# Optimal Division LeetCode Solution

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.

# List of all 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

## Optimal Division LeetCode SolutionC++

``````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;
}
};```Code language: PHP (php)```

## Optimal Division SolutionJava

``````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();
}
}```Code language: JavaScript (javascript)```

## Optimal Division SolutionJavaScript

``````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
}
}
};```Code language: JavaScript (javascript)```

## Optimal Division SolutionPython

``````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:]]) + ")"``````
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