Given an array of non-negative integers, we aim to find the smallest subarray where the bitwise OR of all elements is at least a given integer k. This problem involves bit manipulation and sliding window techniques, which are essential for efficient solutions!

Let’s dive in to understand the problem, explore a basic solution, optimize it, and walk through examples with emojis along the way! 🚀

📝 Problem Statement

We are given:

1. An array nums of non-negative integers.
2. An integer k.

A subarray is called special if the bitwise OR of all elements in the subarray is at least k. We want to find the length of the shortest special subarray. If no such subarray exists, return -1.

Constraints

• \[1 \leq \text{nums.length} \leq 200,000\]
• \[0 \leq \text{nums}[i] \leq 10^9\]
• \[0 \leq k \leq 10^9\]

🔍 Examples

Example 1:

• Input: nums = [1, 2, 3], k = 2
• Output: 1
• Explanation: The subarray [3] has a bitwise OR of 3, which is greater than 2, so we return 1 as the shortest length.

Example 2:

• Input: nums = [2, 1, 8], k = 10
• Output: 3
• Explanation: The subarray [2, 1, 8] has a bitwise OR of 11, meeting the condition with a length of 3.

Example 3:

• Input: nums = [1, 2], k = 0
• Output: 1
• Explanation: The subarray [1] has an OR of 1, which meets the requirement since 1 >= 0.

🧠 Solution Approach

To solve this problem, we can use a sliding window approach paired with bitwise operations.

🚀 Basic Solution and Intuition

Key Observations:

1. Bitwise OR: The OR operation between numbers keeps the bits set in both numbers, but also sets any bits that are 1 in either number. This property means that once a bit is set, it will remain set, making the OR value non-decreasing within a window.
2. Sliding Window: We can attempt to keep a growing window and shrink it from the left once we meet or exceed the OR requirement. This helps find the minimum subarray length satisfying the condition.

Steps:

1. Expand Window: Start expanding the window by including elements to the OR operation.
2. Shrink Window: As soon as the OR of the window is greater than or equal to k, try to shrink it from the left while still meeting the requirement.
3. Track Minimum Length: Each time the condition is met, update the minimum length.

🔍 Bitwise Walkthrough Example: nums = [2, 1, 8], k = 10

Initial Setup

• Goal: Find the shortest subarray where the OR of elements is at least 10 (binary 1010).
• Binary for k = 10: 1010

Let’s keep a bitwise OR tracker s (initially 0) and update it as we expand the window. Additionally, we have an array cnt of size 32 (to track each bit position) initialized to 0.

Step-by-Step Execution

1. First Element: 2
   • Binary of 2: 0010
   • Update s by OR-ing it with 2: s = s | 2 = 0 | 2 = 2 (binary 0010)
   • Update cnt:
     • For 2, only the 1st bit (from the right, 0010) is set, so increment cnt[1] by 1.
     • Updated cnt: [0, 1, 0, 0, ..., 0]
   • Current OR of the subarray [2]: 2 (binary 0010)
   • Check: 2 < 10 (binary 0010 < 1010), so the subarray doesn’t meet the requirement. Continue expanding.
2. Second Element: 1
   • Binary of 1: 0001
   • Update s by OR-ing it with 1: s = s | 1 = 2 | 1 = 3 (binary 0011)
   • Update cnt:
     • For 1, only the 0th bit (from the right, 0001) is set, so increment cnt[0] by 1.
     • Updated cnt: [1, 1, 0, 0, ..., 0]
   • Current OR of the subarray [2, 1]: 3 (binary 0011)
   • Check: 3 < 10 (binary 0011 < 1010), so the subarray doesn’t meet the requirement. Continue expanding.
3. Third Element: 8
   • Binary of 8: 1000
   • Update s by OR-ing it with 8: s = s | 8 = 3 | 8 = 11 (binary 1011)
   • Update `cnt:
     • For 8, only the 3rd bit (from the right, 1000) is set, so increment cnt[3] by 1.
     • Updated cnt: [1, 1, 0, 1, ..., 0]
   • Current OR of the subarray [2, 1, 8]: 11 (binary 1011)

   • Check: 11 >= 10 (binary 1011 >= 1010), so this subarray meets the requirement!

   • Update Result: The length of [2, 1, 8] is 3, so set ans = 3.
4. Shrink the Window to Minimize Length
   • We have found a valid subarray [2, 1, 8]. Now, we try to minimize the length by moving the start pointer i.

   Remove nums[i] = 2:

   • Adjust s by removing the effect of 2 (binary 0010).
   • Decrement cnt[1] by 1 (because 2 contributes to the 1st bit).
   • Since cnt[1] becomes 0, we remove the 1st bit from s using XOR: s = s ^ (1 << 1).

   • Updated s: 9 (binary 1001)

   • Check: s = 9 < 10, so the subarray [1, 8] doesn’t meet the requirement anymore. Stop shrinking.

   Final result for this example is 3, which is the length of the shortest subarray [2, 1, 8] with OR at least 10.

