Problem Description

You have an integer array nums where the elements will be divided sequentially from left to right using floating-point division.

For example, given nums = [2,3,4], the default evaluation would be 2/3/4, which calculates as (2/3)/4 = 0.667/4 = 0.167.

You can add parentheses anywhere in the expression to change the order of operations. The goal is to strategically place parentheses to maximize the final result.

For instance, with nums = [2,3,4]:

• Default: 2/3/4 = 0.167
• With parentheses: 2/(3/4) = 2/0.75 = 2.667

Your task is to return the expression as a string that produces the maximum possible value. The expression should not contain unnecessary parentheses.

Key observations:

• Division operations are performed left to right by default
• Adding parentheses changes the evaluation order
• The result should be a valid mathematical expression in string format
• Redundant parentheses should be avoided

The solution recognizes that to maximize the result of a division chain, you want to minimize the denominator. Since a/(b/c/d/...) equals a*c*d*...)/b, the optimal strategy is to place all elements after the first one in parentheses (except for edge cases with 1 or 2 elements).

Intuition

Let's think about what happens when we divide numbers sequentially. When we have a/b/c, this evaluates as (a/b)/c = a/(b*c). The key insight is that division by a product makes the result smaller.

To maximize our result, we want to make the denominator as small as possible. Consider a/b/c/d/...:

• Without parentheses: a/(b*c*d*...)
• If we group everything after b in parentheses: a/(b/(c/d/...))

When we have a/(b/(c/d)), this actually equals a*(c/d)/b = (a*c)/(b*d). Notice how some terms moved to the numerator!

Let's trace through a concrete example with [6,2,3,4]:

• Default: 6/2/3/4 = ((6/2)/3)/4 = (3/3)/4 = 1/4 = 0.25
• Optimal: 6/(2/3/4) = 6/((2/3)/4) = 6/(2/12) = 6*12/2 = 36

The pattern becomes clear: by putting parentheses around everything after the first division, we effectively flip many divisions into multiplications. Specifically, a/(b/c/d/...) becomes a*c*d*...)/b, maximizing our result.

This leads to the simple strategy:

• For 1 number: just return it
• For 2 numbers: a/b (no choice in arrangement)
• For 3+ numbers: a/(b/c/d/...) where we group all numbers after the first into parentheses

The mathematical reasoning is that we're converting as many division operations as possible into multiplication operations in the numerator, keeping only one division by the second number.

Learn more about Math and Dynamic Programming patterns.

Solution Approach

The implementation is straightforward once we understand the optimal strategy. We handle three distinct cases based on the length of the input array:

Case 1: Single element (n == 1)

• Simply return the number as a string
• No division operations needed
• Example: [5] returns "5"

Case 2: Two elements (n == 2)

• Return the expression with a single division
• Format: "a/b"
• Example: [6,2] returns "6/2"

Case 3: Three or more elements (n >= 3)

• Apply the optimal parenthesization strategy
• Place the first number as the numerator
• Group all remaining numbers in parentheses as the denominator
• Format: "a/(b/c/d/...)"
• Example: [6,2,3,4] returns "6/(2/3/4)"

The implementation uses Python's string formatting and the join() method for clean string construction:

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:]))})'

The key components:

• str(nums[0]): Converts the single element to string
• f'{nums[0]}/{nums[1]}': Uses f-string formatting for two elements
• "/".join(map(str, nums[1:])): Converts all elements after the first to strings and joins them with "/"
• The outer parentheses in the third case ensure proper grouping

This solution runs in O(n) time complexity for string construction and O(n) space complexity for the output string.

Example Walkthrough

Let's walk through the solution with nums = [12, 4, 2] to illustrate why our approach maximizes the result.

Step 1: Understand the default evaluation Without parentheses, division evaluates left to right:

• 12/4/2 = (12/4)/2 = 3/2 = 1.5

Step 2: Explore alternative parenthesizations We could place parentheses in different ways:

• Option A: 12/(4/2) = 12/2 = 6 ✓
• Option B: (12/4)/2 = 3/2 = 1.5 (same as default)

Step 3: Apply our solution strategy Since we have 3 elements (n ≥ 3), we use the pattern a/(b/c/...):

• Place the first number (12) as the numerator
• Group remaining numbers (4/2) in parentheses as the denominator
• Result: "12/(4/2)"

Step 4: Verify this is optimal Let's see why 12/(4/2) gives the maximum value:

• 4/2 = 2 (evaluate the denominator first)
• 12/2 = 6 (divide the numerator by the result)

Mathematically, 12/(4/2) is equivalent to 12 * 2/4 = 24/4 = 6. Notice how the last number (2) effectively moved to the numerator through the parenthesization!

