Solution Approach

The implementation uses enumeration combined with binary search to efficiently find all valid pairs.

Step-by-step implementation:

1. Initialize the result list: Create an empty list ans to store all valid [x, y] pairs.

2. Enumerate x values: Loop through all possible x values from 1 to z:

   for x in range(1, z + 1):

3. Binary search for y: For each fixed x, use binary search to find a y value where f(x, y) = z. The code uses Python's bisect_left function with a custom key:

   y = 1 + bisect_left(
       range(1, z + 1), z, key=lambda y: customfunction.f(x, y)
   )

   Here's what happens:

   • range(1, z + 1) creates the search space for y values
   • bisect_left searches for the leftmost position where we could insert z to maintain sorted order
   • key=lambda y: customfunction.f(x, y) tells the binary search to compare based on function values, not the y values themselves
   • We add 1 to the result because bisect_left returns an index (0-based) but our y values start from 1

4. Check if exact match found: After binary search, verify if the found position gives us exactly z:

   if customfunction.f(x, y) == z:
       ans.append([x, y])

   The binary search finds where f(x, y) >= z. If f(x, y) == z, we have a valid pair. If f(x, y) > z, it means no y value for this x produces exactly z.

5. Return all pairs: After checking all possible x values, return the collected pairs.

Time Complexity: O(z * log z) - We iterate through z values of x, and for each x, we perform a binary search over at most z values of y.

Space Complexity: O(1) if we don't count the output array, as we only use a constant amount of extra space for variables.

Example Walkthrough

Let's walk through a concrete example where we have a hidden function and need to find all pairs (x, y) where f(x, y) = 5.

Suppose the hidden function is f(x, y) = x + y (we don't know this, but it helps visualize the process).

Target: z = 5

We'll iterate through x values from 1 to 5 and use binary search to find corresponding y values:

When x = 1:

• We binary search for y in range [1, 5] where f(1, y) = 5
• Binary search process:
  • Check middle: y = 3, f(1, 3) = 4 < 5 (go right)
  • Check y = 4, f(1, 4) = 5 = 5 ✓
• Found pair: [1, 4]

When x = 2:

• Binary search for y in range [1, 5] where f(2, y) = 5
• Binary search process:
  • Check middle: y = 3, f(2, 3) = 5 = 5 ✓
• Found pair: [2, 3]

When x = 3:

• Binary search for y in range [1, 5] where f(3, y) = 5
• Binary search process:
  • Check middle: y = 3, f(3, 3) = 6 > 5 (go left)
  • Check y = 1, f(3, 1) = 4 < 5 (go right)
  • Check y = 2, f(3, 2) = 5 = 5 ✓
• Found pair: [3, 2]

When x = 4:

• Binary search for y in range [1, 5] where f(4, y) = 5
• Binary search process:
  • Check middle: y = 3, f(4, 3) = 7 > 5 (go left)
  • Check y = 1, f(4, 1) = 5 = 5 ✓
• Found pair: [4, 1]

When x = 5:

• Binary search for y in range [1, 5] where f(5, y) = 5
• Binary search process:
  • Check middle: y = 3, f(5, 3) = 8 > 5 (go left)
  • Check y = 1, f(5, 1) = 6 > 5 (go left further)
  • No valid y exists (all values > 5)
• No pair found

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

The key insight is that for each fixed x, the function values f(x, 1), f(x, 2), f(x, 3)... form an increasing sequence, allowing us to efficiently binary search for the target value rather than checking every possibility.

Time and Space Complexity

Time Complexity: O(n log n), where n is the value of z.

The outer loop iterates from x = 1 to x = z, which gives us O(n) iterations. For each value of x, we perform a binary search using bisect_left on the range [1, z] to find the appropriate value of y. The binary search operation takes O(log n) time since we're searching through a range of size z. The key function lambda y: customfunction.f(x, y) is called O(log n) times during each binary search. Since we perform this binary search for each of the n values of x, the total time complexity is O(n) × O(log n) = O(n log n).

Space Complexity: O(1) (excluding the output array).

The algorithm uses only a constant amount of extra space for variables x and y. The bisect_left function operates on a range object and uses O(1) additional space for its internal operations. The ans list stores the output, but this is typically not counted in space complexity analysis as it's required for the return value. Therefore, the auxiliary space complexity is O(1).

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

Common Pitfalls

1. Using Wrong Binary Search Template Variant

The most critical mistake is using while left < right with right = mid instead of the standard template.

Wrong template:

while left < right:
    mid = (left + right) // 2
    if customfunction.f(x, mid) >= z:
        right = mid  # Wrong pattern
    else:
        left = mid + 1
# Then check if f(x, left) == z

Correct template:

first_true_index = -1
while left <= right:
    mid = (left + right) // 2
    if customfunction.f(x, mid) >= z:
        first_true_index = mid
        right = mid - 1
    else:
        left = mid + 1
# Then check if f(x, first_true_index) == z

2. Inefficient Search Range for y

Pitfall: The code searches for y in the range [1, z] for every x value, but this can be optimized using two pointers.

Optimized solution using two-pointer approach (O(z) instead of O(z log z)):

def findSolution(self, customfunction: 'CustomFunction', z: int) -> List[List[int]]:
    result = []
    x, y = 1, z

    while x <= z and y >= 1:
        current = customfunction.f(x, y)
        if current == z:
            result.append([x, y])
            x += 1
            y -= 1
        elif current < z:
            x += 1
        else:
            y -= 1

    return result

3. Early Termination Opportunity Missed

Pitfall: The solution continues searching even when it's impossible to find more solutions.

Solution: Add an early termination check:

for x in range(1, z + 1):
    # Early termination: if f(x, 1) > z, no valid y exists
    if customfunction.f(x, 1) > z:
        break
    # Binary search for y...

4. Not Checking first_true_index Before Using

Pitfall: Accessing f(x, first_true_index) without checking if first_true_index != -1.

Solution: Always check if first_true_index was set before using it:

if first_true_index != -1 and customfunction.f(x, first_true_index) == z:
    result.append([x, first_true_index])