553. Optimal Division

Problem Description

The task involves an integer array nums, where each adjacent pair of integers performs a floating-point division in the given order. For instance, if nums is [2, 3, 4], it represents the expression 2/3/4. The challenge is to add parentheses to the expression in a way that maximizes the result of the expression. Parentheses can change the order of operations; hence changing the expression's value. The goal is to find where to place these parentheses to get the maximum value expression and return this expression as a string. It's important to note that the final expression must not include unnecessary parentheses.


When dividing a set of numbers, placing the larger numbers in the numerator and the smaller ones in the denominator maximizes the result. With division, the key to maximizing the outcome is to divide by the largest possible number. Based on this logic, we want to minimize the denominator as much as possible to maximize the result of the entire expression.

For three or more numbers, the maximum result is achieved when we divide the first number by the result of dividing all the remaining numbers. Therefore, for nums = [a, b, c, ..., n], the optimal division is a/(b/c/.../n). This ensures that a is divided by the smallest possible value obtained from the division of the rest of the numbers. In this case, we only need parentheses around all the numbers starting from the second number in the array.

For nums with only one element, there is no division to perform, so we return the single number. When nums has exactly two elements, say [a, b], the expression is simply a/b, as there's no other way to place parentheses that can alter the result. The solution naturally emerges from these observations:

  1. If nums contains only one element, that is the expression.
  2. If nums contains two elements, the expression is "a/b".
  3. For three or more elements, we use parentheses to ensure that the first element is divided by the smallest possible value obtained from the division of the rest of the numbers: "a/(b/c/.../n)".

The provided Python solution implements these insights through conditional logic and string formatting.

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Solution Approach

The provided Python solution implements a direct approach without the need for any complex algorithms or data structures. Here's how the code accomplishes the task:

  1. The solution first checks the length of nums:

    • If the array has only one element (n == 1), it returns that element as a string.
    • If the array has two elements (n == 2), it returns the division of the first element by the second as a string formatted "a/b".
  2. For arrays with three or more elements (n >= 3), the solution constructs an expression that places the first number in the numerator and the rest in the denominator within a pair of parentheses. This is based on the intuition that to maximize the value of the division, you want to divide the first number by the result of the subsequence divisions, thus minimizing the denominator.

  3. The code uses Python's string formatting and string join method to construct the final expression. The part "/".join(map(str, nums[1:])) joins all elements of nums starting from the second element with a division sign between them. It then formats the entire string by placing the first element of nums outside and the joined string inside the parentheses.

No additional data structures are required since the final expression is constructed directly from the input list. The solution takes advantage of Python's string manipulation capabilities to format the output as per the problem's requirements.

The simplicity of the solution comes from the mathematical insight that the highest value for the entire expression is obtained when the first element of nums is divided by the smallest possible value from the division of the subsequent elements. This is efficiently implemented in the given Python function optimalDivision.

Here's the implementation logic step by step in the solution:

1class Solution:
2    def optimalDivision(self, nums: List[int]) -> str:
3        n = len(nums)
4        if n == 1:
5            return str(nums[0])                # Case for single element, return as string
6        if n == 2:
7            return f'{nums[0]}/{nums[1]}'      # Case for two elements, return "a/b"
8        # Case for three or more elements, return "a/(b/c/.../n)"
9        return f'{nums[0]}/({"/".join(map(str, nums[1:]))})'

This concise implementation elegantly solves the problem by directly translating the mathematical insight of division into a formatted string, which is exactly what the problem statement is asking for.

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Example Walkthrough

Let's go through a small example using the solution approach to understand how the algorithm maximizes the expression by correctly placing the parentheses.

Suppose we have the array nums = [8, 2, 3, 4], which represents the expression 8/2/3/4. The task is to add parentheses to maximize this expression's value.

Following the solution approach:

  1. We first check the length of nums. Since n = 4, we have more than two elements.

  2. For an array with more than two elements, the goal is to maximize the first number's division by the smallest possible result from subsequence divisions. To achieve this, we should divide the first element by the result of dividing all the subsequent numbers. This translates to the expression: 8/(2/3/4).

  3. To create this expression in Python, the code joins all elements of nums, starting from the second element, with a division sign, resulting in the string "2/3/4".

  4. Then, the code formats the entire expression by placing the first element of nums (8) outside and the joined string inside parentheses, resulting in the final expression "8/(2/3/4)".

Using this approach, we have successfully maximized the value of the original expression. In the absence of parentheses, the expression would calculate as follows: (8/2)/3/4 = 4/3/4 = 1.33. However, by placing parentheses as 8/(2/3/4), we get: 8/(0.166...) ≈ 48, which is indeed the maximum value possible for this expression.

