Frequency of Numbers in a 1D Array - c++

Question

its been 6 hours since I have been writing the code but to no avail, I don't no where I am making the mistake but I am making some. Its a frequency output program and output should be like this:
array[8] = {6,1,7,8,6,6,1,9}
Output:
6:3
1:2
7:1
8:1
9:1
But its repeating the same numbers in my code. Any help would be much appreciable.
int array[8] = {6,1,7,8,6,6,1,9};
int store[8];
int a =0;
int b =0;
int c=0;
int d = 0;
store[d] = array[b];
for (d = 0; d < 8; d++){
int count=0;
c = d;
b = d;
for (int e = 0; e < d; e++){
if (array[b] == store[e]){
store[d] = array[b];
b++;
e = 0;
}
else
{
store[d] = array[b];
break;
}
}
for ( int z = 0; z < 7; z++){
if (store[d] == array[z])
{
count++;
}
}
cout << store[d] << ":" << count << endl;
}

Answer 1

You may use a map first to store num->frequency and then a multimap to store freqeuncy => num.
Here is the working solution.
#include <map>
#include <algorithm>
#include <iostream>
int main()
{
int array[8] = {6,1,7,8,6,6,1,9};
// A map to store num => freq
std::map <int, int> freq;
// A map to store freq(can be duplicate) => num
std::multimap <int, int> freqCounts;
// Store num => frequency
for (int i = 0 ; i < 8; i++)
{
freq[array[i]] += 1;
}
// Now Store freq => num
for(auto const & iter : freq)
{
freqCounts.insert (std::pair<int,int>(iter.second, iter.first));
}
// Print in reverse order i.e. highest frequency first
for (std::multimap<int,int>::reverse_iterator rit=freqCounts.rbegin(); rit!=freqCounts.rend(); ++rit)
{
std::cout << rit->second << " : " << rit->first << '\n';
}
return 0;
}

Answer 2

You never seem to update the counters. Try this:
int array[8] = {6,1,7,8,6,6,1,9};
unsigned int store[10] = {}; // large enough to hold the largest array value,
// initialized to zero
for (int n : array) ++store[n]; // update counts
for (int i = 0; i != 10; ++i)
{
std::cout << "Frequency of int " << i << " is " << store[i] << "\n";
}
If the set of values that occur is sparse, or includes negatives, or simply does not fit into a dense range of integers nicely, you can replace unsigned int[10] with an associative container, e.g.:
std::map<int, unsigned int> store;
// algorithm as before
for (auto const & p : store)
{
std::cout << "Frequency of " << p.first << " is " << p.second << "\n";
}

Answer 3

I'm not sure what you are trying to do with the arrays. I have tried to follow the logic, but it's hard to see it with all the anonymous variable names. It seems like you are trying to look for duplicates earlier in the array, but the variable e never gets any other value than 0, so you will only be comparing with the first item in the array.
You can just look in the array itself for previous occurances, and once you know that the number is the first occurance, you only need to look for more occurances after it in the array:
int array[8] = {6,1,7,8,6,6,1,9};
for (int i = 0; i < 8; i++) {
// look to the left in the array if the number was used before
int found = 0;
for (int j = 0; j < i; j++) {
if (array[i] == array[j]) found++;
}
// go on if it's the first occurance
if (found == 0) {
// we know of one occurance
int count = 1;
// look to the right in the array for other occurances
for (int j = i + 1; j < 8; j++) {
if (array[i] == array[j]) count++;
}
cout << array[i] << ":" << count << endl;
}
}

Answer 4

I wanted to submit my solution which I think it´s a more easy one:
#include <iostream>
using namespace std;
int main() {
int n; //number of Elements in the vector
cin>>n;
int vec[n]; //vector with all elements
int v[n]; //vector with Elements without repetition
int c[n]; // vector which stores the frequency of each element
for(int i=0; i<n; i++)
cin>>vec[i];
int k=0; // number of Elements of the vector without Repetition, in the begining, it has 0 Elements
int j=0; //logic Counter, could be replaced with bool
for(int i=0; i<n; i++) {
for(int h=0; h<=k; h++) {
if(vec[i]==v[h]) {
c[h]++;
j=1;
break;
}
}
//if element i of the original vector is equal to element h of the second vector, then increment the frequency of this element
if(j==0) { //else if the element is not equal to any of the second vector, the Position of the 2nd vector is filled with the element, which in this case is the first of ist Kind.
v[k]=vec[i];
c[k]=1;
k++;
} //the number of Elements is increased by one to store another element;
else {
j=0;
}
}
cout<<endl<<endl;
for(int i=0; i<k; i++)
cout<<v[i]<<":"<<c[i]<<endl;
return 0;
}

Answer 5

#include<iostream>
#include<conio.h>
using namespace std;
main()
{ int count[10],key[10],n=10,m;
int i,j,k,temp;
cout<<"Enter The Size Of Array:-\n";
cin>>n;
int a[n];
cout<<"Enter The Elements in Array:-\n";
for(i=0; i<n; i++)
cin>>a[i];
for(i=0; i<n; i++)
for(j=0; j<n-1; j++)
{ if(a[j]>a[j+1])
{ temp=a[j];
a[j]=a[j+1];
a[j+1]=temp;
}
}
for(i=0; i<n; i++)
cout<<a[i]<<"\t";
for(i=0,k=0; i<n; k++)
{ count[k]=0;
key[k]=a[i];
for(j=i; j<n; j++)
{ if(a[i]==a[j])
count[k]++;
}
i=i+count[k];
}
for(i=0; i<k; i++)
cout<<endl<<key[i]<<" Occurred "<<count[i]<<" Times\n";
getch();
}

Answer 6

/**
* The methods counts the frequency of each element in an array.
*
* Approach: The method checks if the element is already present in the <strong>Map of frequency</strong>.
* If it is not present, add it to the map with the frequency 1 else put it in the map with
* an increment by one of it's existing frequency.
*
* #param arr list of elements
* #return frequency of each elements
*/
public static Map<Integer, Integer> countFrequency(int[] arr) {
Map<Integer, Integer> frequency= new HashMap<Integer, Integer>();
for(int i = 0; i < arr.length; i++) {
if(frequency.get(arr[i])==null) {
frequency.put(arr[i], 1);
}
else {
frequency.put(arr[i],frequency.get(arr[i])+1);
}
}
System.out.println("\nMap: "+frequency);
return frequency;
}