Find Positive Integer Solution for a Given Equation

#1399 Most AskedMedium

Find Positive Integer Solution for a Given Equation #1399 Most Asked

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Two PointersBinary Search

Given a callable function f(x, y) with a hidden formula and a value z, reverse engineer the formula and return all positive integer pairs x and y where f(x,y) == z. You may return the pairs in any order.

While the exact formula is hidden, the function is monotonically increasing, i.e.:

f(x, y) < f(x + 1, y)
f(x, y) < f(x, y + 1)

The function interface is defined like this:

interface CustomFunction {
public:
  // Returns some positive integer f(x, y) for two positive integers x and y based on a formula.
  int f(int x, int y);
};

We will judge your solution as follows:

The judge has a list of 9 hidden implementations of CustomFunction, along with a way to generate an answer key of all valid pairs for a specific z.
The judge will receive two inputs: a function_id (to determine which implementation to test your code with), and the target z.
The judge will call your findSolution and compare your results with the answer key.
If your results match the answer key, your solution will be Accepted.

Example 1:

Input: function_id = 1, z = 5
Output: [[1,4],[2,3],[3,2],[4,1]]
Explanation: The hidden formula for function_id = 1 is f(x, y) = x + y.
The following positive integer values of x and y make f(x, y) equal to 5:
x=1, y=4 -> f(1, 4) = 1 + 4 = 5.
x=2, y=3 -> f(2, 3) = 2 + 3 = 5.
x=3, y=2 -> f(3, 2) = 3 + 2 = 5.
x=4, y=1 -> f(4, 1) = 4 + 1 = 5.

Example 2:

Input: function_id = 2, z = 5
Output: [[1,5],[5,1]]
Explanation: The hidden formula for function_id = 2 is f(x, y) = x * y.
The following positive integer values of x and y make f(x, y) equal to 5:
x=1, y=5 -> f(1, 5) = 1 * 5 = 5.
x=5, y=1 -> f(5, 1) = 5 * 1 = 5.

Constraints:

1 <= function_id <= 9
1 <= z <= 100
It is guaranteed that the solutions of f(x, y) == z will be in the range 1 <= x, y <= 1000.
It is also guaranteed that f(x, y) will fit in 32 bit signed integer if 1 <= x, y <= 1000.

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the range of possible values for `z`, `x`, and `y`?
If no solution exists, what should I return (e.g., an empty list, null)?
Are there any constraints on the function `f(x, y)` beyond being monotonically increasing in both x and y?
Is there a preferred order for the pairs `[x, y]` in the output list? For example, should they be sorted by `x` or `y`?
Can I assume that the given function `f(x, y)` is efficient to compute, or should I consider its computational cost when designing my algorithm?

Brute Force Solution

Approach

The brute force approach involves trying every possible combination of positive integer values for the given equation. We systematically check each pair of numbers to see if they satisfy the equation. If they do, we've found a solution.

Here's how the algorithm would work step-by-step:

Start with the smallest possible positive integer value for the first variable, which is 1.
Then, for that value, try every possible positive integer value for the second variable, starting with 1.
Plug both values into the given equation and check if the equation holds true.
If the equation is true, we have found a valid solution, so remember it.
If the equation is not true, try the next value for the second variable.
Keep testing all possible values for the second variable until you reach a point where it's clear no larger value will work (the result becomes too large or small depending on the equation).
Once you've exhausted all possibilities for the second variable, increase the first variable to the next integer and repeat the process.
Continue this process of trying all possible pairs of positive integers until you have tested all possibilities within a reasonable range (again, where it is clear that values outside the range will not work).
Finally, return all the solutions you remembered that satisfy the equation.

Code Implementation

def find_solution(custom_function):    solutions = []
    max_value = 10 # Limit search space
    for first_variable in range(1, max_value + 1):
        for second_variable in range(1, max_value + 1):
            # Evaluate the function for the current pair
            result = custom_function.f(first_variable, second_variable)

            # Check if the equation is satisfied, target is assumed to be zero implicitly.
            if result == 0:
                solutions.append([first_variable, second_variable])

    return solutions

Big(O) Analysis

Time Complexity

O(n^2) – The brute force approach iterates through possible values for two variables. For the first variable (x), we try values from 1 to n. For each value of x, we iterate through possible values of the second variable (y) from 1 up to a maximum value dependent on x and the equation's target value (which can also be considered proportional to n in the worst case). This results in nested loops where the outer loop iterates up to n and the inner loop iterates up to approximately n for each value of the outer loop variable. Thus, the total number of operations is proportional to n * n, resulting in a time complexity of O(n^2).

Space Complexity

O(K) – The auxiliary space is primarily determined by the storage of valid solutions. The plain English explanation mentions remembering the solutions that satisfy the equation, implying a data structure like a list or array is used. Let's denote K as the number of solutions found. Therefore, the space complexity depends on the number of solutions, resulting in O(K) auxiliary space for storing the solution set.

