LeetCode 553: Optimal Division Solution in Python – A Step-by-Step Guide

Imagine you’re given a list of numbers—like [1000, 100, 10, 2]—and your task is to craft a division expression using all of them, adding parentheses anywhere, to get the largest possible result, such as turning it into "1000/(100/10/2)" to maximize the value to 2000. That’s the clever challenge of LeetCode 553: Optimal Division, a medium-level problem that’s a fantastic way to practice mathematical intuition and string manipulation in Python. We’ll explore two solutions: the Best Solution, a greedy string construction that’s efficient and insightful, and an Alternative Solution, a recursive division with evaluation that’s thorough but less practical. With detailed examples, clear code, and a friendly tone—especially for the greedy approach—this guide will help you optimize that division, whether you’re new to coding or leveling up. Let’s divide those numbers and start maximizing!

What Is LeetCode 553: Optimal Division?

In LeetCode 553: Optimal Division, you’re given an array of positive integers nums, and your task is to return a string representing a division expression that uses all numbers exactly once, with optional parentheses, to yield the maximum possible value when evaluated. For example, with nums = [1000, 100, 10, 2], the optimal expression is "1000/(100/10/2)" = 2000, far better than "1000/100/10/2" = 0.5. This problem builds on LeetCode 227: Basic Calculator II for expression handling but focuses on maximizing through strategic grouping.

Problem Statement

Input: nums (List[int])—array of positive integers.
Output: String—division expression yielding maximum value.
Rules: Use each number once; add parentheses to maximize result; division is left-associative.

Constraints

2 <= nums.length <= 10
2 <= nums[i] <= 1000

Examples

Input: nums = [1000,100,10,2]

Output: "1000/(100/10/2)"
Evaluates to 1000 / (100 / 10 / 2) = 1000 / (100 / 5) = 1000 / 20 = 2000.

Input: nums = [2,3,4]

Output: "2/(3/4)"
Evaluates to 2 / (3 / 4) = 2 / 0.75 = 2.666..., best possible.

Input: nums = [2,3]

Output: "2/3"
No parentheses needed, evaluates to 2 / 3 ≈ 0.666....

Understanding the Problem: Maximizing Division

To solve LeetCode 553: Optimal Division in Python, we need to construct a division expression using all numbers in nums that maximizes the result, leveraging parentheses to alter the order of operations. Division is left-associative (e.g., 1000/100/10 = (1000/100)/10), and without parentheses, the value decreases as more numbers divide the first. With n up to 10, a brute-force approach checking all parenthesizations (exponential) is feasible but unnecessary once we recognize a key insight: maximizing the result means making the numerator as large as possible (first number) and the denominator as small as possible (all others grouped). We’ll explore:

Best Solution (Greedy String Construction): O(n) time, O(n) space—fast and optimal.
Alternative Solution (Recursive Division with Evaluation): O(2ⁿ) time, O(n) space—thorough but slow.

Let’s dive into the greedy solution with a friendly breakdown!

Best Solution: Greedy String Construction

Why Greedy Wins

The greedy string construction is the best for LeetCode 553 because it achieves the maximum value in O(n) time and O(n) space by using a simple, optimal strategy: keep the first number as the numerator and group all remaining numbers in parentheses as the denominator (e.g., "x₁/(x₂/x₃/.../xₙ)"), minimizing the denominator’s value efficiently without needing to evaluate all possibilities. It’s like setting up a division race where the leader stays ahead by shrinking the competition!

How It Works

Think of this as a division optimizer:

Step 1: Insight:

For nums = [x₁, x₂, ..., xₙ], without parentheses, result = x₁/x₂/.../xₙ (small).
With parentheses, group all but first: x₁ / (x₂/x₃/.../xₙ), making denominator smallest.

Step 2: Handle Base Cases:

n = 2: "x₁/x₂" (no parentheses needed).
n = 1: "x₁" (invalid per constraints, but handled).

Step 3: Construct String:

First number, then "/" + "(" + remaining numbers joined with "/" + ")".

