Minimum Array Sum

You are given an integer array nums and three integers k, op1, and op2. You can perform two types of operations on nums:

1. 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.
2. 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.

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.

What is the most efficient algorithm to solve this problem?

def solve():
    nums = [2, 8, 3, 19, 3]
    k = 3
    op1 = 1
    op2 = 1
    
    n = len(nums)
    
    def calculate_sum(arr):
        return sum(arr)

    def apply_operations(arr, k, op1, op2):
        import copy
        min_sum = float('inf')

        # Iterate through all possible combinations of operations
        for i in range(3**n):  # Each num can have op1, op2, or none
            temp_arr = copy.deepcopy(arr)
            current_op1 = 0
            current_op2 = 0
            
            op_sequence = []
            temp = i
            for _ in range(n):
                op_sequence.append(temp % 3)
                temp //= 3
            
            for index in range(n):
                op_type = op_sequence[index]

                if op_type == 1 and current_op1 < op1:
                    temp_arr[index] = (temp_arr[index] + 1) // 2 # Rounding up division
                    current_op1 += 1
                elif op_type == 2 and current_op2 < op2 and temp_arr[index] >= k:
                    temp_arr[index] -= k
                    current_op2 += 1
            
            min_sum = min(min_sum, calculate_sum(temp_arr))

        return min_sum

    result = apply_operations(nums, k, op1, op2)
    print(result)


    nums2 = [2, 4, 3]
    k2 = 3
    op1_2 = 2
    op2_2 = 1

    def apply_operations_dp(nums, k, op1, op2):
        n = len(nums)
        dp = {}

        def solve(index, remaining_op1, remaining_op2, current_nums):
            if index == n:
                return sum(current_nums)

            state = (index, remaining_op1, remaining_op2, tuple(current_nums))
            if state in dp:
                return dp[state]

            ans = float('inf')

            # Option 1: No operation
            ans = min(ans, solve(index + 1, remaining_op1, remaining_op2, current_nums))

            # Option 2: Operation 1 (divide by 2 rounding up)
            if remaining_op1 > 0:
                new_nums = current_nums[:]
                new_nums[index] = (new_nums[index] + 1) // 2
                ans = min(ans, solve(index + 1, remaining_op1 - 1, remaining_op2, new_nums))

            # Option 3: Operation 2 (subtract k)
            if remaining_op2 > 0 and current_nums[index] >= k:
                new_nums = current_nums[:]
                new_nums[index] -= k
                ans = min(ans, solve(index + 1, remaining_op1, remaining_op2 - 1, new_nums))

            dp[state] = ans
            return ans

        return solve(0, op1, op2, nums[:])


    result2 = apply_operations_dp(nums2, k2, op1_2, op2_2)
    print(result2)



solve()

Explanation:

The problem asks to find the minimum possible sum of an array after applying two types of operations:

1. Divide an element by 2, rounding up (at most op1 times, once per index).
2. Subtract k from an element if it's greater or equal to k (at most op2 times, once per index).

Brute Force Approach:

The initial apply_operations function explores all possible combinations of applying or not applying each operation to each element. It generates all possible combinations of operations for each number (operation 1, operation 2, or no operation) using base-3 representation. Then it calculates sum of the array after performing each such combination of operations and updates a minimum sum value throughout. However, this approach has exponential time complexity.

Dynamic Programming Approach:

To optimize, apply_operations_dp uses dynamic programming. The function solve recursively explores all possible operations but stores the results in dp to avoid recomputation. The state is defined by the index, remaining operations of type 1, remaining operations of type 2, and the current state of array nums. The base case is when the index reaches the end of the array.

Time and Space Complexity:

Brute Force Approach:

• Time Complexity: O(3ⁿ), where n is the number of elements in the array. This is because, for each element, there are three possibilities: apply operation 1, apply operation 2, or apply no operation.
• Space Complexity: O(n) due to the recursive call stack (in the worst case).

Dynamic Programming Approach:

• Time Complexity: O(n * op1 * op2 * product of nums), where n is the number of elements in the array, op1 is the number of operation 1, op2 is the number of operation 2.
• Space Complexity: O(n * op1 * op2 * product of nums) due to the memoization table dp.

Edge Cases:

• Empty array: Should return 0.
• k = 0: Only operation 1 can be applied.
• op1 = 0 and op2 = 0: No operations can be applied, return the original sum.
• Large k: Operation 2 might not be applicable to many elements.
• op1 and op2 are much larger than n: The number of effective operations is limited to n for each type.