#3171

Hard

auto_awesomeBinary search over the valid answer space

LeetCode Problem Workspace

Find Subarray With Bitwise OR Closest to K

Find a subarray with bitwise OR closest to a given value k with minimal absolute difference.

array binary search bit manipulation segment tree

Problem Statement

Given an array of integers nums and a target value k, the goal is to identify a contiguous subarray of nums such that the absolute difference between k and the bitwise OR of all its elements is minimized. You need to select the subarray nums[l..r], where the bitwise OR of the elements within this range produces the smallest possible absolute difference from k.

The challenge is to efficiently compute this minimal absolute difference, considering the array size can be as large as 100,000 elements. The problem is rooted in bit manipulation and binary search, with dynamic programming providing an approach to solve it more effectively.

Examples

Example 1

Input: nums = [1,2,4,5], k = 3

Output: 0

The subarray nums[0..1] has OR value 3, which gives the minimum absolute difference |3 - 3| = 0 .

Example 2

Input: nums = [1,3,1,3], k = 2

Output: 1

The subarray nums[1..1] has OR value 3, which gives the minimum absolute difference |3 - 2| = 1 .

Example 3

Input: nums = [1], k = 10

Output: 9

There is a single subarray with OR value 1, which gives the minimum absolute difference |10 - 1| = 9 .

Constraints

1 <= nums.length <= 105
1 <= nums[i] <= 109
1 <= k <= 109

Solution Approach

Binary Search Over Valid Answer Space

This problem can be tackled by binary search over the possible bitwise OR values. By narrowing down the space of possible solutions, we reduce the time complexity significantly. We evaluate subarrays dynamically and apply the binary search for the optimal result.

Dynamic Programming with Bitwise OR

Let dp[i] represent the set of bitwise OR values for all subarrays ending at index i. This allows efficient tracking of all possible OR values, helping to identify the subarray with the minimal absolute difference from k.

Efficient Range Queries

Efficiently querying the bitwise OR for ranges can be achieved using segment trees or similar data structures. This optimization speeds up the process of computing the OR value for subarrays, leading to a faster solution.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity is mainly driven by the binary search over the possible OR values, combined with dynamic programming or segment tree queries, making the solution much faster than brute force approaches. Space complexity depends on the data structures used for storing OR values and dynamic programming states.

What Interviewers Usually Probe

The candidate uses binary search effectively to narrow down the possible range of OR values.
They demonstrate a solid understanding of bitwise operations and their implications for array manipulation.
The candidate proposes an efficient way to track the OR values for multiple subarrays, possibly with dynamic programming or a segment tree.

Common Pitfalls or Variants

Common pitfalls

Not considering the need for efficient range queries can lead to suboptimal performance.
Failing to recognize the potential optimization of binary search over the OR values might result in slower solutions.
Misunderstanding the problem's constraints may lead to an approach that doesn't scale with larger arrays.

Follow-up variants

Modify the problem to find the subarray that minimizes the sum of bitwise OR values across subarrays.
Introduce multiple target values for k, requiring the solution to handle multiple queries in the same problem.
Optimize the solution further by reducing the space complexity, potentially eliminating the need for dynamic programming.

FAQ

How does binary search apply to the problem of finding a subarray with the bitwise OR closest to k?

Binary search can be used to narrow down the range of possible OR values, optimizing the search for the subarray that minimizes the absolute difference from k.

What is the significance of dynamic programming in this problem?

Dynamic programming helps track all bitwise OR values of subarrays efficiently, which is essential for finding the minimal absolute difference from k in large arrays.

What are the main challenges in solving the Find Subarray With Bitwise OR Closest to K problem?

The main challenges include efficiently computing the bitwise OR for subarrays and managing large input sizes with optimal time complexity.

What optimizations can be applied to improve the performance of this problem?

Using binary search over the possible OR values and efficient range queries with dynamic programming or segment trees can significantly improve performance.

How do you handle the large constraints in the Find Subarray With Bitwise OR Closest to K problem?

Handling the large constraints requires optimizing the solution using binary search, dynamic programming, or segment trees to ensure the solution scales well with the input size.

terminal

Solution

Solution 1: Two Pointers + Bitwise Operations

According to the problem description, we need to calculate the result of the bitwise OR operation of elements from index $l$ to $r$ in the array $\textit{nums}$, that is, $\textit{nums}[l] \lor \textit{nums}[l + 1] \lor \cdots \lor \textit{nums}[r]$, where $\lor$ represents the bitwise OR operation.

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

21

22

class Solution:
    def minimumDifference(self, nums: List[int], k: int) -> int:
        m = max(nums).bit_length()
        cnt = [0] * m
        s = i = 0
        ans = inf
        for j, x in enumerate(nums):
            s |= x
            ans = min(ans, abs(s - k))
            for h in range(m):
                if x >> h & 1:
                    cnt[h] += 1
            while i < j and s > k:
                y = nums[i]
                for h in range(m):
                    if y >> h & 1:
                        cnt[h] -= 1
                        if cnt[h] == 0:
                            s ^= 1 << h
                i += 1
                ans = min(ans, abs(s - k))
        return ans

Solution 2: Hash Table + Enumeration

According to the problem description, we need to calculate the result of the bitwise OR operation of elements from index $l$ to $r$ in the array $nums$, that is, $nums[l] \lor nums[l + 1] \lor \cdots \lor nums[r]$. Here, $\lor$ represents the bitwise OR operation.

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

21

22

class Solution:
    def minimumDifference(self, nums: List[int], k: int) -> int:
        m = max(nums).bit_length()
        cnt = [0] * m
        s = i = 0
        ans = inf
        for j, x in enumerate(nums):
            s |= x
            ans = min(ans, abs(s - k))
            for h in range(m):
                if x >> h & 1:
                    cnt[h] += 1
            while i < j and s > k:
                y = nums[i]
                for h in range(m):
                    if y >> h & 1:
                        cnt[h] -= 1
                        if cnt[h] == 0:
                            s ^= 1 << h
                i += 1
                ans = min(ans, abs(s - k))
        return ans

Continue Practicing

#3209 number of subarrays with and value of k #3117 minimum sum of values by dividing array #3161 block placement queries