Times Array Has Been Rotated in C++

Times Array Has Been Rotated in C++ (also known as Find Rotation Count in Rotated Sorted Array) is a classic problem frequently tested in technical interviews at top-tier engineering organizations such as Amazon, Microsoft, Adobe, Apple, and Goldman Sachs. You are given an array of unique integers of length N that was originally sorted in ascending order and subsequently rotated clockwise (rightward) by an unknown integer k times. The algorithmic challenge is to determine the exact rotation count k in strict O(log N) time complexity and O(1) auxiliary space. Because every single clockwise rotation shifts every element forward by one index (with the tail wrapping to index 0), the rotation count k is mathematically equivalent to the index of the minimum element (the pivot inflection point). In this comprehensive masterclass, we will deconstruct the problem from first principles: moving from an O(N) Brute Force Linear Drop Scan, through a Subarray Monotonicity Binary Search, to the Optimal Modified Upper-Boundary Binary Search (nums[mid] vs nums[high]). We will also analyze left vs. right rotation mappings, handle duplicates (LeetCode 154 variant), examine CPU branch predictor interactions, run a 10,000,000-element benchmark lab, and provide hands-on “Try it yourself” exercises after every section.

🔹 Mathematical Foundations & Rotation Invariants

To establish an intuitive mental model, let us observe how rightward (clockwise) and leftward (counter-clockwise) rotations displace array elements. Consider an initial sorted array A = [1, 2, 3, 4, 5, 6, 7] of length N = 7:

Original Array (k = 0):
Index:    0    1    2    3    4    5    6
Value:   [1,   2,   3,   4,   5,   6,   7]  → Min value 1 is at index 0 (k = 0)

Rotate Right 1 time (k = 1):
Index:    0    1    2    3    4    5    6
Value:   [7,   1,   2,   3,   4,   5,   6]  → Min value 1 is at index 1 (k = 1)

Rotate Right 2 times (k = 2):
Index:    0    1    2    3    4    5    6
Value:   [6,   7,   1,   2,   3,   4,   5]  → Min value 1 is at index 2 (k = 2)

Rotate Right 4 times (k = 4):
Index:    0    1    2    3    4    5    6
Value:   [4,   5,   6,   7,   1,   2,   3]  → Min value 1 is at index 4 (k = 4)

Plotting the Cartesian topology of values against indices exposes two strictly ascending piecewise continuous ramps separated by a single catastrophic cliff drop (the inflection point):

Value
  ^
7 |              /|  (Index 3, Value 7)
6 |            /  |
5 |          /    |
4 |  (0,4) /      |
  |               |
3 |               |                   / (6,3)
2 |               |                 /
1 |               |               /
0 |               |______(4,1) ← MINIMUM ELEMENT INDEX = 4 = ROTATION COUNT (k)
  +------------------------------------> Index
     0   1   2   3   4   5   6
     [ Left Segment ] [ Right Segment ]

From this geometric topology, we derive the core mathematical theorems governing rotated arrays:

• Theorem 1: Equivalence of Rotation Count and Pivot Index
  For any array rotated clockwise k times where 0 ≤ k < N, the minimum element of the original array (which started at index 0) is shifted exactly k positions to the right. Therefore:

  Rotation Count (Clockwise) k = index_of(min(nums))

• Theorem 2: Left vs. Right Rotation Duality
  If the problem statement specifies that the array was rotated leftwards (counter-clockwise), the relationship is:

  Left Rotations k_left = (N - index_of(min(nums))) % N

  For example, if nums = [4, 5, 6, 7, 1, 2, 3] with N = 7, the minimum index is 4. Clockwise rotations = 4. Counter-clockwise rotations = (7 - 4) % 7 = 3 (rotating left by 3 shifts [4, 5, 6, 7] to the front).
• Theorem 3: The Inflection Invariant
  The minimum element nums[p] is the unique element in the entire array where nums[p] < nums[p - 1] (with circular wrapping). For all other indices i, nums[i] > nums[i - 1].

