Problem Description

In this problem, you're presented with an array nums of non-negative integers. Your task is to determine, for each index i from 0 to n - 1, the length of the smallest subarray starting at i that produces the maximum bitwise OR value compared to all possible subarrays starting at that same index i. We define the bitwise OR of an array as the result of applying the bitwise OR operation to all numbers in it.

To clarify with an example, if the array was [1,2,3], the subarrays starting from index 0 would be [1], [1,2], and [1,2,3]. If the maximum bitwise OR achievable is with subarray [1,2], and that value cannot be achieved with a smaller subarray from that index, then the length you should report for index 0 would be the length of [1,2] which is 2. You need to perform this computation efficiently for every index of the array.

Your final output should be an integer array answer, where answer[i] is the length of the smallest subarray starting at index i with the maximum bitwise OR value.

Intuition

The solution to this problem uses a bit-manipulation trick. For each index, we know that the maximum bitwise OR subarray must extend far enough to include any bits that wouldn't otherwise be set to 1 without it. What's less intuitive is figuring out how far that is without checking every subarray.

Bitwise OR operation has an 'accumulative' property; once a bit is set to 1, it remains 1 irrespective of the OR operations that follow. This leads us to a realization: we don't need to continue our subarray beyond the point where all bits that can be set to 1, are set.

The insight is to iterate backwards from the end of the array, keeping track of the most recent index where each bit was set to 1. By doing this, when we're considering a given index i, we can look at our collection of indices, and figure out the furthest one we need to reach - this gives us the length of the minimum subarray with the maximum bitwise OR.

The process involves a 32-length array (for each bit in the integer) where we keep updating our 'latest' indices as we move from the end to the start of the array. For each number, we consider every bit, update the latest index if the bit is 1, and find the distance to the farthest index we need to include if the bit is not set. Hence, we derive the minimum length for the maximum OR subarray starting at that index. The solution brilliantly captures the bit-level details and aggregates them into a subarray problem.

Learn more about Binary Search and Sliding Window patterns.

Solution Approach

The given Python solution first initializes two arrays: ans which will contain our result - the smallest subarray sizes, and f to keep track of the most recent index for each of the 32 bits (representing an integer in the array).

1. Iterate through each element in nums in reverse order (starting from the last element).
2. For the current element nums[i], initialize a temporary variable t with the value 1 as the smallest subarray size to consider is always at least 1 (the element itself).
3. Iterate through each of the 32 bits (0 to 31) to check which bits are set to 1 in the current element.
   • If the bit at position j in nums[i] is set to 1, update f[j] to be i. This step guarantees that f[j] always points to the latest index i where the j-th bit was set.
   • If the bit at position j is not set in nums[i] (the value is 0), and f[j] isn't -1 (indicating that we've seen the j-th bit set in some higher index), then we must extend our subarray to include this bit. Thus, t is updated to be the maximum of itself and f[j] - i + 1 which represents the length needed to include the j-th bit.
4. After checking all bits, ans[i] is updated with the value in t, which now represents the minimum subarray size starting at index i that maximizes the bitwise OR value.
5. Once the loop is finished, we return the ans array which now contains the desired subarray sizes for each index in the original nums array.

The solution employs bit manipulation and array indexing in an elegant way to avoid checking every possible subarray, reducing what would be a potentially quadratic-time algorithm to a linear-time solution since it iterates through the array and each bit position just once.

Example Walkthrough

Let's walk through a simple example to illustrate the solution approach. Imagine we have the array nums = [3, 11, 7]. The binary representations of the numbers in the array are 0011, 1011, and 0111 respectively.

1. Initialize ans array to hold the result and f array to keep track of the most recent indices where each bit is set to 1. Both arrays have a size of 3, the length of nums, and f is initialized with -1 to indicate that no indices have been visited yet:

   • ans = [0, 0, 0] (to be filled with results)
   • f = [-1, -1, -1, -1] (one entry for each bit)

2. Start iterating through nums in reverse order:

   • i=2 (Last element, nums[2] = 0111):
     • Initialize t = 1 (the smallest subarray always includes at least nums[i] itself).
     • Check each bit:
       • For bit 0: It is set, so update f[0] = 2.
       • For bit 1: It is set, so update f[1] = 2.
       • For bit 2: It is set, so update f[2] = 2.
       • For bit 3: It is not set, and f[3] is -1 (no previous 1's).
     • Update ans[2] with t = 1.

   At this point:

   • ans = [0, 0, 1]

   • f = [2, 2, 2, -1]

   • i=1 (Second element, nums[1] = 1011):

     • Reset t = 1.
     • Check each bit:
       • For bit 0: It is set, so update f[0] = 1.
       • For bit 1: It is set, so update f[1] = 1.
       • For bit 2: It is not set, but f[2] is 2, so update t to 2 - 1 + 1 = 2.
       • For bit 3: It is set, so update f[3] = 1.
     • Update ans[1] with t = 2.

   At this point:

   • ans = [0, 2, 1]

   • f = [1, 1, 2, 1]

   • i=0 (First element, nums[0] = 0011):

     • Reset t = 1.
     • Check each bit:
       • For bit 0: It is set, so update f[0] = 0.
       • For bit 1: It is set, so update f[1] = 0.
       • For bit 2: It is not set, and f[2] is 2, so update t to the maximum of t and (2 - 0 + 1) = 3.
       • For bit 3: It is not set, and f[3] is 1, so update t to the maximum of t and (1 - 0 + 1) = 2.
     • Update ans[0] with the maximum t = 3.

   At this point:

   • ans = [3, 2, 1]
   • f = [0, 0, 2, 1]

3. After iterating through all indices, ans now contains the correct answer. The final array ans = [3, 2, 1] represents the smallest subarray sizes starting at each index i that achieves the maximum bitwise OR value.

In this example, the smallest subarray starting at index 0 with the maximum bitwise OR is [3, 11, 7], at index 1 is [11, 7], and at index 2 is [7]. The algorithm efficiently identifies these subarrays without having to examine every possible subarray, showcasing the power of bit manipulation and dynamic programming.

Solution Implementation

Time and Space Complexity

The given Python code is designed to find the smallest subarrays that contain all of the bits that appear at least once to the right of each element, including the element itself.

Time Complexity

The time complexity of the given code is O(n * b), where n is the length of the nums array, and b is the number of bits used to represent the numbers. Since the input numbers are integers and Python's integers have a fixed number of bits (which is 32 bits for standard integers), the value of b can be considered a constant, thus simplifying the time complexity further to O(n).

This complexity is derived from the outer loop running n times (from n - 1 to 0) and the inner loop over b bits running a constant 32 times. The algorithms only perform a fixed set of operations within the inner loop, which does not depend on n.

Space Complexity

The space complexity of the code is O(b), since we are maintaining a fixed-size array f of size 32, corresponding to the 32 bits in an integer. Space is also used for the output list ans of size n, but when talking about space complexity we generally do not count the space required for output, or if we do, we state the complexity as additional space beyond the required output.

If one includes the space taken by the output, then it would be O(n). However, since the space for the output is typically not counted, and the auxiliary space used is only the fixed-size list f, the space complexity remains O(b), which reduces to O(1) since the number of bits is constant and does not grow with the input.

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