553. Optimal Division

1. Description

You are given an integer array nums. The adjacent integers in nums will perform the float division.
For example, for nums = [2,3,4], we will evaluate the expression “2/3/4”.
However, you can add any number of parenthesis at any position to change the priority of operations. You want to add these parentheses such the value of the expression after the evaluation is maximum.
Return the corresponding expression that has the maximum value in string format.
Note: your expression should not contain redundant parenthesis.

2. Example

Example 1:
Input: nums = [1000,100,10,2]
Output: “1000/(100/10/2)”
Explanation: 1000/(100/10/2) = 1000/((100/10)/2) = 200
However, the bold parenthesis in “1000/((100/10)/2)” are redundant since they do not influence the operation priority.
So you should return “1000/(100/10/2)”.
Other cases:
1000/(100/10)/2 = 50
1000/(100/(10/2)) = 50
1000/100/10/2 = 0.5
1000/100/(10/2) = 2

Example 2:
Input: nums = [2,3,4]
Output: “2/(3/4)”
Explanation: (2/(3/4)) = 8/3 = 2.667
It can be shown that after trying all possibilities, we cannot get an expression with evaluation greater than 2.667

3. Constraints

• 1 <= nums.length <= 10
• 2 <= nums[i] <= 1000
• There is only one optimal division for the given input.

4. Solutions

Math

n = nums.size()
Time complexity: O(n)
Space complexity: O(1)

// all numbers are positive, so this is a stupid problem
class Solution {
public:
    string optimalDivision(const vector<int> &nums) {
        if (nums.size() == 1) {
            return to_string(nums[0]);
        } else if (nums.size() == 2) {
            return to_string(nums[0]) + "/" + to_string(nums[1]);
        } else {
            string result = to_string(nums[0]) + "/(" + to_string(nums[1]);
            for (int i = 2; i < nums.size(); ++i) {
                result = move(result + "/" + to_string(nums[i]));
            }
            result.push_back(')');

            return result;
        }
    }
};