Step 5: Code execution

nums = [12, 4, 2]
n = len(nums)  # n = 3
# Since n >= 3, we use the third case:
result = f'{nums[0]}/({"/".join(map(str, nums[1:]))})'
# nums[0] = 12
# nums[1:] = [4, 2]
# "/".join(map(str, [4, 2])) = "4/2"
# result = "12/(4/2)"

The solution correctly returns "12/(4/2)" which evaluates to 6, the maximum possible value.

Solution Implementation

Time and Space Complexity

Time Complexity: O(n)

• The algorithm performs a constant amount of work for checking the length n of the input list.
• For n = 1 or n = 2, it returns immediately with O(1) operations.
• For n > 2, the main operation is "/".join(map(str, nums[1:])):
  • nums[1:] creates a slice containing n-1 elements: O(n-1)
  • map(str, nums[1:]) converts each of the n-1 elements to strings: O(n-1)
  • "/".join(...) concatenates n-1 strings with "/" separators: O(n-1)
• The string formatting operation combines the results: O(n)
• Overall time complexity: O(n)

Space Complexity: O(n)

• The slice nums[1:] creates a new list with n-1 elements: O(n-1)
• The map operation and join create intermediate string representations: O(n-1)
• The final result string has length proportional to the number of elements and their digit representations: O(n)
• Overall space complexity: O(n)

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Attempting Complex Dynamic Programming or Recursion

Many developers instinctively reach for dynamic programming when they see "maximize" in the problem statement, leading to unnecessarily complex solutions that try to evaluate all possible parenthesization combinations.

Why it's a pitfall: The mathematical insight that a/(b/c/d...) = (a*c*d...)/(b) means there's always one optimal solution pattern, making DP overkill.

Solution: Recognize the mathematical pattern first. Since we're maximizing a division chain with positive numbers, we want to minimize the denominator, which is achieved by the specific parenthesization pattern a/(b/c/d...).

2. Adding Unnecessary Parentheses

A common mistake is adding parentheses in cases where they're not needed, such as:

• Wrapping single numbers: "(5)" instead of "5"
• Adding parentheses for two-element arrays: "6/(2)" instead of "6/2"
• Over-parenthesizing: "(6)/((2/3/4))" instead of "6/(2/3/4)"

Why it's a pitfall: The problem explicitly states to avoid redundant parentheses, and unnecessary parentheses make the expression harder to read.

Solution: Only add parentheses when they change the evaluation order:

if n == 2:
    return f"{nums[0]}/{nums[1]}"  # No parentheses needed
if n >= 3:
    return f"{nums[0]}/({'/'.join(map(str, nums[1:]))})"  # Parentheses required

3. Incorrect String Concatenation

Developers might incorrectly build the string, especially when dealing with the denominator part:

# Wrong approaches:
result = str(nums[0]) + "/(" + str(nums[1:]) + ")"  # Outputs array representation
result = f"{nums[0]}/({nums[1:]})"  # Outputs list format, not division chain

Why it's a pitfall: These approaches don't properly format the array elements as a division chain.

Solution: Use join() with map() to properly convert and format:

denominator_part = "/".join(map(str, nums[1:]))  # Converts each number to string and joins with "/"

4. Forgetting Edge Cases

Not handling arrays with 1 or 2 elements properly, attempting to apply the general pattern to all cases:

# Wrong: Applying the same pattern to all cases
def optimalDivision(nums):
    return f"{nums[0]}/({'/'.join(map(str, nums[1:]))})"
    # This fails for single elements and adds unnecessary parentheses for two elements

Why it's a pitfall: Single-element arrays have no division, and two-element arrays don't benefit from parentheses.

Solution: Explicitly handle different array sizes:

n = len(nums)
if n == 1:
    return str(nums[0])
if n == 2:
    return f"{nums[0]}/{nums[1]}"
# Only then apply the parenthesization pattern