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Minimize OR of Remaining Elements Using Operations

Difficulty: Hard

Problem Description

You are given a 0-indexed integer array nums and an integer k. In one operation, you can pick any index i of nums such that 0 <= i < nums.length - 1 and replace nums[i] and nums[i + 1] with a single occurrence of nums[i] & nums[i + 1]. Your task is to return the minimum possible value of the bitwise OR of the remaining elements of nums after applying at most k operations.

Key Insights

  • The bitwise AND operation tends to reduce the values of the elements involved.
  • The bitwise OR of an array is minimized when the individual elements are as small as possible.
  • The number of operations allowed (k) can be used strategically to combine elements that will lead to a smaller OR.
  • The problem can be approached using a greedy strategy to choose pairs that will yield the smallest results when combined.

Space and Time Complexity

Time Complexity: O(n log n) - due to the need to sort or prioritize operations based on the potential outcome.
Space Complexity: O(n) - to store intermediate results or data structures.

Solution

The solution involves iterating through the nums array and performing a greedy choice to combine elements using the AND operation. We will maintain a priority queue or a sorted list of potential results after each operation, ensuring we always consider the smallest possible outcomes first. The goal is to apply at most k operations in such a way that the final OR of the array is minimized.

Code Solutions

def minimizeOR(nums, k):
    import heapq
    
    # Initialize a min-heap with the initial elements
    heapq.heapify(nums)
    
    for _ in range(k):
        # Pop the two smallest elements
        first = heapq.heappop(nums)
        second = heapq.heappop(nums)
        
        # Perform the AND operation and push the result back
        new_value = first & second
        heapq.heappush(nums, new_value)
    
    # Calculate the final OR of the remaining elements
    result = 0
    for num in nums:
        result |= num
    
    return result
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