Java – Factorial of cache using static int arrays

static int count = 0;
static long[] cache = new long[3000];

static {
    cache[0] = 1;
}
public static void main(String args[]) {
    for (int i = 0; i < 100; i++){
        factorial_with_cache(2999);
    }
    System.out.println(count);        
}

public static long factorial_with_cache (int n){
    if (cache[n] != 0){
        return cache[n];
    }
    cache[n] = n * factorial_with_cache(n - 1);
    count++;
    return cache[n];
}

I built a function that calculates factorials using caching (ignoring overflows).

But its runtime isn’t much better compared to the non-caching feature, and I found caching not working.

Because I expected the variable “count” to be 2999 after the loop, but I found that it was 293465 and much more than that. (There is no loop, it prints 2999).

What’s wrong with this function?

Because the range of the long data type:

long 8 bytes (-9,223,372,036,854,775,808 to 9,223,372,036,854,775,807)

Before you search for the factorial of 25, your factorial gives a positive value.
After 25, the calculated factorial value is negative (which means you overflow for a long time) and your count will work as expected until 65 factorial is calculated (until negative) and then the value of factorial 66 reaches 0…

Just print the factorial below:

for (int i = 0; i < 100; i++) {
          long fact=factorial_with_cache(66);
          System.out.println(fact);
        }
        System.out.println(count);

The count is printed as

(forLoopCount*(number-65))+65 , In your case (100*(2999-65))+65 is 293465

Because the factorial value traced back in the cache is not zero, it is the 65th element (at index 64).