Optimal Solution

Approach

The core idea is to leverage the equation's increasing nature. We can start from opposite ends and move inwards, intelligently eliminating possibilities that cannot be solutions.

Here's how the algorithm would work step-by-step:

Start with 'x' at its minimum possible value (1) and 'y' at its maximum possible value.
Evaluate the equation with these 'x' and 'y' values.
If the result is greater than the target value, it means 'y' is too big. Decrease 'y' by 1, because the function increases as 'y' increases.
If the result is less than the target value, it means 'x' is too small. Increase 'x' by 1, because the function increases as 'x' increases.
If the result equals the target value, you've found a solution! Record the 'x' and 'y' values.
Continue adjusting 'x' and 'y' until 'x' becomes greater than the maximum possible value of 'x' or 'y' becomes smaller than the minimum possible value of 'y'.
During each adjustment, if we find a valid solution, save that x,y pair.
The process effectively eliminates entire regions of the search space, making it much faster than trying every possible combination of 'x' and 'y'.

Code Implementation

class Solution:
    def findSolution(self, customfunction, z_value):
        x_value = 1
        y_value = 1000
        solution_list = []

        # Iterate until x exceeds its max or y falls below its min
        while x_value <= 1000 and y_value >= 1:
            result = customfunction.f(x_value, y_value)

            # If result is greater than target, decrement y
            if result > z_value:
                y_value -= 1

            # If result is less than target, increment x
            elif result < z_value:
                x_value += 1

            # Found a solution
            else:
                solution_list.append([x_value, y_value])
                x_value += 1
                # Need to increment x and continue search

        return solution_list

Big(O) Analysis

Time Complexity

O(n) – The algorithm starts with x = 1 and y = n (where n is the upper bound for both x and y). In each iteration, either x increases or y decreases. The while loop continues until x > n or y < 1. Therefore, in the worst-case scenario, the algorithm iterates a number of times proportional to n because x and y are moving towards each other from opposite ends. Thus, the time complexity is O(n).

Space Complexity

O(N) – The algorithm stores the valid (x, y) pairs in a list to return as the solution. In the worst-case scenario, where every possible x and y combination results in a valid solution to the equation equaling the target, we would need to store all N potential solution pairs (x, y) where N is the upper bound of either x or y. Therefore, the auxiliary space grows linearly with the size of the maximum input (N). This results in a space complexity of O(N).

Edge Cases

Case	How to Handle
z is less than the minimum possible value of f(1,1)	Return an empty list immediately, as no solution can exist for z values that are too small.
f(x, y) always returns a value greater than z for all possible x and y within reasonable bounds	The binary search or two-pointer approach should naturally terminate and return an empty list when no solutions are found.
f(x, y) encounters integer overflow during calculation	Check for potential overflow and return an empty list if overflow occurs, or use a language feature to allow for larger number storage.
Maximum possible value of x or y is reached before finding a solution	Set a reasonable upper bound on x and y, and terminate the search if these bounds are reached without finding a solution.
z is a very large value that requires large x and y to satisfy the function	The search algorithm should handle large z efficiently by using binary search or a two-pointer approach, rather than a naive exhaustive search.
f(x, y) has multiple solutions. x and y can switch values but results in the same answer.	The algorithm must correctly identify and return all unique pairs of [x, y] that satisfy the equation without duplicates or omissions.
The function f(x,y) is computationally expensive	Employ a binary search or two-pointer approach to minimize the number of calls to f(x, y).
z is equal to the maximum possible value of f(x,y) within the given bounds.	The algorithm must correctly identify the solution at the boundary and return it without going out of bounds.

z is less than the minimum possible value of f(1,1)

How to Handle:

Return an empty list immediately, as no solution can exist for z values that are too small.

f(x, y) always returns a value greater than z for all possible x and y within reasonable bounds

How to Handle:

The binary search or two-pointer approach should naturally terminate and return an empty list when no solutions are found.

f(x, y) encounters integer overflow during calculation

How to Handle:

Check for potential overflow and return an empty list if overflow occurs, or use a language feature to allow for larger number storage.

Maximum possible value of x or y is reached before finding a solution

How to Handle:

Set a reasonable upper bound on x and y, and terminate the search if these bounds are reached without finding a solution.

z is a very large value that requires large x and y to satisfy the function

How to Handle:

The search algorithm should handle large z efficiently by using binary search or a two-pointer approach, rather than a naive exhaustive search.

f(x, y) has multiple solutions. x and y can switch values but results in the same answer.

How to Handle:

The algorithm must correctly identify and return all unique pairs of [x, y] that satisfy the equation without duplicates or omissions.

The function f(x,y) is computationally expensive

How to Handle:

Employ a binary search or two-pointer approach to minimize the number of calls to f(x, y).

z is equal to the maximum possible value of f(x,y) within the given bounds.

How to Handle:

The algorithm must correctly identify the solution at the boundary and return it without going out of bounds.