Optimal Division

Medium

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ArraysStringsDynamic ProgrammingGreedy Algorithms

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.

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the range of values for the integers in the input array? Can they be negative, zero, or non-integer?
What should I return if the input array is empty or null?
Are we guaranteed that the input array will always have at least two numbers?
If there are multiple ways to achieve the maximum result, is any valid expression acceptable?
Is the goal to minimize the number of parentheses used in the expression while still maximizing the result, or is the absolute maximum the only concern?

Brute Force Solution

Approach

The core idea is to explore all possible ways to group the numbers and apply divisions. This will involve creating all possible parenthesized expressions and then evaluating each expression to find the maximum result. In essence, we try every grouping, compute its value, and then choose the best one.

Here's how the algorithm would work step-by-step:

Consider all possible ways to put parentheses around the given numbers and division operators.
For each arrangement of parentheses, calculate the result of the division.
Keep track of the largest result you've seen so far.
After trying all possible arrangements, return the arrangement that resulted in the largest calculated result.

Code Implementation

def optimal_division_brute_force(numbers):
    # Function to calculate the value of an expression string
    def calculate_expression(expression):
        try:
            return eval(expression)
        except ZeroDivisionError:
            return float('-inf')

    def generate_expressions(current_expression, remaining_numbers):
        if not remaining_numbers:
            return [current_expression]

        expressions = []
        for i in range(len(remaining_numbers)): 
            # Iterate through the possible splits.
            number = str(remaining_numbers[i])
            new_remaining_numbers = remaining_numbers[i+1:]
            if current_expression == "":
                new_expression = number
            else:
                new_expression = current_expression + "/" + number

            expressions.extend(generate_expressions(new_expression, new_remaining_numbers))

        return expressions

    # Generate all possible expressions.
    all_expressions = generate_expressions("", numbers)

    max_value = float('-inf')
    optimal_expression = ""

    # Evaluate each expression and keep track of the maximum value.
    for expression in all_expressions:
        value = calculate_expression(expression)
        if value > max_value:
            # Update max value if current value is higher
            max_value = value
            optimal_expression = expression

    return optimal_expression

Big(O) Analysis

Time Complexity

O(n!) – The algorithm considers all possible parenthesized expressions for n numbers and division operators. The number of ways to parenthesize an expression with n terms is given by the Catalan number, which is approximately 4^n / (n * sqrt(n)). For each parenthesized expression, we need to evaluate it which takes O(n) time. Since we are exploring all possible groupings which relate to permutations of how we apply the divisions, we can approximate the total time complexity by the number of permutations which is n!. Therefore, the overall time complexity can be considered as O(n!).

Space Complexity

O(N^2) – The provided solution, based on exploring all possible parenthesized expressions, implicitly relies on a recursive or iterative approach that builds intermediate results. Generating all possible groupings of N numbers can lead to storing a significant number of intermediate expressions and their calculated values. In the worst-case scenario, the number of such groupings could grow polynomially with N, potentially requiring storing results of all subproblems to find the optimal division. Therefore, the auxiliary space complexity can be approximated as O(N^2) due to the storage of intermediate results and/or recursive call stack, particularly if memoization is used.

Optimal Solution

Approach

The goal is to maximize the result of a series of divisions. The best way to do this is to realize that we want to divide the first number by as small a number as possible. We achieve this by grouping all the subsequent numbers together as a divisor.

Here's how the algorithm would work step-by-step:

The first number will always be the numerator.
Create a fraction where the first number is divided by the result of dividing all other numbers together.
If there are only two numbers, simply divide the first by the second.
If there are more than two numbers, group all numbers after the first one in parentheses to force their multiplication before dividing the first number.
This structure guarantees the smallest possible divisor for the first number, thereby maximizing the overall result.

Code Implementation

def optimalDivision(numbers):
    number_of_elements = len(numbers)

    # Handle cases with 1 or 2 numbers.
    if number_of_elements == 1:
        return str(numbers[0])
    if number_of_elements == 2:
        return str(numbers[0]) + "/" + str(numbers[1])

    # For more than 2 numbers, use parentheses.
    result = str(numbers[0]) + "/("

    # Construct the string representation.
    for i in range(1, number_of_elements):
        result += str(numbers[i])
        if i != number_of_elements - 1:
            result += "/"

    result += ")"
    return result

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the input array of size n once to construct the string representation of the optimal division. Constructing the string involves appending elements and parenthesis, which take constant time per element. Therefore, the overall time complexity is directly proportional to the number of elements in the input array, resulting in O(n) time complexity.

Space Complexity

O(1) – The algorithm's space complexity is O(1) because it primarily constructs a string representation of the division expression. It does not utilize any auxiliary data structures like arrays, lists, or hash maps whose size scales with the input array's length, N. The string construction happens in place, or the string builder is constant in size and independent of the number of integers being divided. Therefore, the memory usage remains constant, regardless of the input size.

Edge Cases

Case	How to Handle
Null or empty input array	Return an empty string if the input array is null or empty to prevent errors.
Array with a single element	Return the single element as a string since no division is possible.
Array with two elements	Return the two elements divided as a string (e.g., 'a/b').
Array with three or more elements; large values.	Parenthesize the denominator to maximize the result: a/(b/c/d...) and be mindful of integer overflow when constructing the string.
Maximum size of the input array.	Consider using StringBuilder for string concatenation to optimize performance for large inputs.
Array containing zero.	The division can create zero result, so no special check is needed as the standard approach handles it.
Array containing negative numbers.	The presence of negative numbers doesn't break mathematical rules so treat as standard.
Floating-point precision issues	The problem does not ask for the floating point number so no special handling is necessary related to the floating-point representation.

Null or empty input array

How to Handle:

Return an empty string if the input array is null or empty to prevent errors.

Array with a single element

How to Handle:

Return the single element as a string since no division is possible.

Array with two elements

How to Handle:

Return the two elements divided as a string (e.g., 'a/b').

Array with three or more elements; large values.

How to Handle:

Parenthesize the denominator to maximize the result: a/(b/c/d...) and be mindful of integer overflow when constructing the string.

Maximum size of the input array.

How to Handle:

Consider using StringBuilder for string concatenation to optimize performance for large inputs.

Array containing zero.

How to Handle:

The division can create zero result, so no special check is needed as the standard approach handles it.

Array containing negative numbers.

How to Handle:

The presence of negative numbers doesn't break mathematical rules so treat as standard.

Floating-point precision issues

How to Handle:

The problem does not ask for the floating point number so no special handling is necessary related to the floating-point representation.