Solution Approach

We use the standard binary search template to find the first bad version. The feasible function is isBadVersion(mid) which directly gives us the false, false, ..., true, true pattern.

Template Implementation:

left, right = 1, n
first_true_index = -1

while left <= right:
    mid = (left + right) // 2
    if isBadVersion(mid):  # feasible condition
        first_true_index = mid
        right = mid - 1  # search for earlier bad version
    else:
        left = mid + 1

return first_true_index

How it works:

1. Initialize left = 1, right = n, and first_true_index = -1
2. Use while left <= right loop (NOT left < right)
3. Check isBadVersion(mid) as the feasible condition
4. If true (bad version found), record first_true_index = mid and search left with right = mid - 1
5. If false (good version), search right with left = mid + 1
6. Return first_true_index as the first bad version

The key difference from other template variants:

• We use first_true_index to track the answer (NOT just returning left at the end)
• We use while left <= right (NOT while left < right)
• We use right = mid - 1 when feasible (NOT right = mid)

This approach guarantees we'll find the first bad version in O(log n) time complexity.

Example Walkthrough

Let's walk through finding the first bad version when n = 8 and version 5 is the first bad one, using the binary search template.

Initial setup: versions [1, 2, 3, 4, 5, 6, 7, 8] where versions 1-4 are good (false) and 5-8 are bad (true).

Initial State:

• left = 1, right = 8
• first_true_index = -1

Iteration 1:

• left = 1, right = 8
• mid = (1 + 8) // 2 = 4
• isBadVersion(4) → false (version 4 is good)
• Update: left = mid + 1 = 5

Iteration 2:

• left = 5, right = 8
• mid = (5 + 8) // 2 = 6
• isBadVersion(6) → true (version 6 is bad)
• Update: first_true_index = 6, right = mid - 1 = 5

Iteration 3:

• left = 5, right = 5
• mid = (5 + 5) // 2 = 5
• isBadVersion(5) → true (version 5 is bad)
• Update: first_true_index = 5, right = mid - 1 = 4

Termination:

• left = 5 > right = 4, exit loop
• Return first_true_index = 5

We found the first bad version (5) with only 3 API calls. The key observation is that we tracked first_true_index to remember the earliest bad version found, while continuing to search left for potentially earlier bad versions.

Time and Space Complexity

The time complexity is O(log n), where n is the input parameter representing the total number of versions. This is because the algorithm uses binary search, which repeatedly divides the search space in half. In each iteration, the algorithm eliminates half of the remaining search space by checking the middle element with isBadVersion(mid). The number of iterations required is logarithmic with respect to n, specifically ⌈log₂(n)⌉ iterations in the worst case.

The space complexity is O(1), as the algorithm only uses a constant amount of extra space. The variables l, r, and mid are the only additional memory allocations, and they don't depend on the size of the input n. The algorithm operates in-place without using any recursive calls or additional data structures that would scale with the input size.

Common Pitfalls

1. Using Wrong Binary Search Template Variant

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

# WRONG - different template variant
while left < right:
    mid = (left + right) // 2
    if isBadVersion(mid):
        right = mid
    else:
        left = mid + 1
return left

Solution: Use the standard binary search template with first_true_index:

# CORRECT - standard template
left, right = 1, n
first_true_index = -1

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

return first_true_index

2. Integer Overflow in Middle Calculation

Pitfall: When calculating the middle point using mid = (left + right) // 2, if both left and right are very large integers, their sum could overflow in some languages.

Solution: Use the overflow-safe formula:

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

3. Starting Search from Wrong Index

Pitfall: Starting with left = 0 instead of left = 1, forgetting that versions are numbered from 1 to n.

Solution: Always initialize with left = 1, right = n to match the problem's version numbering.

4. Returning left Instead of first_true_index

Pitfall: Using the template but then returning left at the end instead of first_true_index, which can give incorrect results in edge cases.

Solution: Always return first_true_index which tracks the actual first bad version found during the search.