Problem Description

You are given an array nums of non-negative integers and an integer k.

An array is called special if the bitwise OR of all of its elements is at least k.

Return the length of the shortest special non-empty subarray of nums, or return -1 if no special subarray exists.

Intuition

To solve this problem, the two pointers technique offers an efficient way to find the shortest special subarray. The idea is to maintain a window, represented by two pointers i and j, that slides over the array. Here's the detailed thought process:

1. Understanding the 'Bitwise OR': The OR operation on binary numbers results in a new number where each bit is the OR of the corresponding bits of the operands. Once a bit in the result has been set to 1, it cannot revert to 0 by adding more numbers. Hence, when moving the right endpoint j to the right, the bitwise OR value never decreases.

2. Initialize Variables: We start with both pointers i (left) and j (right) at the beginning of nums. We use a variable s to track the current bitwise OR of the subarray defined by the two pointers. Additionally, a cnt array of size 32 keeps track of how many numbers in the current subarray contribute to each bit position, allowing us to efficiently adjust s when the left pointer moves.

3. Expanding the Right Pointer: Increment j and update s to include nums[j]. This involves setting bits in s based on the bits in nums[j].

4. Contracting the Left Pointer: Once s becomes greater than or equal to k, attempt to shrink the window by moving the left pointer i to the right. During this, adjust s by removing the effect of nums[i], decrementing the corresponding bit counts in cnt. If a bit count drops to zero, it indicates that the bit must be cleared from s.

5. Recording the Shortest Length: Each time the condition s >= k is met, compare the current window size (j - i + 1) with the previously recorded shortest length and update it if a new shorter length is found.

6. Result Handling: After iterating through the array, if no subarray satisfied the condition, return -1; otherwise, return the recorded shortest length.

This approach efficiently finds the shortest subarray by leveraging the non-decreasing nature of the OR operation and limiting unnecessary operations to the smallest necessary subarrays.

Learn more about Sliding Window patterns.

Solution Approach

The solution employs a Two Pointers technique combined with Counting to efficiently find the shortest special subarray. Here’s a detailed breakdown of the approach:

1. Initialization:

   • Let n be the length of nums.
   • Create an array cnt of size 32 to count the number of active bits in the current window.
   • Initialize ans to n + 1 as a placeholder for the minimum length result. This value ensures that if no valid subarray is found, it can be identified by the comparison to n.
   • Use variables s and i to track the current bitwise OR of the subarray and the left pointer, respectively.

2. Iterating with the Right Pointer:

   • Use a for loop with j iterating over each element x in nums.
   • Update s with the bitwise OR of x.
   • For each bit in x that is set, increment the corresponding count in cnt.

3. Adjusting the Left Pointer:

   • If s meets or exceeds k, attempt to shrink the window by incrementing i.
   • Inside this inner while loop, update ans to the minimum of its current value and the window size j - i + 1.
   • For the element nums[i] being removed, decrement its corresponding bits in cnt. If a bit count reaches zero, remove that bit from s using the XOR operation.
   • Increment i to continue shrinking the subarray.

4. Determine the Result:

   • After the loop completes, return -1 if ans is still greater than n, indicating no valid subarray was found. Otherwise, return ans, which represents the shortest length of the special subarray.

The algorithm leverages the properties of bitwise operations to keep the time complexity efficient and uses additional space for counting bits but remains optimal for practical input sizes.

Example Walkthrough

Let's use an example to illustrate the solution approach. Suppose nums = [1, 2, 4, 6] and k = 7.

1. Initialization:

   • Length of nums, n = 4.
   • cnt initialized as an array of size 32, all elements set to 0.
   • Set ans = n + 1 = 5, s = 0, and left pointer i = 0.

2. Iterating with the Right Pointer:

   • Start with j = 0, nums[j] = 1. Update s with s = s | 1 = 1. Increment bit counts in cnt for bits set in 1.
   • Move to j = 1, nums[j] = 2. Update s with s = s | 2 = 3. Update cnt for bits set in 2.
   • Move to j = 2, nums[j] = 4. Update s with s = s | 4 = 7. Update cnt for bits set in 4.

3. Adjusting the Left Pointer:

   • Now s = 7, which meets k = 7. Enter the inner loop to try and minimize the subarray:
     • Current window size j - i + 1 = 3. Update ans = min(5, 3) = 3.
     • Update s by removing nums[i] = 1: s = s ^ 1 = 6. Decrement bit counts in cnt, and increment i to 1.
   • Now s = 6 is less than k, exit the inner loop.

4. Continue with the Right Pointer:

   • Move to j = 3, nums[j] = 6. Update s with s = s | 6 = 6.
   • Since s < 7, do not enter the inner loop.

5. Determine the Result:

   • The traversal ends. The shortest special subarray length found is ans = 3. Thus, return 3, which is the length of the subarray [2, 4, 6].

In this example, the method successfully identifies and returns the length of the shortest subarray [2, 4, 6] that meets the OR requirement of being at least k = 7.

Solution Implementation

Time and Space Complexity

• Time Complexity: The time complexity is O(n \times 32), where n is the length of the nums array. Each element might need up to 32 operations (bit checks and modifications), corresponding to the number of bits in an integer.

• Space Complexity: The space complexity is O(32), which comes from the cnt list used to store counts for up to 32 bits.

Learn more about how to find time and space complexity quickly.