Try it yourself

• What is the rotation count for nums = [15, 18, 2, 3, 6, 12]?
• If an array of size 6 is rotated right 15 times, what is its effective rotation count k? (Hint: 15 % 6 = 3).

🔹 Approach 1: Brute Force (Linear Scan for Inflection Drop)

Intuition & Reasoning: In an unrotated sorted array, elements increase strictly monotonically: nums[i] > nums[i - 1] for all 1 ≤ i < N. When rotated, there is exactly one adjacent pair where this order is violated: nums[i] < nums[i - 1]. By scanning linearly from left to right, the very first index i where this drop occurs is guaranteed to be the index of the minimum element, which equals the rotation count. If the loop completes without finding any drop, the array was rotated 0 times (or an exact multiple of N).

#include <iostream>
#include <vector>
#include <cassert>

using namespace std;

// Approach 1: Brute Force Linear Scan for Inflection Drop
// Time Complexity: O(N), Space Complexity: O(1)
int count_rotations_brute_force(const vector<int>& nums) {
    if (nums.empty()) return 0;
    
    const size_t n = nums.size();
    for (size_t i = 1; i < n; ++i) {
        // The first element that is smaller than its predecessor is the pivot
        if (nums[i] < nums[i - 1]) {
            return static_cast<int>(i);
        }
    }
    
    // If no drop is found, the array is strictly sorted (0 rotations)
    return 0;
}

int main() {
    vector<int> a1 = {4, 5, 6, 7, 0, 1, 2};
    vector<int> a2 = {15, 18, 2, 3, 6, 12};
    vector<int> a3 = {1, 2, 3, 4, 5};
    vector<int> a4 = {2, 1};

    cout << "--- Approach 1: Brute Force Linear Drop Scan ---\n";
    cout << "Array [4, 5, 6, 7, 0, 1, 2]  → Rotations: " << count_rotations_brute_force(a1) << "\n";
    cout << "Array [15, 18, 2, 3, 6, 12]  → Rotations: " << count_rotations_brute_force(a2) << "\n";
    cout << "Array [1, 2, 3, 4, 5]        → Rotations: " << count_rotations_brute_force(a3) << "\n";
    cout << "Array [2, 1]                  → Rotations: " << count_rotations_brute_force(a4) << "\n";

    assert(count_rotations_brute_force(a1) == 4);
    assert(count_rotations_brute_force(a2) == 2);
    assert(count_rotations_brute_force(a3) == 0);
    assert(count_rotations_brute_force(a4) == 1);
    cout << "All assertions passed successfully!\n";
    return 0;
}

Output

--- Approach 1: Brute Force Linear Drop Scan ---
Array [4, 5, 6, 7, 0, 1, 2]  → Rotations: 4
Array [15, 18, 2, 3, 6, 12]  → Rotations: 2
Array [1, 2, 3, 4, 5]        → Rotations: 0
Array [2, 1]                  → Rotations: 1
All assertions passed successfully!

Step by Step State Trace

Tracing count_rotations_brute_force on nums = [15, 18, 2, 3, 6, 12] (length N = 6):

Index i = 1: nums[1] = 18, nums[0] = 15 → 18 > 15 (Ascending, continue)
Index i = 2: nums[2] = 2,  nums[1] = 18 → 2 < 18  (DROP DETECTED!)
             → Immediately return i = 2.
Total comparisons: 2. Rotation count = 2.

Complexity Analysis

• Time Complexity (TC):
  • Worst-Case: O(N): When the array is unrotated (e.g. [1, 2, 3, 4, 5]) or the minimum is at the very last index ([2, 3, 4, 5, 1]), the loop evaluates N – 1 comparisons.
  • Best-Case: O(1): When rotated by 1 (e.g. [5, 1, 2, 3, 4]), the drop is found on the very first comparison at i = 1.
• Space Complexity (SC): O(1): Requires only a single scalar loop index variable.
• Algorithmic Bottleneck: Inspecting elements one by one fails to exploit the global sorting invariants of the two halves. For large datasets (e.g. N = 10,000,000), a linear scan requires millions of CPU instructions and cache line transfers.

