LeetCode Optimal Division

原题链接在这里：https://leetcode.com/problems/optimal-division/description/

题目：

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:

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

题解：

backtracking. a/b 最大需要尽可能最大化a 同时最小化b.

dfs的终止条件是只有一个或者两个元素.

Time Complexity: exponential.

Space: O(nums.length). stack space.

AC Java:

class Solution {
    class Result{
        double val;
        String str;
    }
    public String optimalDivision(int[] nums) {
        return getMax(nums, 0, nums.length-1).str;
    }
    
    private Result getMax(int [] nums, int l, int r){
        Result result = new Result();
        result.val = Double.MIN_VALUE;
        if(l == r){
            result.val = nums[l];
            result.str = String.valueOf(nums[l]);
            return result;
        }
        
        if(l+1 == r){
            result.val = (double)nums[l] / (double)nums[r];
            result.str = nums[l] + "/" + nums[r];
            return result;
        }
        
        for(int i = l; i<r; i++){
            Result r1 = getMax(nums, l, i);
            Result r2 = getMin(nums, i+1, r);
            if(r1.val/r2.val > result.val){
                result.val = r1.val/r2.val;
                result.str = r1.str + "/" + (r-(i+1) == 0 ? r2.str : "("+r2.str+")");
            }
        }
        return result;
    }
    
    private Result getMin(int [] nums, int l, int r){
        Result result = new Result();
        result.val = Double.MAX_VALUE;
        if(l == r){
            result.val = nums[l];
            result.str = String.valueOf(nums[l]);
            return result;
        }
        
        if(l+1 == r){
            result.val = (double)nums[l] / (double)nums[r];
            result.str = nums[l] + "/" + nums[r];
            return result;
        }
        
        for(int i = l; i<r; i++){
            Result r1 = getMin(nums, l, i);
            Result r2 = getMax(nums, i+1, r);
            if(r1.val/r2.val < result.val){
                result.val = r1.val/r2.val;
                result.str = r1.str + "/" + (r-(i+1) == 0 ? r2.str : "("+r2.str+")");
            }
        }
        return result;
    }
}

a/b最大时a要最大, b要最小. 

每个数都大于1, 所以除个数肯定不比自己本上大. 所以a就是第一个数.

越除越小, 所以每个数都不能放过. b就是剩下的每个数以此相除.

Time Complexity: O(nums.length).

Space: O(nums.length).

AC Java:

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