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Minimum Array Sum

Difficulty: Medium

Problem Description

You are given an integer array nums and three integers k, op1, and op2. You can perform two operations on nums with specific constraints to minimize the sum of the array elements. Operation 1 allows you to divide an element by 2 (rounding up), and Operation 2 allows you to subtract k from an element, provided it is greater than or equal to k. Each operation can be performed at most once per index and has a limited number of uses.

Key Insights

  • Operation 1 reduces an element significantly if its value is large.
  • Operation 2 can reduce an element to zero if it is sufficiently large compared to k.
  • Prioritizing which operation to use on which elements is crucial for minimizing the overall sum.
  • The operations can be mixed on the same index, but each can only be used once per index.

Space and Time Complexity

Time Complexity: O(n log n), where n is the number of elements in the array. This accounts for sorting for optimal operation selection. Space Complexity: O(1), as we use a constant amount of space for calculations.

Solution

To solve the problem, we can follow these steps:

  1. Calculate the potential outcomes for each element based on the two operations:
    • For Operation 1, compute the value after dividing by 2 (rounding up).
    • For Operation 2, compute the value after subtracting k (if applicable).
  2. Sort the potential outcomes based on their effectiveness in reducing the overall sum.
  3. Apply the operations in a way that maximizes their effect, prioritizing the largest reductions first, until we exhaust the operations.

This approach leverages sorting to help decide the best elements to apply the operations on, ensuring we minimize the total sum effectively.

Code Solutions

def minArraySum(nums, k, op1, op2):
    operations = []
    
    for num in nums:
        # Calculate the potential values after each operation
        op1_value = (num + 1) // 2  # Round up division by 2
        op2_value = max(num - k, 0)  # Subtraction, ensuring non-negative
        
        # Store potential values with their operation types
        operations.append((num, op1_value, op2_value))
    
    # Sort operations based on the effectiveness of reductions
    operations.sort(key=lambda x: min(x[1], x[2]), reverse=True)
    
    total_sum = sum(nums)
    
    # Apply Operation 1
    for i in range(min(op1, len(operations))):
        total_sum -= (operations[i][0] - operations[i][1])
    
    # Apply Operation 2
    for i in range(min(op2, len(operations))):
        total_sum -= (operations[i][0] - operations[i][2])
    
    return total_sum
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