How to Solve Find Positive Integer Solution for a Given Equation

This problem requires identifying all positive integer pairs (x, y) such that a hidden monotonic function f(x, y) equals a target z. Since f(x, y) is strictly increasing in both arguments, we can loop over possible values efficiently using binary search or two pointers. The solution systematically explores valid pairs, ensuring correctness while avoiding unnecessary computations over the full 1 to 1000 range.

Problem Statement

You are given a callable function f(x, y) with an unknown formula and a positive integer target z. The function is guaranteed to be monotonically increasing with respect to both x and y. Your task is to return all pairs of positive integers (x, y) where f(x, y) equals z. You may return the pairs in any order.

The interface provides the function f through a hidden implementation and limits x and y to the range 1 to 1000. You should design an approach that efficiently checks potential pairs without iterating all 1,000,000 possibilities. Consider using binary search over the answer space or a two-pointer strategy to leverage the monotonic property.

Examples

Example 1

Input: See original problem statement.

Output: See original problem statement.

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); };

Example 2

Input: function_id = 1, z = 5

Output: [[1,4],[2,3],[3,2],[4,1]]

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 3

Input: function_id = 2, z = 5

Output: [[1,5],[5,1]]

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.

Solution Approach

Loop with Two Pointers

Start with x = 1 and y = 1000, then iteratively check f(x, y). If f(x, y) is greater than z, decrement y; if less, increment x. This takes advantage of the monotonic property and avoids a full O(n^2) search.

Binary Search per x

For each x from 1 to 1000, perform a binary search on y in the range 1 to 1000 to find y such that f(x, y) == z. This reduces unnecessary checks and ensures logarithmic exploration in y.

Combine Both Strategies

Use two pointers to move across the potential solution space while applying binary search where the function grows steeply. This hybrid approach balances iteration and search, minimizing total function calls.

Complexity Analysis

Time complexity ranges from O(n log n) using binary search per x to O(n) with a two-pointer approach, where n = 1000. Space complexity is O(1) aside from storing the result pairs.

What Interviewers Usually Probe

• Are you considering how the monotonic property can eliminate unnecessary checks?
• Can you optimize function calls instead of brute forcing all pairs?
• How will you ensure that all valid pairs are returned without duplicates?

Common Pitfalls or Variants

Common pitfalls

• Brute forcing all 1,000,000 x-y pairs without leveraging monotonicity.
• Assuming the function can be directly inverted without checking integer constraints.
• Failing to handle edge cases where x or y equals 1 or 1000.

Follow-up variants

• Find pairs for a non-monotonic hidden function requiring full enumeration.
• Return solutions where f(x, y) is within a tolerance range instead of exact z.
• Optimize for very large ranges, e.g., 1 <= x, y <= 10^6, using modified search strategies.

How GhostInterview Helps

• GhostInterview identifies the binary search and two-pointer pattern for efficient pair detection.
• It highlights failure modes like brute force over 1 million possibilities and guides optimizations.
• It ensures your solution respects the problem constraints and maximizes correctness with minimal function calls.

Topic Pages

• Math
• Two Pointers
• Binary Search
• Interactive

Related GhostInterview Pages

• LeetCode Interview Copilot - Use GhostInterview as a live solver when you want direct help with LeetCode-style coding questions.
• Coding Interview Assistant - See how GhostInterview supports array, string, linked list, graph, and tree interview workflows.
• How GhostInterview Works - Review the screenshot, reasoning, and answer flow before using the solver in a live interview.

FAQ

What is the main strategy to solve 'Find Positive Integer Solution for a Given Equation'?

The key is to exploit the monotonic property with two pointers or binary search, checking pairs efficiently instead of brute force.

Why can’t we iterate over all x and y values directly?

Iterating all 1,000,000 combinations is inefficient; leveraging monotonicity drastically reduces the number of required function calls.

How does binary search help in this problem?

Binary search narrows down the possible y for each x quickly by exploiting the increasing nature of f(x, y).

Are duplicates possible in the output pairs?

No, each valid (x, y) pair is unique due to the strictly increasing property of f(x, y).

Can this approach handle maximum constraints efficiently?

Yes, combining two-pointer traversal with binary search ensures it works efficiently for x and y up to 1000.