (int i = 0; i < nums.size(); i++)
{
if (nums.at(i) > first)
{
third = second;
second = first;
first = nums.at(i);
}
else if (nums.at(i) > second)
{
third = second;
second = nums.at(i);
}
else if (nums.at(i) > third)
{
third = nums.at(i);
}
std::cout << "1st: " << first << " 2nd: " << second << " 3rd: " << third << std::endl;
}
return 0;
}
The following solution finds the three largest numbers in O(n) and preserves their relative order:
std::vector<int>::iterator p = std::max_element(vec.begin(), vec.end());
int x = *p;
*p = std::numeric_limits<int>::min();
std::vector<int>::iterator q = std::max_element(vec.begin(), vec.end());
int y = *q;
*q = std::numeric_limits<int>::min();
int z = *std::max_element(vec.begin(), vec.end());
*q = y; // restore original value
*p = x; // restore original value
A general solution for the top N elements of a vector:
Create an array or vector topElements of length N for your top N elements.
Initialise each element of topElements to the value of your first element in your vector.
Select the next element in the vector, or finish if no elements are left.
If the selected element is greater than topElements[0], replace topElements[0] with the value of the element. Otherwise, go to 3.
Starting with i = 0, swap topElements[i] with topElements[i + 1] if topElements[i] is greater than topElements[i + 1].
While i is less than N, increment i and go to 5.
Go to 3.
This should result in topElements containing your top N elements in reverse order of value - that is, the largest value is in topElements[N - 1].