Try it yourself

• What happens if you run this brute force scan on an empty vector?
• Can we use C++ STL std::is_sorted_until to write this in one line? (Hint: distance(nums.begin(), is_sorted_until(nums.begin(), nums.end())) % nums.size()).

🔹 Approach 2: Subarray Monotonicity Binary Search (Tracking Candidate Index)

Intuition & Reasoning: In any rotated sorted array, splitting the range [low, high] at midpoint mid guarantees that at least one half is strictly sorted. We can maintain a global minimum value and its corresponding index (min_val and min_idx):

• If the entire window nums[low] ≤ nums[high] is already sorted, the minimum of this window is at low. We update min_idx = low and terminate early.
• If the left half is sorted (nums[low] ≤ nums[mid]): The smallest element in the left half is nums[low]. We compare nums[low] with min_val, update min_idx, and eliminate the left half by moving low = mid + 1 to search for smaller elements in the unsorted right half.
• If the right half is sorted (nums[mid] ≤ nums[high]): The smallest element in the right half is nums[mid]. We compare nums[mid] with min_val, update min_idx, and eliminate the right half by moving high = mid - 1.

#include <iostream>
#include <vector>
#include <climits>
#include <cassert>

using namespace std;

// Approach 2: Subarray Monotonicity Binary Search with Min Index Tracking
// Time Complexity: O(log N), Space Complexity: O(1)
int count_rotations_subarray_bs(const vector<int>& nums) {
    if (nums.empty()) return 0;
    
    int low = 0;
    int high = static_cast<int>(nums.size()) - 1;
    int min_val = INT_MAX;
    int min_index = 0;

    while (low <= high) {
        // Entire subarray is already strictly sorted
        if (nums[low] <= nums[high]) {
            if (nums[low] < min_val) {
                min_val = nums[low];
                min_index = low;
            }
            break;
        }

        int mid = low + (high - low) / 2;

        // Check if Left Half is sorted
        if (nums[low] <= nums[mid]) {
            if (nums[low] < min_val) {
                min_val = nums[low];
                min_index = low;
            }
            // Pivot must lie in the right unsorted half
            low = mid + 1;
        } else {
            // Right Half is sorted
            if (nums[mid] < min_val) {
                min_val = nums[mid];
                min_index = mid;
            }
            // Pivot must lie in the left unsorted half
            high = mid - 1;
        }
    }

    return min_index;
}

int main() {
    vector<int> a1 = {4, 5, 6, 7, 0, 1, 2};
    vector<int> a2 = {11, 12, 15, 18, 2, 5, 6, 8};
    vector<int> a3 = {1, 2, 3, 4, 5};

    cout << "--- Approach 2: Subarray Monotonicity Binary Search ---\n";
    cout << "Array [4, 5, 6, 7, 0, 1, 2]         → Rotations: " << count_rotations_subarray_bs(a1) << "\n";
    cout << "Array [11, 12, 15, 18, 2, 5, 6, 8]  → Rotations: " << count_rotations_subarray_bs(a2) << "\n";
    cout << "Array [1, 2, 3, 4, 5]               → Rotations: " << count_rotations_subarray_bs(a3) << "\n";

    assert(count_rotations_subarray_bs(a1) == 4);
    assert(count_rotations_subarray_bs(a2) == 4);
    assert(count_rotations_subarray_bs(a3) == 0);
    cout << "All assertions passed successfully!\n";
    return 0;
}

Output

--- Approach 2: Subarray Monotonicity Binary Search ---
Array [4, 5, 6, 7, 0, 1, 2]         → Rotations: 4
Array [11, 12, 15, 18, 2, 5, 6, 8]  → Rotations: 4
Array [1, 2, 3, 4, 5]               → Rotations: 0
All assertions passed successfully!

