Introduction

In this blog post, we will explore how to solve the LeetCode problem "3366" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

You are given an integer array nums and three integers k, op1, and op2. You can perform the following operations on nums: Operation 1: Choose an index i and divide nums[i] by 2, rounding up to the nearest whole number. You can perform this operation at most op1 times, and not more than once per index. Operation 2: Choose an index i and subtract k from nums[i], but only if nums[i] is greater than or equal to k. You can perform this operation at most op2 times, and not more than once per index. Note: Both operations can be applied to the same index, but at most once each. Return the minimum possible sum of all elements in nums after performing any number of operations. Example 1: Input: nums = [2,8,3,19,3], k = 3, op1 = 1, op2 = 1 Output: 23 Explanation: Apply Operation 2 to nums[1] = 8, making nums[1] = 5. Apply Operation 1 to nums[3] = 19, making nums[3] = 10. The resulting array becomes [2, 5, 3, 10, 3], which has the minimum possible sum of 23 after applying the operations. Example 2: Input: nums = [2,4,3], k = 3, op1 = 2, op2 = 1 Output: 3 Explanation: Apply Operation 1 to nums[0] = 2, making nums[0] = 1. Apply Operation 1 to nums[1] = 4, making nums[1] = 2. Apply Operation 2 to nums[2] = 3, making nums[2] = 0. The resulting array becomes [1, 2, 0], which has the minimum possible sum of 3 after applying the operations. Constraints: 1 <= nums.length <= 100 0 <= nums[i] <= 105 0 <= k <= 105 0 <= op1, op2 <= nums.length

Explanation

Here's a breakdown of the approach, complexity, and the Python code:

• Key Idea: The core idea is to prioritize operations to minimize the sum. Applying operation 2 (subtracting k) greedily when possible generally leads to a larger reduction compared to operation 1 (dividing by 2). However, we must decide which elements to apply these operations to.
• Greedy with Sorting: Sort the array. We can apply op2 operations on elements from largest to smallest, which are >= k. Then, we apply op1 operations, again from largest to smallest after the op2 operations.

• Simplified Logic: We will only consider apply the op2 operation if the element nums[i] is larger than k.

• Complexity:

  • Runtime: O(n log n) due to sorting.
  • Space: O(1) excluding the input array.

Code

    def minSumAfterOperations(nums, k, op1, op2):
    """
    Calculates the minimum sum of the array after performing operations.

    Args:
        nums: The input integer array.
        k: The value to subtract in operation 2.
        op1: The maximum number of times operation 1 can be performed.
        op2: The maximum number of times operation 2 can be performed.

    Returns:
        The minimum possible sum of the elements in nums after performing operations.
    """

    n = len(nums)
    nums.sort()  # Sort the array in ascending order

    # Apply operation 2 (subtract k) to the largest elements first, if possible
    for i in range(n - 1, -1, -1):
        if op2 > 0 and nums[i] >= k:
            nums[i] -= k
            op2 -= 1

    nums.sort()  # Sort again after applying op2 for optimal op1 application

    # Apply operation 1 (divide by 2)
    for i in range(n - 1, -1, -1):
        if op1 > 0:
            nums[i] = (nums[i] + 1) // 2
            op1 -= 1

    return sum(nums)