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BidhanN
Asked: 2022-08-13T14:21:02+05:45 In:

WAP to find the frequency of numbers present in a list. Solve in C and C++

eg: 1,2,2,3,10,90,88,47,58,83
Output: 0-10=5

11-20=0

c++
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    3 Answers

    1. Explainer Hey you! Yes you, Let’s build something awesome together.
      This answer was edited.

      Using C

      

      

#include <stdio.h>

int main()
{
//Initialize array
int arr[] = {1, 2, 2, 3, 10, 90, 88, 47, 58, 83};
int i, j;
//Calculate length of array arr
int length = sizeof(arr)/sizeof(arr[0]);
//Array fr will store frequencies of element
int fr[length];
int visited = -1;
for(i = 0; i < length; i++){
int count = 1;
for(j = i+1; j < length; j++){
if(arr[i] == arr[j]){
count++;
//To avoid counting same element again
fr[j] = visited;
}
}
if(fr[i] != visited)
fr[i] = count;
}
//Displays the frequency of each element present in array
printf("---------------------n");
printf(" Element | Frequencyn");
printf("---------------------n");
for(i = 0; i < length; i++){
if(fr[i] != visited){
printf(" %d", arr[i]);
printf(" | ");
printf(" %dn", fr[i]);
}
}
printf("---------------------n");
return 0;
}


      

      

      

      

      

      

        

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This answer was edited. 

      

      Using c++
      

      #include <bits/stdc++.h>
using namespace std;

// Main function to run the program
int main() 
{ 
    int arr[] = {10, 30, 10, 20, 10, 20, 30, 10}; 
    int n = sizeof(arr)/sizeof(arr[0]); 

    int visited[n];

    for(int i=0; i<n; i++){

        if(visited[i]!=1){
           int count = 1;
           for(int j=i+1; j<n; j++){
              if(arr[i]==arr[j]){
                 count++;
                 visited[j]=1;
              }
            }

            cout<<arr[i]<<" occurs at "<<count<<" times "<<endl;
         }
     }

    return 0; 
}
      

      

      

      

      

      

        

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Suti 
      


      

      

      

      

      

      Using C
      

      #include <stdio.h>

void main()
{
    int arr1[100], fr1[100];
    int n, i, j, ctr;


       printf("nnCount frequency of each element of an array:n");
       printf("------------------------------------------------n");    

       printf("Input the number of elements to be stored in the array :");
       scanf("%d",&n);

       printf("Input %d elements in the array :n",n);
       for(i=0;i<n;i++)
            {
          printf("element - %d : ",i);
          scanf("%d",&arr1[i]);
          fr1[i] = -1;
        }
    for(i=0; i<n; i++)
    {
        ctr = 1;
        for(j=i+1; j<n; j++)
        {
            if(arr1[i]==arr1[j])
            {
                ctr++;
                fr1[j] = 0;
            }
        }

        if(fr1[i]!=0)
        {
            fr1[i] = ctr;
        }
    }
    printf("nThe frequency of all elements of array : n");
    for(i=0; i<n; i++)
    {
        if(fr1[i]!=0)
        {
            printf("%d occurs %d timesn", arr1[i], fr1[i]);
        }
    }
}
      

      

      

      

      

      

        

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