Answer to Question #199307 in C for Prasanna

Question

From a set of array elements, find the number of occurrences of each element present in the given array.

Input :

5

2

4

2

5

Output:

2 occurs 2 time

4 occurs 1 time

5 occurs 2 time

Answer 1

#include <stdio.h> 
#include <stdlib.h> 
#include <string.h> 


void selectionSort(int numbers[],int size) {
	int i, j, indexmin;
	for(i = 0; i<size-1; i++) {
		indexmin = i;  
		for(j = i+1; j<size; j++)
			if(numbers[j] < numbers[indexmin])
				indexmin = j;


		int temp;
		temp = numbers[i];
		numbers[i] = numbers[indexmin];
		numbers[indexmin] = temp;
	}
}


int main() {
	int numbers[1000];
	int usedNumbers[12];
	int i=0,j;
	int times = 0;
	int usedNumbersCounter = 0;
	int occurs=0;
	char input[10];
	int totalNumberOfNumbers=0;


	//Loop until empty line
	while(1){
		gets(input);
		if (strcmp(input, "\0") == 0){
			break;
		}else{
			numbers[totalNumberOfNumbers]=atoi(input);
			totalNumberOfNumbers++;	
		}
	}
	//sort values
    selectionSort(numbers,totalNumberOfNumbers);
	for (i=0; i<totalNumberOfNumbers; i++){
		times = 0;
		for (j=0; j<totalNumberOfNumbers; j++){
			if (numbers[i] == numbers[j]){
				times++;
			}
		}
		occurs=0;
		for (j=0; j<totalNumberOfNumbers; j++){
			if (numbers[i] == usedNumbers[j]){
				occurs=1;
				break;
			}
		}
		if(occurs==0){
			printf("%d occurs %d time\n",numbers[i],times);
			usedNumbers[usedNumbersCounter]=numbers[i];
			usedNumbersCounter++;
		}
	}

	getchar();
	getchar();
	return 0;
}