Solution Approach

The solution implements binary search using Python's bisect.bisect() function with a custom key function.

Implementation Details:

1. Range Setup: We create a range from 1 to n+1 using range(1, n + 1), which represents all possible numbers we need to search through.

2. Using bisect.bisect(): The bisect.bisect() function performs binary search on the range. It takes three main parameters:

   • The sequence to search: range(1, n + 1)
   • The value to search for: 0
   • A key function: lambda x: -guess(x)

3. Key Function Transformation: The key function lambda x: -guess(x) is crucial. For each number x in the range:

   • If x is less than the picked number, guess(x) returns 1, so -guess(x) returns -1
   • If x equals the picked number, guess(x) returns 0, so -guess(x) returns 0
   • If x is greater than the picked number, guess(x) returns -1, so -guess(x) returns 1

4. How bisect Works: The bisect.bisect() function searches for the insertion point of value 0 in the transformed sequence. Since the negated guess values create a sequence that goes from -1 (for numbers below target) to 0 (at target) to 1 (above target), bisect will find the exact position where 0 appears, which is our answer.

5. Return Value: The function directly returns the result of bisect.bisect(), which is the index (1-based in this case due to our range starting at 1) where the picked number is located.

Time Complexity: O(log n) - Binary search divides the search space in half with each iteration.

Space Complexity: O(1) - The range object is a generator that doesn't store all values in memory, and we only use constant extra space for the binary search operation.

Example Walkthrough

Let's walk through a concrete example where n = 10 and the picked number is 6.

Initial Setup:

• Search range: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
• Target: Find the picked number (which is 6, but we don't know this yet)
• We'll use bisect.bisect(range(1, 11), 0, key=lambda x: -guess(x))

How Binary Search Proceeds:

Step 1: Bisect starts by checking the middle of the range

• Middle position: x = 5
• Call guess(5): returns 1 (5 is too low, since picked = 6)
• Key function: -guess(5) = -1
• Since -1 < 0, bisect knows the answer is in the upper half

Step 2: Search narrows to [6, 7, 8, 9, 10]

• Middle position: x = 8
• Call guess(8): returns -1 (8 is too high, since picked = 6)
• Key function: -guess(8) = 1
• Since 1 > 0, bisect knows the answer is in the lower half

Step 3: Search narrows to [6, 7]

• Check position: x = 6
• Call guess(6): returns 0 (correct!)
• Key function: -guess(6) = 0
• Since we found exactly 0, this is our insertion point

Step 4: Bisect continues to verify the exact insertion position

• Check position: x = 7
• Call guess(7): returns -1 (too high)
• Key function: -guess(7) = 1
• This confirms that 6 is the last position where the key function returns ≤ 0

Result: bisect.bisect() returns 6, which is the correct answer.

Visualizing the Transformation:

Number x: 1 2 3 4 5 6 7 8 9 10 guess(x): 1 1 1 1 1 0 -1 -1 -1 -1 -guess(x): -1 -1 -1 -1 -1 0 1 1 1 1  ↑ ↑  all negative target (returns 0)

The bisect function finds where to insert 0 in this transformed sequence, which is exactly at position 6 - our answer!

Time and Space Complexity

The time complexity is O(log n), where n is the upper limit given in the problem. This is because bisect.bisect performs a binary search on the range [1, n]. Binary search works by repeatedly dividing the search space in half, requiring at most log₂(n) comparisons to find the target element. Even though the code creates a range object and uses a lambda function as a key, the bisect module intelligently performs binary search without materializing the entire range, making only O(log n) calls to the guess function.

The space complexity is O(1). While the code creates a range(1, n + 1) object, Python's range objects are lazy iterators that don't store all values in memory - they only store the start, stop, and step values. The bisect.bisect function uses constant extra space for its binary search implementation (just a few variables to track indices). The lambda function and its execution also use constant space.

Common Pitfalls

Pitfall 1: Using Wrong Binary Search Template Variant

The Problem: Using while left < right with right = mid instead of the standard template with while left <= right and right = mid - 1.

Wrong Implementation:

while left < right:  mid = (left + right) // 2  if guess(mid) <= 0:  right = mid # WRONG - should be mid - 1  else:  left = mid + 1 return left # WRONG - should track first_true_index

Solution: Use the standard template with first_true_index to track the answer:

first_true_index = -1 while left <= right:  mid = (left + right) // 2  if guess(mid) <= 0:  first_true_index = mid  right = mid - 1  else:  left = mid + 1 return first_true_index

Pitfall 2: Misunderstanding the Guess API Return Values

The guess() function returns:

• -1 if your guess is higher than the picked number
• 1 if your guess is lower than the picked number
• 0 if your guess is correct

Wrong Interpretation:

# WRONG: Inverting the meaning of -1 and 1 if guess(mid) == 1: # Thinking 1 means "too high"  right = mid - 1

Solution: The feasible condition is guess(mid) <= 0, which means mid is greater than or equal to the picked number.

Pitfall 3: Integer Overflow in Mid Calculation

In languages with fixed integer sizes, (left + right) / 2 can overflow.

Wrong Implementation:

int mid = (left + right) / 2; // Can overflow when left + right > Integer.MAX_VALUE

Solution: Use left + (right - left) / 2 to avoid overflow:

int mid = left + (right - left) / 2;

Pitfall 4: Off-by-One Errors with Search Range

Using incorrect initial boundaries can miss the answer.

Wrong Implementation:

left, right = 0, n # WRONG: 0 is not a valid guess # or left, right = 1, n - 1 # WRONG: might miss n

Solution: Always use left = 1 and right = n since the picked number is between 1 and n inclusive.