Step by Step State Trace

Tracing count_rotations_subarray_bs on nums = [11, 12, 15, 18, 2, 5, 6, 8] (length N = 8):

Initial: low = 0, high = 7, min_val = ∞, min_index = 0

Iteration 1:
  Window: [11, 12, 15, 18, 2, 5, 6, 8]
  nums[0] (11) <= nums[7] (8) is FALSE.
  mid = 0 + (7 - 0)/2 = 3 (nums[mid] = 18).
  nums[low] (11) <= nums[mid] (18) is TRUE → Left half [11..18] is sorted!
  nums[low] = 11 < min_val (∞) → min_val = 11, min_index = 0.
  low = mid + 1 = 4.

Iteration 2:
  Window: indices [4..7] → [2, 5, 6, 8]
  nums[low] (2) <= nums[high] (8) is TRUE → Window is completely sorted!
  nums[low] = 2 < min_val (11) → min_val = 2, min_index = 4.
  break loop.

Final Result: min_index = 4 (Rotations = 4).
Total iterations: 2!

Complexity Analysis

• Time Complexity (TC): O(log N): At each step, half of the search space is eliminated, requiring at most ⌈log₂ N⌉ iterations.
• Space Complexity (SC): O(1): Only four integer variables (low, high, min_val, min_index) are maintained.
• Inefficiency: Although asymptotically optimal, this approach requires tracking both min_val and min_index alongside multiple conditional branches (nums[low] ≤ nums[high] and nums[low] ≤ nums[mid]). We can simplify the decision logic to a single boundary comparison.

Try it yourself

• Why is the early exit if (nums[low] ≤ nums[high]) safe and correct?
• What happens if nums[low] == nums[mid] when all elements are distinct? Under what array sizes can this happen?

🔹 Approach 3: Optimal Modified Binary Search (Upper-Boundary Comparison)

Intuition & Structural Invariant: We can eliminate all extra tracking variables by comparing nums[mid] directly against the upper boundary nums[high]. Let us formalize the two invariant cases:

Case 1: nums[mid] > nums[high]
  [ ... Left Sorted Segment (Large Values) ... mid ... ]  [ ... Right Sorted Segment ... high ]
  Since nums[mid] is strictly larger than nums[high], mid MUST belong to the Left Portion.
  The pivot (minimum element) MUST lie strictly to the right of mid!
  → low = mid + 1;

Case 2: nums[mid] <= nums[high]
  [ ... Left Sorted Segment ... ]  [ ... mid ... Right Sorted Segment (Small Values) ... high ]
  Since nums[mid] is less than or equal to nums[high], mid belongs to the Right Portion.
  The minimum could be nums[mid] itself, or lie to its left.
  → high = mid; (Never mid - 1, to avoid discarding the minimum!)

The loop invariant terminates when low == high, converging precisely on the index of the minimum element.

#include <iostream>
#include <vector>
#include <cassert>

using namespace std;

// Approach 3: Optimal Modified Binary Search (Direct Upper-Boundary Invariant)
// Time Complexity: O(log N), Space Complexity: O(1)
int count_rotations_optimal(const vector<int>& nums) {
    if (nums.empty()) return 0;

    int low = 0;
    int high = static_cast<int>(nums.size()) - 1;

    // Invariant: The minimum element is always within [low, high]
    while (low < high) {
        int mid = low + (high - low) / 2;

        if (nums[mid] > nums[high]) {
            // Midpoint is in the left (larger) portion; pivot is strictly to the right
            low = mid + 1;
        } else {
            // Midpoint is in the right (smaller) portion; pivot is at mid or to the left
            high = mid;
        }
    }

    // When low == high, both pointers have converged on the minimum element's index
    return low;
}

int main() {
    vector<int> a1 = {4, 5, 6, 7, 0, 1, 2};
    vector<int> a2 = {11, 12, 15, 18, 2, 5, 6, 8};
    vector<int> a3 = {1, 2, 3, 4, 5};
    vector<int> a4 = {2, 3, 4, 5, 1};

    cout << "--- Approach 3: Optimal Modified Binary Search ---\n";
    cout << "Array [4, 5, 6, 7, 0, 1, 2]         → Rotations: " << count_rotations_optimal(a1) << "\n";
    cout << "Array [11, 12, 15, 18, 2, 5, 6, 8]  → Rotations: " << count_rotations_optimal(a2) << "\n";
    cout << "Array [1, 2, 3, 4, 5]               → Rotations: " << count_rotations_optimal(a3) << "\n";
    cout << "Array [2, 3, 4, 5, 1]               → Rotations: " << count_rotations_optimal(a4) << "\n";

    assert(count_rotations_optimal(a1) == 4);
    assert(count_rotations_optimal(a2) == 4);
    assert(count_rotations_optimal(a3) == 0);
    assert(count_rotations_optimal(a4) == 4);
    cout << "All assertions passed successfully!\n";
    return 0;
}

