Solution Approach

We use the standard binary search template to find the first index where nums[i] <= nums[n-1] is true.

Initial Setup:

left, right = 0, len(nums) - 1
first_true_index = -1

Initialize the search boundaries and a variable to track the first position where the feasible condition is true.

Feasible Function: For any index mid, the feasible condition is: nums[mid] <= nums[n-1]

This creates a pattern like [false, false, false, true, true, true] where we want to find the first true.

Binary Search Loop: Using while left <= right:

1. Calculate the middle index: mid = (left + right) // 2

2. Check if feasible (nums[mid] <= nums[n-1]):

   • If true: This could be the minimum or the minimum could be further left. Record first_true_index = mid, then search left with right = mid - 1
   • If false: The minimum must be to the right. Search right with left = mid + 1

Return Result: After the loop, first_true_index holds the index of the minimum element. Return nums[first_true_index].

Edge Case: If the array is not rotated (or rotated n times), the entire array satisfies the feasible condition, so first_true_index will be 0, correctly returning the first element.

Time Complexity: O(log n) - We eliminate half of the search space in each iteration. Space Complexity: O(1) - We only use a constant amount of extra space for the pointers.

Example Walkthrough

Let's trace through the algorithm with the array [4,5,6,7,0,1,2].

Setup:

• Array: [4,5,6,7,0,1,2]
• left = 0, right = 6
• first_true_index = -1
• Feasible condition: nums[mid] <= nums[6] (i.e., nums[mid] <= 2)

Iteration 1:

• left = 0, right = 6
• mid = (0 + 6) // 2 = 3
• Check nums[3] = 7 <= 2? No (not feasible)
• Update: left = mid + 1 = 4

Iteration 2:

• left = 4, right = 6
• mid = (4 + 6) // 2 = 5
• Check nums[5] = 1 <= 2? Yes (feasible)
• Update: first_true_index = 5, right = mid - 1 = 4

Iteration 3:

• left = 4, right = 4
• mid = (4 + 4) // 2 = 4
• Check nums[4] = 0 <= 2? Yes (feasible)
• Update: first_true_index = 4, right = mid - 1 = 3

Loop ends (left > right)

Result:

• first_true_index = 4
• Return nums[4] = 0

The algorithm correctly identifies that the minimum element 0 is at index 4, which is exactly where the rotation break occurs (between 7 and 0).

Feasibility pattern for this array:

Index:     0      1      2      3      4     5     6
Value:     4      5      6      7      0     1     2
<= 2?:   false  false  false  false  true  true  true

Time and Space Complexity

Time Complexity: O(log n), where n is the length of the input array nums.

The algorithm uses binary search to find the minimum element in a rotated sorted array. In each iteration of the while loop, the search space is reduced by half. The number of iterations is logarithmic with respect to the array size, as we divide the search range by 2 in each step until left >= right. The initial check if nums[0] <= nums[-1] takes O(1) time, and all operations inside the loop (comparison and index calculations) are O(1). Therefore, the overall time complexity is O(log n).

Space Complexity: O(1)

The algorithm uses only a constant amount of extra space. It maintains three integer variables (left, right, and mid) regardless of the input size. No additional data structures are created, and the algorithm operates directly on the input array without recursion (which would use stack space). Thus, the space complexity is constant.

Common Pitfalls

1. Using Wrong Binary Search Template Variant

The Pitfall: Using while left < right with right = mid instead of the standard template:

# INCORRECT variant
while left < right:
    mid = (left + right) // 2
    if nums[mid] > nums[right]:
        left = mid + 1
    else:
        right = mid  # Wrong variant!
return nums[left]

Correct Template Implementation:

first_true_index = -1
while left <= right:
    mid = (left + right) // 2
    if nums[mid] <= nums[n - 1]:
        first_true_index = mid
        right = mid - 1
    else:
        left = mid + 1
return nums[first_true_index]

2. Incorrect Feasible Condition

The Pitfall: Using nums[0] as the reference instead of nums[n-1]:

# INCORRECT - comparing with nums[0]
if nums[mid] < nums[0]:  # This works but requires different logic
    first_true_index = mid
    right = mid - 1

Why This Is Problematic: Using nums[0] requires handling the unrotated array case separately, since all elements satisfy nums[mid] >= nums[0]. Using nums[n-1] creates a cleaner feasible pattern that works for all cases.

Solution: Use nums[n-1] as reference:

if nums[mid] <= nums[n - 1]:
    first_true_index = mid
    right = mid - 1

3. Forgetting first_true_index Initialization

The Pitfall: Not properly handling the case where the array is already sorted (no rotation):

# If array is [1, 2, 3, 4, 5]
# All elements satisfy nums[mid] <= nums[n-1]
# first_true_index will be set on the first iteration

For non-rotated arrays, every element satisfies the feasible condition, so first_true_index will correctly be set to 0 (the index of the minimum).

4. Integer Overflow in Mid Calculation

The Pitfall: Using (left + right) / 2 in languages with fixed integer sizes:

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

Solution: Use overflow-safe calculation:

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