Question

(Count occurrence of numbers) Write a program that reads some integers between 1 and 100 and counts the occurrences of each. Here is a sample run of the program: Enter integers between 1 and 100: 2 occurs 2 times 3 occurs 1 time 4 occurs 1 time 5 occurs 2 times 6 occurs 1 time 23 occurs 1 time 43 occurs 1 time 2 5 6 5 4 3 23 43 2 Note that if a number occurs more than one time, the plural word “times” is used in the output.

Answer 1

Consider the following C program written in Dev C++ :

#include <stdio.h>
#include <stdlib.h> //Libraries used.

/* run this program using the console pauser or add your own getch, system("pause") or input loop */

int main(int argc, char *argv[]) {
     
int arr[100],length,i=0,j=0 ; //declaring array and length variable.
  
printf("Enter the size of the array : ");
scanf("%d",&length);
  
printf("\nEnter integers between 1 and 100 : ");
for(i=0;i<length;i++)
scanf("%d",&arr[i]); //getting the array as an input.
  
//Array frequency will store frequencies of element in array.
int frequency[length];
int visited = -1;
  
for(i = 0; i < length; i++){
int count = 1; //for counting purpose.
for(j = i+1; j < length; j++){
if(arr[i] == arr[j]){
count++; //incrementing the count variable.
//To avoid counting same element again
frequency[j] = visited;
}
}
if(frequency[i] != visited)
frequency[i] = count;
}
  
//Displays the frequency of each element present in array

for(i = 0; i < length; i++){
if(frequency[i] != visited){
if(frequency[i]>1)
printf("\n %d occurs %d times.",arr[i],frequency[i]);
else
printf("\n %d occurs %d time.",arr[i],frequency[i]);
}
}
  
   return 0;
}

The output of the above code is :