Output

--- Approach 3: Optimal Modified Binary Search ---
Array [4, 5, 6, 7, 0, 1, 2]         → Rotations: 4
Array [11, 12, 15, 18, 2, 5, 6, 8]  → Rotations: 4
Array [1, 2, 3, 4, 5]               → Rotations: 0
Array [2, 3, 4, 5, 1]               → Rotations: 4
All assertions passed successfully!

Step by Step State Transition Trace

Tracing count_rotations_optimal on nums = [4, 5, 6, 7, 0, 1, 2] (length N = 7):

Iter 1: low = 0, high = 6
        mid = 0 + (6 - 0) / 2 = 3 → nums[mid] = 7, nums[high] = 2
        Condition: nums[mid] > nums[high] (7 > 2) is TRUE
        Action: low = mid + 1 = 4. Window shrinks to [4..6].

Iter 2: low = 4, high = 6
        mid = 4 + (6 - 4) / 2 = 5 → nums[mid] = 1, nums[high] = 2
        Condition: nums[mid] > nums[high] (1 > 2) is FALSE
        Action: high = mid = 5. Window shrinks to [4..5].

Iter 3: low = 4, high = 5
        mid = 4 + (5 - 4) / 2 = 4 → nums[mid] = 0, nums[high] = 1
        Condition: nums[mid] > nums[high] (0 > 1) is FALSE
        Action: high = mid = 4. Window shrinks to [4..4].

Termination: low == high (4 == 4).
Result: Returns index 4 (Value 0). Rotation count = 4!

Complexity Analysis & Correctness Proof

• Time Complexity (TC): O(log N): In each step, the search window [low, high] is strictly halved. The number of iterations is exactly ⌊log₂ N⌋.
• Space Complexity (SC): O(1): Requires exactly two pointer variables (low and high) allocated on the stack.
• Loop Invariant Proof:
  • Initialization: The minimum element is initially guaranteed to lie in the range [0, N - 1].
  • Maintenance: If nums[mid] > nums[high], the pivot cannot be at or before mid, so discarding [low..mid] preserves the minimum in [mid + 1, high]. If nums[mid] ≤ nums[high], the pivot cannot be strictly after mid, so setting high = mid preserves the minimum in [low..mid].
  • Termination: Since each iteration strictly reduces the interval length (high - low) and never discards the minimum, the loop terminates when low == high at the exact pivot index.

Try it yourself

• Why does the loop condition use low < high instead of low ≤ high? What happens if you use ≤ with high = mid? (Hint: Infinite loop when low == high!).
• Test the code on an array of length 2: [2, 1] and [1, 2]. Verify how the pointers shrink.

🔹 Advanced Systems Topic: Duplicate Elements & Branch Prediction

In production systems and follow-up interview questions (e.g. LeetCode 154), arrays may contain duplicate values (e.g., [2, 2, 2, 0, 2] or [3, 3, 1, 3, 3, 3]).

1. Handling Duplicates with Boundary Trimming

When duplicates exist, we encounter a third condition: nums[mid] == nums[high]. In this case, mid and high hold identical values, making it impossible to determine whether the pivot is in the left or right half with a single comparison:

Example A: [3, 3, 1, 3, 3] → mid = 2 (val 1 != 3), but if mid = 1 (val 3 == high val 3), pivot is to the RIGHT.
Example B: [3, 1, 3, 3, 3] → mid val 3 == high val 3, pivot is to the LEFT.

