How do you solve Minimum Impossible OR on LeetCode?

Minimum Impossible OR (LeetCode #2568) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(1) space complexity. Key insight: The smallest impossible OR is often the first power of 2 not represented in the array.

What is the time complexity of Minimum Impossible OR?

The optimal solution for Minimum Impossible OR runs in O(n log n) time and uses O(1) space. The time complexity is O(n log n) due to the sorting step, while the space complexity is O(1) as we only use a few extra variables.

What is the brute force solution for Minimum Impossible OR?

The brute force approach for Minimum Impossible OR is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves generating all possible subsequences of the array and calculating their bitwise OR. We then check which positive integers can be formed and identify the smallest integer that cannot be formed. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Minimum Impossible OR?

Minimum Impossible OR uses the following concepts: Array, Bit Manipulation, Brainteaser. The recommended patterns to study are: Bit Manipulation, Sorting.

What are the interview tips for Minimum Impossible OR?

Always clarify the problem constraints and edge cases. Think about the properties of bitwise operations when dealing with OR. Practice breaking down the problem into smaller parts before coding.

What are common mistakes when solving Minimum Impossible OR?

Not considering the case where the smallest integer is greater than all elements in the array. Forgetting to sort the array before processing.

#2568

Minimum Impossible OR

Medium

ArrayBit ManipulationBrainteaserBit ManipulationSorting

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n)
Space	O(1)	O(1)

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Intuition

Time O(n log n)Space O(1)

The optimal approach leverages the properties of bitwise OR and the fact that we can track the smallest integer not expressible by iterating through the sorted array. By maintaining a running total of expressible integers, we can efficiently find the minimum impossible integer.

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Algorithm

5 steps

1Step 1: Sort the input array.
2Step 2: Initialize a variable 'smallest' to 1.
3Step 3: Iterate through the sorted array. For each number, if it is greater than 'smallest', break the loop.
4Step 4: If the number is less than or equal to 'smallest', update 'smallest' to 'smallest | num'.
5Step 5: Return 'smallest' as the result.

solution.py8 lines

1def minImpossibleOR(nums):
2    nums.sort()
3    smallest = 1
4    for num in nums:
5        if num > smallest:
6            break
7        smallest |= num
8    return smallest

ℹ

Complexity note: The time complexity is O(n log n) due to the sorting step, while the space complexity is O(1) as we only use a few extra variables.

1The smallest impossible OR is often the first power of 2 not represented in the array.
2Sorting helps in efficiently tracking the smallest non-expressible integer.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.