How many times a digit show up - c

Question

Please help me. I don't get why the out put is in "occurs" is "30", instead of '3'.. It's as if I'm multiplying the answer with '10', but I'm not.. Maybe the answer to my problem is right in to my code but Can someone explain why and how? please.. Thank you very much in advance..
Please take a look at my code.
#include <stdio.h>
int main(){
int arr[10] = {7, 7, 3, 2, 9, 8, 5, 1, 7, 9};
int occur[10] = {NULL};
int max = 0;
int most;
for(int i = 0; i < 10; i++)
{
for(int j = 0; j < 10; j++)
{
occur[arr[j]]++;
if(occur[arr[j]] > max)
{
max = occur[arr[j]];
most = arr[j];
}
}
}
printf("Most frequent: %d\ occurs: %d\n", most, max);
return 0;
}
I am getting the correct answer in "Most Frequent". But the "occurs" is 30, instead of just 3 because 7 occurs 3 times.

Answer 1

It becomes 30 because there is an outer loop which executes 10 times.
I'm guessing that you want to get the most frequent number in the array and how many times it occurred that's why you have an outer loop. This will not work if you have a number in your array that is greater than 9 which will result in index out of bounds problem in occur array. You should change your implementation to this:
#include <stdio.h>
int main(){
int arr[10] = {7, 7, 3, 2, 9, 8, 5, 1, 7, 9};
int max = 0;
int most;
for(int i = 0; i < 10; i++)
{
int tmp = arr[i], count = 0;
// if the current number is the current max number then skip
if(tmp == max)
continue;
for(int j = 0; j < 10; j++)
{
// increment count if number in index j is equal to tmp number
count += arr[j] == tmp ? 1 : 0;
}
// [this condition will depend on the requirement.]
// replace max and most if the count of tmp number is greater than your
// current max
if(count > max){
max = count;
most = tmp;
}
}
printf("Most frequent: %d\ occurs: %d\n", most, max);
return 0;
}
This is not tested so if there are any problems, please feel free to edit.

Answer 2

You ARE multiplying max by 10 since you are doing everything 100 times (instead of 10) because of your totally redundant for i loop.
Specifically your problem is you are incrementing the values in occurs 10 times (instead of once). Since most doesn't use the incremented values it doesn't have problem.

Answer 3

The faster, O(2n-1) complexity solution
#include <stdio.h>
int main(){
int arr[10] = {7, 7, 3, 2, 9, 8, 5, 1, 7, 9};
int occur[10] = {NULL};
int max = 0;
for(int i = 0; i < 10; ++i)
++occur[arr[i]];
for (int i = 1; i < 10; ++i)
if (occur[i] > occur[max])
max = i;
printf("Most frequent: %d\ occurs: %d\n", max, occur[max]);
return 0;
}
Yet faster, in O(n)... I had a feeling that..
int main(){
int arr[10] = {7, 7, 3, 2, 9, 8, 5, 1, 7, 9};
int occur[10] = {NULL};
int max = 0;
for(int i = 0; i < 10; ++i)
if (++occur[arr[i]] > occur[max])
max = arr[i];
printf("Most frequent: %d\ occurs: %d\n", max, occur[max]);
return 0;
}

Answer 4

I will not argue that your O(n^2) operations algorithm is not the ideal way to do the task.
But moving one line of code will fix your code.
Your loop:
for(int i = 0; i < 10; i++)
{
for(int j = 0; j < 10; j++)
{
occur[arr[j]]++;
Fix:
for(int i = 0; i < 10; i++)
{
occur[arr[i]]++;
for(int j = 0; j < 10; j++)
{
I'll let you figure out how you can do this in O(2n) operations or less...