Solution: Safely decrement --high. Because nums[mid] == nums[high], discarding high does not lose the minimum value (as nums[mid] still preserves it). This maintains correctness while gracefully degrading worst-case time complexity to O(N) for pathological inputs like [1, 1, 1, 1, 1].

// Duplicate-Handling Rotation Count (LeetCode 154 Variant)
// Average: O(log N), Worst-Case: O(N)
int count_rotations_with_duplicates(const vector<int>& nums) {
    if (nums.empty()) return 0;
    int low = 0, high = static_cast<int>(nums.size()) - 1;

    while (low < high) {
        int mid = low + (high - low) / 2;
        if (nums[mid] > nums[high]) {
            low = mid + 1;
        } else if (nums[mid] < nums[high]) {
            high = mid;
        } else {
            // Indeterminate state: decrement upper boundary safely
            --high;
        }
    }
    return low;
}

2. CPU Branch Prediction & Cache Behavior

In the optimal binary search loop, the condition nums[mid] > nums[high] jumps irregularly across disparate memory pages during the first few iterations (stride N/2, N/4). This induces L1/L2 cache misses on cold data. However, because each iteration performs only one branch evaluation and logarithmic steps (at most 24 memory reads for 10M elements), the total latency is dominated by DRAM random access (tens of nanoseconds) rather than CPU arithmetic.

🔹 Lab: High Performance Benchmark Across 10,000,000 Elements

The following benchmark executable compares all 3 approaches across 1,000,000 repeated rotation queries on an array of 10,000,000 integers using std::chrono::high_resolution_clock:

#include <iostream>
#include <vector>
#include <chrono>

using namespace std;
using namespace std::chrono;

int count_rotations_brute(const vector<int>& nums) {
    for (size_t i = 1; i < nums.size(); ++i) {
        if (nums[i] < nums[i - 1]) return static_cast<int>(i);
    }
    return 0;
}

int count_rotations_opt(const vector<int>& nums) {
    int low = 0, high = static_cast<int>(nums.size()) - 1;
    while (low < high) {
        int mid = low + (high - low) / 2;
        if (nums[mid] > nums[high]) low = mid + 1;
        else high = mid;
    }
    return low;
}

int main() {
    const size_t N = 10'000'000;
    const size_t ROTATIONS = 3'750'000;
    cout << "Generating 10,000,000-element rotated array (k = " << ROTATIONS << ")...\n";

    vector<int> data(N);
    for (size_t i = 0; i < N; ++i) {
        data[i] = static_cast<int>((i + N - ROTATIONS) % N);
    }

    // Benchmark 1: Single full pass of Brute Force Linear Scan
    auto t0 = high_resolution_clock::now();
    int r1 = count_rotations_brute(data);
    auto t1 = high_resolution_clock::now();
    auto ms_brute = duration_cast<milliseconds>(t1 - t0).count();
    cout << "Brute Force (1 query on 10M elements):    " << ms_brute << " ms (Result: " << r1 << ")\n";

    // Benchmark 2: 1,000,000 queries of Optimal Binary Search
    volatile int sink = 0;
    auto t2 = high_resolution_clock::now();
    for (int q = 0; q < 1'000'000; ++q) {
        sink += count_rotations_opt(data);
    }
    auto t3 = high_resolution_clock::now();
    auto ms_opt = duration_cast<milliseconds>(t3 - t2).count();
    cout << "Optimal BS (1,000,000 queries on 10M):     " << ms_opt << " ms (Avg per query: " 
         << (ms_opt * 1000.0 / 1'000'000.0) << " µs)\n";

    return 0;
}

Benchmark Results

Generating 10,000,000-element rotated array (k = 3750000)...
Brute Force (1 query on 10M elements):    12 ms (Result: 3750000)
Optimal BS (1,000,000 queries on 10M):     18 ms (Avg per query: 0.018 µs)

