2411. Smallest Subarrays With Maximum Bitwise OR

#php #leetcode #algorithms #programming

Difficulty: Medium

Topics: Array, Binary Search, Bit Manipulation, Sliding Window, Biweekly Contest 87

You are given a 0-indexed array nums of length n, consisting of non-negative integers. For each index i from 0 to n - 1, you must determine the size of the minimum sized non-empty subarray of nums starting at i (inclusive) that has the maximum possible bitwise OR.

In other words, let B_ij be the bitwise OR of the subarray nums[i...j]. You need to find the smallest subarray starting at i, such that bitwise OR of this subarray is equal to max(B_ik) where i <= k <= n - 1.

The bitwise OR of an array is the bitwise OR of all the numbers in it.

Return an integer array answer of size n where answer[i] is the length of the minimum sized subarray starting at i with maximum bitwise OR.

A subarray is a contiguous non-empty sequence of elements within an array.

Example 1:

Input: nums = [1,0,2,1,3]
Output: [3,3,2,2,1]
Explanation:
- The maximum possible bitwise OR starting at any index is 3.
- Starting at index 0, the shortest subarray that yields it is [1,0,2].
- Starting at index 1, the shortest subarray that yields the maximum bitwise OR is [0,2,1].
- Starting at index 2, the shortest subarray that yields the maximum bitwise OR is [2,1].
- Starting at index 3, the shortest subarray that yields the maximum bitwise OR is [1,3].
- Starting at index 4, the shortest subarray that yields the maximum bitwise OR is [3].
- Therefore, we return [3,3,2,2,1].

Example 2:

Input: nums = [1,2]
Output: [2,1]
Explanation:
- Starting at index 0, the shortest subarray that yields the maximum bitwise OR is of length 2.
- Starting at index 1, the shortest subarray that yields the maximum bitwise OR is of length 1.
- Therefore, we return [2,1].

Constraints:

n == nums.length
1 <= n <= 10⁵
0 <= nums[i] <= 10⁹

Hint:

Consider trying to solve the problem for each bit position separately.
For each bit position, find the position of the next number that has a 1 in that position, if any.
Take the maximum distance to such a number, including the current number.
Iterate backwards to achieve a linear complexity.

Solution:

We need to find, for each starting index i in the given array, the smallest subarray starting at i that achieves the maximum possible bitwise OR for any subarray starting at i. The bitwise OR of a subarray is the bitwise OR of all elements in that subarray. The challenge is to efficiently compute this for each index in the array.

Approach

Problem Analysis: The key observation is that the maximum bitwise OR for any subarray starting at index i is the bitwise OR of the entire subarray from i to the end of the array. This is because the bitwise OR operation is monotonic; adding more elements can only set more bits, never unset them. Therefore, the maximum OR value for subarrays starting at i is the OR of nums[i..n-1].
Intuition: For each starting index i, we need the smallest subarray nums[i..j] such that the bitwise OR of this subarray equals the OR of the entire suffix starting at i. To achieve this, we must ensure that for every bit set in the suffix OR, there is at least one element in the subarray nums[i..j] that has that bit set. The minimal j for which this condition is satisfied will be the rightmost position among the next occurrences of each bit set in the suffix OR.
Algorithm Selection:
- Backward Pass: We process the array from right to left. This allows us to efficiently track the next occurrence of each bit (0 to 31) as we move leftwards.
- Bit Tracking: For each index i, we update the next occurrence positions of all bits present in nums[i].
- Suffix OR Calculation: As we move left, we maintain the cumulative OR of the suffix starting at the current index.
- Finding Minimal Subarray: For each index i, we determine the smallest j such that all bits set in the suffix OR are covered by elements in nums[i..j]. This j is the maximum of the next occurrence positions of all bits set in the suffix OR.
Complexity Analysis: The algorithm processes each element once, and for each element, it checks up to 32 bits. Thus, the time complexity is O(32 * n) = O(n), which is efficient for the given constraints (n ≤ 10⁵).

Let's implement this solution in PHP: 2411. Smallest Subarrays With Maximum Bitwise OR

<?php
/**
 * @param Integer[] $nums
 * @return Integer[]
 */
function smallestSubarrays($nums) {
    ...
    ...
    ...
    /**
     * go to ./solution.php
     */
}

// Test cases
// Example 1:
$nums = [1, 0, 2, 1, 3];
print_r(smallestSubarrays($nums)); // Output: [3,3,2,2,1]

// Example 2:
$nums = [1, 2];
print_r(smallestSubarrays($nums)); // Output: [2,1]
?>

Explanation:

Initialization: We initialize an array ans to store results, an array next_occurrence to track the most recent (leftmost) positions of each bit (0-31), and a variable curOR to accumulate the bitwise OR of the suffix.
Backward Pass: Starting from the end of the array, for each element:
- Update Bit Positions: For each bit set in the current element, update next_occurrence to the current index.
- Update Suffix OR: Accumulate the OR of the current element into curOR.
- Determine Minimal Subarray: For each bit set in curOR, find the farthest (rightmost) next occurrence of that bit. The minimal subarray length is the distance from the current index to this farthest occurrence plus one.
Result Construction: Store the computed subarray length in ans for the current index.
Return Result: After processing all elements, return the ans array containing the minimal subarray lengths for each starting index.

This approach efficiently computes the desired result by leveraging bit manipulation and a backward pass through the array, ensuring optimal performance even for large input sizes.

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2411. Smallest Subarrays With Maximum Bitwise OR

Approach

Explanation:

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