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Java-based LeetCode algorithm problem solutions, regularly updated
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3366\. Minimum Array Sum
Medium
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`
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