Sometimes you can optimize by solving for one of the variables in terms of the others, so that you don't need to waste time computing it. A common type of "equation" problem is the 2-sum, 3-sum, k-sum, k-sum-smaller, etc problem. Here's a problem that shows the core idea. Print out all positive integers `a`, `b`, and `c` that sum to the number 10. You might think to do a brute force solution like this: ``` for a in range(1,11): for b in range(1,11): for c in range(1,11): if a + b + c == 10: prin

Equation Problems

Sometimes you can optimize by solving for one of the variables in terms of the others, so that you don't need to waste time computing it.

A common type of "equation" problem is the 2-sum, 3-sum, k-sum, k-sum-smaller, etc problem. Here's a problem that shows the core idea.

-----problem-----
Print out all positive integers `a`, `b`, and `c` that sum to the number 10.


$$
\large \texttt{a + b + c = 10}
$$


-----solution-----

You might think to do a brute force solution like this:

```
for a in range(1,11):
    for b in range(1,11):
        for c in range(1,11):

            if a + b + c == 10:
                print(a,b,c)
```

\tc $O(n^3)$. We have a triple for-loop.

\sc $O(1)$.

But you can actually do much better than brute-force. 
Whenever you see a problem like this, you can always write one of the variables using the other variables. Here, you can subtract a and b from both sides of the equation to solve for c, like this:
$$
\large \texttt{c = 10 - a - b}
$$
This equation tells you that the value of `c` is completely determined by the values of `a` and `b`. This is because `c = 10 - a - b`. This means we don't need to loop over it. Here's the above code, but where we removed the for-loop over c:

```
for a in range(1, 11):
    for b in range(1, 11):

        c = 10 - a - b # `c` is determined by a and b

        if c > 0: # c has to be positive
            print(a,b,c)

```



\tc $O(n^2)$. We have a double for-loop.

\sc $O(1)$.



Iteration

Equation Problems

Print Integers that Sum to 10