1237. Find Positive Integer Solution for a Given Equation

Problem Description

In this problem, we are given a special callable function f(x, y) that has a hidden formula. The function f is known to be monotonically increasing in both its arguments x and y. This means f(x, y) is always less than f(x+1, y) and f(x, y+1). Our task is to identify all pairs of positive integers (x, y) such that when the function f is applied to these pairs, the result equals a given target value z. Note that x and y must be positive integers. The final answer can be returned in any order.

Since we don't know the exact formula of f, we cannot directly calculate the values. Therefore, we need to use the properties of the function to systematically find all pairs (x, y) that satisfy f(x, y) == z.

Intuition

The key to solving this problem lies in understanding two main points:

• The function f is monotonically increasing which means as we increase x or y, the value of f(x, y) increases.
• The function f outputs positive integers, and we are looking for positive integer inputs (x, y) as well. This means there will be a distinct and finite number of solutions.

Given these properties, one efficient way to find all pairs (x, y) that satisfy f(x, y) == z is by using a two-pointer technique. We start with the smallest possible value for x (which is 1) and the largest possible value for y within a reasonable range (for example, 1000) and evaluate f(x, y):

• If f(x, y) < z, it means that with the current pair (x, y), the result is too small. Since f is monotonically increasing and we cannot decrease y further (as we started with the maximum possible value), the only way to increase f(x, y) is by increasing x. Hence, we increment x.
• If f(x, y) > z, the result is too large. We decrease y since decreasing y is certain to decrease the value of f(x, y).
• If f(x, y) == z, we found a valid pair and we add it to our answer list. We then increment x and decrement y to find other possible pairs.

The algorithm continues this process until we exhaust all combinations up to the maximum value for both x and y. This approach leverages the function's monotonicity and avoids testing impossible combinations, leading to an efficient solution.

Learn more about Math, Two Pointers and Binary Search patterns.

Solution Approach

The implementation is based on the two-pointer technique, as mentioned in the intuition. Here's a detailed walkthrough:

1. Initialize two pointers, x and y. Set x to 1, as we want to start with the smallest positive integer for x, and set y to 1000, assuming that the function f doesn't have x or y values greater than 1000. This is a reasonable assumption for these types of problems unless specified otherwise.

2. Use a while loop to iterate as long as x is less than or equal to 1000 and y is greater than 0. This ensures we consider all possible combinations of x and y within the given bounds.

3. Inside the loop, call customfunction.f(x, y) with the current x and y variables to get the output of f.

4. Check if the value of f(x, y) is less than the target value z. If it is, we need to increase x (since increasing x will increase the value of f(x, y) due to the monotonic nature of the function).

5. If f(x, y) is greater than z, then our current output is too high, and we need to reduce it. We do this by reducing y, as decreasing y will decrease f(x, y).

6. If f(x, y) equals z, we've found a valid pair, and we add [x, y] to the list ans. We then move both pointers—incrementing x and decrementing y—to continue searching for other possible solutions.

7. The loop exits once we either exceed the maximum value for x or y reaches 0. At this point, we would have tested all reasonable x and y combinations.

8. Return the list ans containing all pairs [x, y] that satisfy f(x, y) == z.

The algorithm effectively navigates the search space using the two-pointer technique, which exploits the monotonically increasing property of the function f. In terms of data structures, the implementation uses a list ans to store the pairs that fulfill the condition. No advanced data structures are needed since the problem is tackled primarily with algorithmic logic.

This approach is quite efficient, as it avoids unnecessary calculations for x and y pairs that obviously do not meet the condition. By carefully increasing and decreasing x and y, it homes in on the correct pairs without checking every possible combination.

Note: The approach described as "Binary search" in the provided reference solution approach appears to be a mislabel, as the actual implemented technique is, in fact, the two-pointer technique. Binary search would involve repeatedly halving the search space to find a solution, which isn't the case here.

Example Walkthrough

Let's illustrate the solution approach with an example. Suppose we have the special function f(x, y) and our target value z is 14. We don't know the hidden formula within f, but we are given that f is monotonically increasing in both x and y, and we are tasked to find all positive integer pairs (x, y) that satisfy the equation f(x, y) == z.

Following our solution approach:

1. We initialize x as 1 and y as 1000 (assuming f's range is within these bounds).

2. We enter our while loop and start evaluating f(x, y) to see if it matches z.

3. In our first iteration, we call f(1, 1000). If we find that f(1, 1000) < 14, we can conclude that we need to increase x because increasing x would increase the output of f. Hence, we move to x = 2.

4. We call f(2, 1000) and suppose we find f(2, 1000) > 14. The result is too high, meaning y must be decreased. Let's say we choose y = 999.

5. We call f(2, 999) and find that f(2, 999) == 14. We've found a matching pair (2, 999), so we add it to our answer list.

6. We continue iterating, incrementing x to 3 and decrementing y to 998, and we call f(3, 998). If f(3, 998) < 14, we must increase x. If f(3, 998) > 14, we must decrease y. If f(3, 998) == 14, we've found another valid pair.

7. The process repeats, traversing various (x, y) pairs and responding accordingly, until x > 1000 or y <= 0.

8. We finish iterating and return the list ans that contains all our valid pairs [x, y].

Through this example, it becomes clear how the two-pointer technique effectively pinpoints the valid (x, y) combinations without redundant checks, by taking advantage of f's monotonicity. As such, assuming our y never needed to decrease below 995 and our x never rose above 10 before we finished, we might end up with a list ans that could look something like [ [2, 999], [4, 997], [8, 995] ]. The solution is efficient because it avoids unnecessarily checking combinations where f(x, y) would be guaranteed to miss the target z.

(Note: The actual results of f(2, 1000), f(2, 999), and f(3, 998) are arbitrary for the purposes of this example since the function f is unknown and merely illustrative to demonstrate the search process.)

Solution Implementation

Time and Space Complexity

Time Complexity

The given code has a while loop that iterates at most min(1000, z) times because:

• The maximum possible value for x and y is 1000 (as per the constraints).
• For every iteration, either x is incremented or y is decremented.
• The loop stops when x exceeds 1000 or y reaches below 1.

Assuming that the custom function f(x, y) is O(1) (since we don't have information about its internal implementation, we consider the worst-case scenario where it takes a constant time), the time complexity of the entire operation is O(min(1000, z)).

Space Complexity

The space complexity mainly depends on the number of valid (x, y) pairs that satisfy the equation f(x, y) = z. In the worst case, all pairs (x, y) where 1 <= x, y <= 1000 could be solutions, so there could be as many as 1000 solutions.

However, since the space used by ans depends on the output, which we do not count towards the space complexity in the analysis, the additional space used by the algorithm (i.e., for variables x, y, t) is constant.

Therefore, the space complexity of the code is O(1).

Learn more about how to find time and space complexity quickly using problem constraints.