PROBLEM LINKS

Practice
Contest

DIFFICULTY

Cakewalk

PREREQUISITES

loops, simple maths

PROBLEM

Find number of $(x, y)$ pairs such that $1 \leq x \leq H$, $1 \leq y \leq W$ and $x * y = K$.

QUICK EXPLANATION

Iterate over $x$ and you can check if there exists a valid $y$ in the desired range satisfying $x \cdot y = K$ or not.

EXPLANATION

$\mathcal{O}(H * W)$ bruteforce solution

Constraints on $H$ and $W$ are quite small. We can exploit these to get a simple bruteforce solution. We iterate over all $(x, y)$ pairs and check whether their product $x * y$ is equal to $K$ or not. We can count such valid pairs in $\mathcal{O}(W * H)$ time.

Pseudo Code:

ans = 0
for x = 1 to H:
  for y = 1 to W:
    if x * y == K:
      ans++

$\mathcal{O}(K log K)$ solution

Let us make a simple change in the last solution? What if we you stop iterating over $y$ when the value of $x * y$ exceeds $K$. Will that improve time complexity?

ans = 0
for x = 1 to H:
  for y = 1 to W:
    if x * y > K:
      break;
    if x * y == K:
      ans++

Yes, it will. Let us estimate it. From a casual look, it looks that its time complexity will still be $\mathcal{O}(H * W)$. But it's not. Let us delve into depth.

For each $x$, the inner loop over $y$ will run at most $\frac{K}{x}$ times. So, total number of iterations the program will be run will be given by

$$\frac{K}{1} + \frac{K}{2} + \dots + \frac{K}{H}$$ $$\leq \frac{K}{1} + \frac{K}{2} + \dots + \frac{K}{K}$$ $$\leq K \cdot (\frac{1}{1} + \frac{1}{2} + \dots + \frac{1}{K})$$

The expression $$\frac{1}{1} + \frac{1}{2} + \dots + \frac{1}{n}$$ is also known as harmonic sum $H_n$. One can prove that $H_n = \mathcal{O}(log \, n)$.

Hence, the time complexity will be $\mathcal{O}(K log K)$.

$\mathcal{O}(H)$ solution

Let us exploit the condition $x * y = K$. For a fixed $x$, do we really need to iterate over all $y$'s to check how many of them satisfy $x * y = K$. It turns out, no. Firstly there will be at most a single $y$. If $K$ is not divisible by $x$, then there can't exist such $y$. Otherwise $y$ will be $\frac{K}{x}$.

In summary, we iterate over only $x$ values and find the corresponding $y$ (if it exists), and check whether the $y$ is $\geq 1$ and $\leq H$.

Time complexity of this method will be $\mathcal{O}(H)$, as are iterating over $x$ values only once.

Factorization based solutions

If $x \cdot y = K$, then both $x$ and $y$ should divide $K$. So we can find all the divisors of the $K$. Let $x$ be one such divisor, then $y$ will be $\frac{K}{x}$.

You can find all the divisors of $K$ in $\mathcal{O}(sqrt(K))$ time.

SETTER'S SOLUTION

Can be found here.

TESTER'S SOLUTION

Can be found here.