Last updated on October 5th, 2024 at 06:02 pm

Here, We see Optimal Division LeetCode Solution. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc. with different approaches.

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Optimal Division LeetCode Solution

Problem Statement

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.

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

1. Optimal Division LeetCode Solution C++

class Solution {
public:
    string optimalDivision(vector<int>& nums) {
        string s="";
       int n=nums.size();
        for(int i=0;i<n;i++){
            if(i!=n-1){
                s+=to_string(nums[i])+'/';
            }
           else
            s+=to_string(nums[i]);
        }
        if(n<3){
            return s;
        }
        int k=s.find("/");
        s.insert(k+1,"(");
        s+=')';
        return s;
    }
};

2. Optimal Division Solution Java

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

3. Optimal Division Solution JavaScript

var optimalDivision = function(nums) {
    const n = nums.length
    switch (n) {
        case 1:
        return nums[0] + ''
        case 2:
        return nums[0] + '/' + nums[1]
        default: {
        let s = nums[0] + '/('
        for (let i = 1; i < n - 1; i++) {
            s += '' + nums[i] + '/'
        }
        s += nums[n - 1] + ')'
        return s
        }
    }    
};

4. Optimal Division Solution Python

class Solution(object):
    def optimalDivision(self, nums):
        if len(nums) == 1:
            return str(nums[0])
        elif len(nums) == 2:
            return str(nums[0]) + "/" + str(nums[1])
        else:
            return str(nums[0]) + "/(" + "/".join([str(num) for num in nums[1:]]) + ")"