What is the brute force solution for Minimize OR of Remaining Elements Using Operations?

The brute force approach for Minimize OR of Remaining Elements Using Operations is Brute Force, with O(n²) time complexity and O(1) space complexity. In the brute-force approach, we try every possible combination of operations to minimize the OR of the remaining elements. This involves simulating all possible pairs of elements we can combine with the AND operation up to k times. The optimal Optimal Solution approach improves this to O(n * 32).

What algorithm or data structure is used to solve Minimize OR of Remaining Elements Using Operations?

Minimize OR of Remaining Elements Using Operations uses the following concepts: Array, Greedy, Bit Manipulation. The recommended patterns to study are: Bit Manipulation, Greedy Algorithms.

What are the interview tips for Minimize OR of Remaining Elements Using Operations?

Always explain your thought process clearly when discussing your approach. Consider edge cases, such as when k is larger than the number of elements. Practice bit manipulation problems to become comfortable with operations like AND and OR.

What are common mistakes when solving Minimize OR of Remaining Elements Using Operations?

Forgetting to consider the impact of AND operations on the bits of numbers. Not properly tracking the number of operations used while checking combinations.

#3022

Minimize OR of Remaining Elements Using Operations

Hard

ArrayGreedyBit ManipulationBit ManipulationGreedy Algorithms

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n * 32)
Space	O(1)	O(1)

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Intuition

Time O(n * 32)Space O(1)

The optimal solution uses a greedy approach and bit manipulation to minimize the OR efficiently. By checking from the most significant bit down to the least significant bit, we can determine which bits can be safely included in the final answer based on the operations allowed.

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Algorithm

4 steps

1Step 1: Initialize a mask to 0, which will hold the bits we want to keep.
2Step 2: Iterate from the most significant bit to the least significant bit (from 31 to 0).
3Step 3: For each bit, temporarily add it to the mask and check if we can achieve a configuration where the remaining OR has that bit set.
4Step 4: If we can achieve that configuration with k operations, keep the bit in the mask; otherwise, remove it.

solution.py15 lines

1# Full working Python code
2def minOR(nums, k):
3    mask = 0
4    for bit in range(31, -1, -1):
5        mask |= (1 << bit)
6        count = 0
7        for num in nums:
8            if (num & mask) == 0:
9                count += 1
10        if count > k:
11            mask ^= (1 << bit)
12    return mask
13
14# Example usage
15print(minOR([3,5,3,2,7], 2))  # Output: 3

ℹ

Complexity note: The time complexity is O(n * 32) because we check each bit position (up to 32 bits) for all n elements. The space complexity is O(1) since we only use a few variables to store the mask and counts.

1Using bit manipulation allows us to efficiently determine which bits can be retained in the final answer.
2Greedy approaches often yield optimal solutions in problems involving minimizing or maximizing values.

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