🔹 Comparison Matrix of All Approaches

┌──────────────────────────────────────┬──────────────────┬──────────────────┬─────────────────────────────┬───────────────────────────┐
│ Approach                             │ Time Complexity  │ Space Complexity │ Invariant Comparison        │ Production Readiness      │
├──────────────────────────────────────┼──────────────────┼──────────────────┼─────────────────────────────┼───────────────────────────┤
│ 1. Brute Force (Linear Scan)         │ O(N)             │ O(1)             │ nums[i] < nums[i - 1]       │ Toy code / small arrays   │
│ 2. Subarray Monotonicity BS          │ O(log N)         │ O(1)             │ nums[low] <= nums[mid]      │ Verbose tracking logic    │
│ 3. Optimal Modified BS (mid vs high) │ O(log N)         │ O(1)             │ nums[mid] vs nums[high]     │ Industry gold standard    │
│ 4. Duplicate-Safe Trimming BS        │ O(log N) avg     │ O(1)             │ --high on equality          │ Required for LeetCode 154 │
└──────────────────────────────────────┴──────────────────┴──────────────────┴─────────────────────────────┴───────────────────────────┘

🔹 Frequently Asked Questions (FAQs)

Q1: Why is the rotation count strictly equal to the index of the minimum element?
In any 0-indexed sorted array without duplicates, the minimum element starts at index 0. A single rightward rotation moves every element one slot to the right (the element at index i shifts to (i + 1) % N). Rotating k times shifts the original index 0 element to index k. Therefore, finding where the minimum element landed directly yields the rotation count k.

Q2: What is the rotation count if the array is rotated N times?
Rotating an array of size N exactly N times returns the array to its original, completely unrotated configuration. The minimum element resides at index 0, so the rotation count is 0 (or N ≡ 0 mod N).

Q3: How do we calculate counter-clockwise (left) rotations instead of right rotations?
If an array of size N has minimum element index min_idx, the number of left rotations is given by k_left = (N - min_idx) % N. For example, in [4, 5, 6, 1, 2, 3] with N = 6 and min_idx = 3, right rotations = 3 and left rotations = (6 - 3) % 6 = 3.

Q4: Why does comparing nums[mid] with nums[low] fail on unrotated arrays?
In an unrotated array like [1, 2, 3, 4, 5], nums[mid] = 3 > nums[low] = 1. If our algorithm assumes that nums[mid] > nums[low] implies the minimum lies in the right half, it would discard index 0 (the true minimum) and search to the right, yielding an incorrect answer. Comparing against nums[high] is completely unambiguous across both rotated and unrotated states.

Q5: Why do we assign high = mid instead of high = mid - 1 in the optimal binary search?
When nums[mid] ≤ nums[high], nums[mid] itself might be the minimum element. If we set high = mid - 1, we would discard the minimum element from our search range. Setting high = mid safely retains the candidate within the active window [low, high].

Q6: Can we use std::min_element in production C++ code?
std::min_element(nums.begin(), nums.end()) computes the correct iterator in O(N) time. For small vectors (e.g. N ≤ 64), std::min_element is often faster due to SIMD vectorization and zero branch mispredictions. However, for large vectors or strict algorithmic constraints, binary search is mathematically superior.

Q7: How does duplicate handling degrade worst-case performance?
When all elements are identical except one (e.g., [1, 1, 1, 0, 1]), nums[mid] == nums[high] forces a linear decrement --high at every step. This reduces the search window by only 1 element per iteration, leading to O(N) comparisons in the worst case.

Q8: What is the risk of integer overflow when computing mid?
Writing (low + high) / 2 can overflow 32-bit signed integers when low + high > 2,147,483,647. Always write low + (high - low) / 2 or use C++20 std::midpoint(low, high) to ensure complete numerical safety.

🔹 Wrapping Up

Determining the Number of Times an Array Has Been Rotated in C++ is a core algorithmic competency that unlocks deeper mastery of binary search boundary invariants. By translating the physical rotation count into the index of the piecewise inflection minimum, we achieve an optimal O(log N) time and O(1) auxiliary space solution. To extend your problem-solving toolkit, be sure to study our companion guides on Find Minimum in Rotated Sorted Array, Single Element in a Sorted Array, and Search in Rotated Sorted Array.