🧩 Additional Example with Bits: nums = [5, 1, 4, 2, 8], k = 7

Initial Setup

• Goal: Find the shortest subarray where the OR is at least 7 (binary 0111).
• Binary for k = 7: 0111

Step-by-Step Execution

1. First Element: 5
   • Binary of 5: 0101
   • Update s = s | 5 = 0 | 5 = 5 (binary 0101)
   • Update cnt:
     • Set bits are in the 0th and 2nd positions, so increment cnt[0] and cnt[2].
     • cnt: [1, 0, 1, 0, ..., 0]
   • Current OR of the subarray [5]: 5 (binary 0101)
   • Check: 5 < 7 (binary 0101 < 0111), continue expanding.
2. Second Element: 1
   • Binary of 1: 0001
   • Update s = s | 1 = 5 | 1 = 5 (binary 0101)
   • Update `cnt:
     • Only the 0th bit is set for 1, so increment cnt[0].
     • cnt: [2, 0, 1, 0, ..., 0]
   • Current OR of [5, 1]: 5 (binary 0101)
   • Check: 5 < 7, continue expanding.
3. Third Element: 4
   • Binary of 4: 0100
   • Update s = s | 4 = 5 | 4 = 5 (binary 0101)
   • Update `cnt:
     • Only the 2nd bit is set for 4, so increment cnt[2].
     • cnt: [2, 0, 2, 0, ..., 0]
   • Current OR of [5, 1, 4]: 5 (binary 0101)
   • Check: 5 < 7, continue expanding.
4. Fourth Element: 2
   • Binary of 2: 0010
   • Update s = s | 2 = 5 | 2 = 7 (binary 0111)
   • Update `cnt:
     • Only the 1st bit is set for 2, so increment cnt[1].
     • cnt: [2, 1, 2, 0, ..., 0]
   • Current OR of [5, 1, 4, 2]: 7 (binary 0111)

   • Check: 7 >= 7 (binary 0111 >= 0111), so we found a valid subarray [5, 1, 4, 2] with OR 7.

   • Update Result: Length of [5, 1, 4, 2] is 4, so set ans = 4.
5. Shrink the Window to Minimize Length
   • Try removing nums[i] = 5 from the OR calculation.

   Remove nums[i] = 5:

   • Update cnt for 5 by decrementing cnt[0] and cnt[2].

   • Since cnt[0] and cnt[2] remain non-zero, s remains 7.

   • Update Result: The new subarray [1, 4, 2] also has an OR of 7 and a length of 3. Update ans = 3.

Final result for this example is 3, the length of [1, 4, 2], the shortest subarray with OR at least 7.

✨ Optimized Solution

The provided solution leverages a sliding window with bitwise tracking. We maintain a bit counter to track set bits within the window and modify the OR result dynamically. This allows efficient updates when expanding or shrinking the window.

Here’s the solution code:

def shortest_subarray_with_or_at_least_k(nums, k):
    n = len(nums)
    cnt = [0] * 32  # Track the count of set bits
    ans = n + 1     # Initialize with a large value
    s = i = 0       # OR value and window start pointer
    
    for j, x in enumerate(nums):
        # Include x in the OR result
        s |= x
        # Update count of set bits
        for h in range(32):
            if x >> h & 1:
                cnt[h] += 1
        
        # Shrink the window while OR is at least k
        while s >= k and i <= j:
            ans = min(ans, j - i + 1)
            y = nums[i]
            for h in range(32):
                if y >> h & 1:
                    cnt[h] -= 1
                    if cnt[h] == 0:
                        s ^= 1 << h
            i += 1
    
    return -1 if ans > n else ans

⏱️ Time Complexity

This solution has a time complexity of O(n) because each element is processed once for expansion and once for contraction in the sliding window. This efficiency is essential for large inputs!

🧐 Example Walkthrough (Detailed)

Let’s go through Example 2 in detail to understand the process step-by-step.

Input:

• nums = [2, 1, 8], k = 10

Step-by-Step Execution:

1. Initialize:
   • s = 0 (current OR)
   • cnt = [0] * 32 (bit count)
   • ans = len(nums) + 1 = 4
2. First Element (2):
   • s |= 2 → s = 2 (OR of 2 itself)
   • Update cnt for the bits of 2 (binary 10):
     • cnt[1] = 1
   • s < k, so continue expanding.
3. Second Element (1):
   • s |= 1 → s = 3
   • Update cnt for the bits of 1 (binary 1):
     • cnt[0] = 1
   • s < k, so continue expanding.
4. Third Element (8):
   • s |= 8 → s = 11 (OR of 2, 1, 8)
   • Update cnt for the bits of 8 (binary 1000):
     • cnt[3] = 1
   • Now, s >= k, so we found a special subarray [2, 1, 8] with OR 11.
5. Shrink Window:
   • Update ans = 3 (length of [2, 1, 8])
   • Try removing 2 from OR:
     • Adjust cnt for 2, removing cnt[1]
     • s = 9, but s < k now, so stop shrinking.

🪖 Edge Cases

1. No Solution:
   • If k is very large, e.g., nums = [1, 2, 4] and k = 1000, there may be no subarray where OR reaches k. The function should return -1.
2. Single Element:
   • nums = [k], k itself: return 1 since [k] meets the condition directly.
3. All Elements are Zero:
   • nums = [0, 0, 0], k = 1: should return -1 because OR will always be zero.

🔚 Conclusion

This problem showcases the importance of bitwise operations and sliding window techniques in handling subarray conditions efficiently. The optimized solution ensures that we meet the problem constraints without excessive time complexity.

By following this approach, you’ll gain valuable insights into how bitwise operations and window techniques complement each other for powerful problem-solving!