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Solution: Count Element Occurrence

Solution: Count Element Occurrence

Here is a detailed analysis of the different ways to count the frequency of a number in a sorted array of integers

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#include <iostream>
using namespace std;
int calcFreq(int arr[], int arrSize, int s) {
int count = 0;
for(int i = 0; i < arrSize; i++) {
if(arr[i] == s)
count++;
}
return count;
}
int main() {
int arr[] = {-5,-3,0,1,3,3,3,3,4,5};
cout << calcFreq(arr, 10, 3) << endl;
}

This is an extremely simple way to solve this problem. We simply initialize a variable to keep count called, count to 0 and then iterate over the array, increasing count by 1 every time the target value is encountered.

Time Complexity

The time complexity of this algorithm is in ...

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