The Python code effectively and efficiently constructs this optimal expression using string formatting without the need for additional computation or complex data structures.

Solution Implementation

1from typing import List
3class Solution:
4    def optimalDivision(self, nums: List[int]) -> str:
5        # Calculate the length of the input list of numbers
6        size = len(nums)
8        # If there is only one number, return it as a string
9        if size == 1:
10            return str(nums[0])
12        # If there are two numbers, return them as a division
13        if size == 2:
14            return f'{nums[0]}/{nums[1]}'
16        # For more than two numbers, the optimal division is to divide the first number
17        # by the result of the division of all remaining numbers to achieve the largest result.
18        # This is done by grouping all but the first number inside parentheses.
19        return f'{nums[0]}/(' + "/".join(map(str, nums[1:])) + ')'
1class Solution {
2    // Method to find the optimal division of integers as a string
3    public String optimalDivision(int[] nums) {
4        // Get the number of elements in the array
5        int arrayLength = nums.length;
7        // If there is only one number, return it as there's nothing to divide
8        if (arrayLength == 1) {
9            return String.valueOf(nums[0]);
10        }
12        // If there are two numbers, return their division
13        if (arrayLength == 2) {
14            return nums[0] + "/" + nums[1];
15        }
17        // If there are more than two numbers, we use parentheses to maximize the result
18        StringBuilder result = new StringBuilder();
20        // Start the string with the first number and an opening parenthesis
21        result.append(nums[0]).append("/(");
23        // Loop through the rest of the array, except for the last element
24        for (int i = 1; i < arrayLength - 1; ++i) {
25            // Add each number followed by a division sign
26            result.append(nums[i]).append("/");
27        }
29        // Add the last element of the array and the closing parenthesis
30        result.append(nums[arrayLength - 1]).append(")");
32        // Return the constructed string
33        return result.toString();
34    }
1class Solution {
3    // This function finds the optimal division of array numbers as a string expression
4    string optimalDivision(vector<int>& nums) {
5        int n = nums.size();
7        // If there is only one number, return it as a string
8        if (n == 1) {
9            return to_string(nums[0]);
10        }
12        // If there are two numbers, return their division
13        if (n == 2) {
14            return to_string(nums[0]) + "/" + to_string(nums[1]);
15        }
17        // For more than two numbers, enclose all but the first number in parentheses
18        // This is to ensure the division is performed correctly for optimal result
19        string answer = to_string(nums[0]) + "/(";
21        // Append all the remaining numbers separated by a division operator
22        for (int i = 1; i < n - 1; i++) {
23            answer.append(to_string(nums[i]) + "/");
24        }
26        // Add the last number and close the parentheses
27        answer.append(to_string(nums[n - 1]) + ")");
29        return answer;
30    }
2 * Returns a string representation of the division of numbers that maximizes the result.
3 * If there are more than two numbers, brackets are added to divide the first number by
4 * the result of the division of all subsequent numbers.
5 * @param {number[]} numbers - An array of numbers to divide.
6 * @returns {string} - The string representation of the optimal division.
7 */
8function optimalDivision(numbers: number[]): string {
9    // Get the number of elements in the array
10    const count = numbers.length;
12    // Combine the numbers into a string separated by slashes
13    const divisionString = numbers.join('/');
15    // If there are more than two numbers, add brackets after the first division
16    if (count > 2) {
17        // Find the index of the first slash to insert the opening bracket
18        const firstSlashIndex = divisionString.indexOf('/') + 1;
19        // Return the string with brackets inserted
20        return `${divisionString.slice(0, firstSlashIndex)}(${divisionString.slice(firstSlashIndex)})`;
21    }
23    // If there are two or fewer numbers no brackets are needed
24    return divisionString;

Time and Space Complexity

Time Complexity

The time complexity of the code primarily depends on the length of the nums array. The map function iterates over nums[1:], which has a complexity of O(n-1) where n is the length of the array. The join operation on a list of strings has a complexity of O(m), where m is the total length of the strings being concatenated. Since in this case we're dealing with string representations of the numbers, the length of each integer after conversion to string can vary, but in the worst case, it will be proportional to the logarithm of the number. However, for the purpose of time complexity analysis, we can consider m to be linearly dependent on n because the number of divisions does not change the overall complexity class. Hence, the join operation is also O(n). The overall time complexity is O(n).

Space Complexity

The space complexity includes the space required for the output and the temporary lists created by map and join. The size of the output string is O(n) since it includes all the elements of nums plus additional characters. The temporary list created by map is also O(n). However, these are not additional spaces in terms of space complexity analysis since they are required for the output. Thus, the space complexity is O(n).

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