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553. Optimal Division

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.

 

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

 

Constraints:

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

Solutions

• Java
• C++
• Python
• Go
• TypeScript
• RenderScript

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

• class Solution {
  public:
      string optimalDivision(vector<int>& nums) {
          int n = nums.size();
          if (n == 1) return to_string(nums[0]);
          if (n == 2) return to_string(nums[0]) + "/" + to_string(nums[1]);
          string ans = to_string(nums[0]) + "/(";
          for (int i = 1; i < n - 1; i++) ans.append(to_string(nums[i]) + "/");
          ans.append(to_string(nums[n - 1]) + ")");
          return ans;
      }
  };
  

• class Solution:
      def optimalDivision(self, nums: List[int]) -> str:
          n = len(nums)
          if n == 1:
              return str(nums[0])
          if n == 2:
              return f'{nums[0]}/{nums[1]}'
          return f'{nums[0]}/({"/".join(map(str, nums[1:]))})'
  
  

• func optimalDivision(nums []int) string {
  	n := len(nums)
  	if n == 1 {
  		return strconv.Itoa(nums[0])
  	}
  	if n == 2 {
  		return fmt.Sprintf("%d/%d", nums[0], nums[1])
  	}
  	ans := &strings.Builder{}
  	ans.WriteString(fmt.Sprintf("%d/(", nums[0]))
  	for _, num := range nums[1 : n-1] {
  		ans.WriteString(strconv.Itoa(num))
  		ans.WriteByte('/')
  	}
  	ans.WriteString(fmt.Sprintf("%d)", nums[n-1]))
  	return ans.String()
  }
  

• function optimalDivision(nums: number[]): string {
      const n = nums.length;
      const res = nums.join('/');
      if (n > 2) {
          const index = res.indexOf('/') + 1;
          return `${res.slice(0, index)}(${res.slice(index)})`;
      }
      return res;
  }
  
  

• impl Solution {
      pub fn optimal_division(nums: Vec<i32>) -> String {
          let n = nums.len();
          match n {
              1 => nums[0].to_string(),
              2 => nums[0].to_string() + "/" + &nums[1].to_string(),
              _ => {
                  let mut res = nums[0].to_string();
                  res.push_str("/(");
                  for i in 1..n {
                      res.push_str(&nums[i].to_string());
                      res.push('/');
                  }
                  res.pop();
                  res.push(')');
                  res
              }
          }
      }
  }
  
  

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