Optimal Division

Problem

Given a list of positive integers, the adjacent integers will perform the float division. For example, [2,3,4] -> 2 / 3 / 4.

However, you can add any number of parenthesis at any position to change the priority of operations. You should find out how to add parenthesis to get the maximum result, and return the corresponding expression in string format. Your expression should NOT contain redundant parenthesis.

Example:
Input: [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 don't 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

Note:

• The length of the input array is [1, 10].
• Elements in the given array will be in range [2, 1000].
• There is only one optimal division for each test case.

Solution

Math Solution: O(n) time, O(n) space

public class Solution {
    public String optimalDivision(int[] nums) {
        if (nums.length == 1) return nums[0] + "";
        if (nums.length == 2) return nums[0] + "/" + nums[1];
        StringBuilder sb = new StringBuilder(nums[0] + "/(" + nums[1]);
        for (int i = 2; i < nums.length; i++) sb.append("/" + nums[i]);
        sb.append(")");
        return sb.toString();
    }
}

Analysis

For there elements a, b, c we have two different divisions
a/b/c or a/(b/c)
For the first division, we simplify it with a/(b*c) For the second one, it becomes a*c/b
Divide a on both of them then they become 1/(c*b) and c/b
It's obvious that c/b >= 1/(c*b) for c >= 1
Because the array contains positive integers (1, 2, 3, ...) we know that second one is always bigger
Therefore, we prove that our solution always returns the max result
To avoid redundant parentheses we divide the nums into three situations