How to find the most repeated value in a array? - c

Question

I'm trying to find the most repeated int number of my vector.
Here is my code:
for (i = 0; i < dim; i++){
temp = vet[i];
for (i = 0; i < dim; i++){
if(vet[i] == temp){
count++;
}
}
if (most < count){
most = count;
elem = vet[i];
}
}
return elem;
}
It's not correct. I hope you can help me. Thank you!

Answer 1

The most obvious problem is that your code uses i in both the inner and the outer loops. The most and count variables are uninitialized in the above code, and count needs to be reset before starting the inner loop each time.
The method used in this code of iterating over the entire array for each element to count appearances is not very efficient. The efficiency could be improved by starting the inner loop from i + 1, instead of from 0. By doing this, the first frequency count for each element will be correct, though the later counts will be low since the earlier indices will not be visited. But this does not matter, since the first count will have set the most variable if possible. The count variable can be set to 1 before the inner loop begins, since the ith element is the test value, and the inner loop will skip this index. This change will substantially reduce the number of array accesses.
Note that this function will return the value of the first element in the array that is also the most frequently appearing.
int get_most_common(int vet[], size_t dim)
{
size_t i, j, count;
size_t most = 0;
int temp, elem;
for(i = 0; i < dim; i++) {
temp = vet[i];
count = 1;
for(j = i + 1; j < dim; j++) {
if(vet[j] == temp) {
count++;
}
}
if (most < count) {
most = count;
elem = vet[i];
}
}
return elem;
}

Answer 2

You can always try the brute force method, count the frequency of each element, then find the maximum one.
To implement a full version of such function with efficiency, you will need special data structure such as hashtable or dictionary.
But the following codes work well if you just need to return the first item that match such condition.
#include <stdio.h>
// returns the first matched most frequent item of a list
// list items of same value must be grouped, for example, a sorted list
// only returns the first matched item
int max_frequent_item_of(int vet[], int dim)
{
int i = 0;
// element and its count of current sublist
int current = vet[0];
int current_count = 0;
// most frequent element and its count of the list so far
int most = vet[0];
int most_count = 0;
while (i < dim) {
// if i-th element still belong to current sublist, run the counter
if (vet[i] == current) {
current_count += 1;
i += 1;
}
// otherwise, the current sublist just ended
else {
// update the most frequent element
if (current_count > most_count) {
most = current;
most_count = current_count;
}
// reset current sublist
current = vet[i];
current_count = 1;
i += 1;
}
}
printf("most frequent item %d, count %d\n", most, most_count);
return most;
}
int main(void)
{
// vet must be in order, sort with `qsort` if necessary
int vet[] = {1, 1, 2, 3, 4, 4, 4, 8, 9, 9};
int size = 10;
int n;
printf("list: ");
for (n = 0 ; n < size; n++)
{
printf("%d ", vet[n]);
}
printf("\n");
max_frequent_item_of(vet, size);
return 0;
}
output
list: 1 1 2 3 4 4 4 8 9 9
most frequent item 4, count 3

Answer 3

I write This Code do what you need and extra if two numbers have the same repeating times, it returns the Bigger one.
int ReturnMostRepeated(int val[], size_t dim)
{
size_t i,j,count;
int temp=0, temp2;
for(i =0;i<dim;i++)
{
count = 1;
for(j=i+1;j<dim;j++)
{
if(val[j]==val[i])
{
count++;
}
}
if (count>temp)
{
temp = count;
temp2 = val[i];
}else if(count == temp) if (val[i]>temp2) temp2 = val[i];
}
return temp2;
}