Step 4: Return Expression:

String yielding max value.

Why It Works:

Minimizing denominator maximizes result (all xᵢ ≥ 2).
Greedy choice is mathematically optimal.

It’s like a division-maximizing strategist!

Step-by-Step Example

Example: nums = [1000, 100, 10, 2]

Step 1: Insight:

Without parens: 1000/100/10/2 = ((1000/100)/10)/2 = 0.5.
Goal: Maximize by grouping denominator.

Step 2: Construct:

First = "1000".
Rest = [100, 10, 2] → "100/10/2".
Expression = "1000/(100/10/2)".

Step 3: Verify:

100/10/2 = (100/10)/2 = 5, then 1000/5 = 2000.
Compare: 1000/100/10/2 = 0.5, much smaller.

Result: "1000/(100/10/2)".

Example: nums = [2, 3]

Step 1: n = 2, no grouping needed beyond simple division.
Step 2: "2/3".
Result: "2/3".

Code with Detailed Line-by-Line Explanation

Here’s the Python code, explained for beginners:

class Solution:
    def optimalDivision(self, nums: List[int]) -> str:
        # Step 1: Handle base cases
        if len(nums) == 1:
            return str(nums[0])
        if len(nums) == 2:
            return f"{nums[0]}/{nums[1]}"

        # Step 2: Construct optimal expression
        first = str(nums[0])
        rest = "/".join(str(num) for num in nums[1:])
        return f"{first}/({rest})"

Lines 4-5: Single number case, return as string.
Lines 6-7: Two numbers, simple division.
Lines 10-12: General case:

First number as numerator.
Join rest with "/" in parentheses.

Time Complexity: O(n)—join operation over array.
Space Complexity: O(n)—string storage.

It’s like a division-string crafter!

Alternative Solution: Recursive Division with Evaluation

Why an Alternative Approach?

The recursive division with evaluation explores all possible ways to parenthesize the division expression, evaluating each to find the maximum, running in O(2ⁿ) time and O(n) space (recursion stack). It’s thorough but impractical for larger n, making it a good alternative for understanding the problem’s full scope or small inputs.

How It Works

Picture this as a division-experiment lab:

Step 1: Recursively split array at each point.
Step 2: Try all parenthesizations, compute values.
Step 3: Track max value and expression.
Step 4: Return optimal string.

It’s like a division-max explorer!

Step-by-Step Example

Example: nums = [2, 3, 4]

DFS:

2/3/4: 0.166...
2/(3/4): 2 / 0.75 = 2.666...
(2/3)/4: 0.0416...

Max: "2/(3/4)" = 2.666....
Result: "2/(3/4)".

Code for Recursive Approach

class Solution:
    def optimalDivision(self, nums: List[int]) -> str:
        def dfs(start, end):
            if start == end:
                return (str(nums[start]), nums[start])
            if start + 1 == end:
                expr = f"{nums[start]}/{nums[end]}"
                return (expr, nums[start] / nums[end])

            max_val = float('-inf')
            max_expr = ""
            for i in range(start, end):
                left_expr, left_val = dfs(start, i)
                right_expr, right_val = dfs(i + 1, end)
                curr_val = left_val / right_val
                curr_expr = f"{left_expr}/{right_expr}"
                if curr_val > max_val:
                    max_val = curr_val
                    max_expr = curr_expr
                # Try grouping right
                if i + 2 <= end:
                    sub_expr, sub_val = dfs(i + 1, end)
                    curr_val = nums[start] / sub_val
                    curr_expr = f"{nums[start]}/({sub_expr})"
                    if curr_val > max_val:
                        max_val = curr_val
                        max_expr = curr_expr

            return (max_expr, max_val)

        result, _ = dfs(0, len(nums) - 1)
        return result

Lines 4-7: Base cases for 1 or 2 numbers.
Lines 10-25: Recursively split, evaluate, track max.
Line 28: Start DFS, return expression.
Time Complexity: O(2ⁿ)—exponential splits.
Space Complexity: O(n)—recursion stack.

It’s a recursive division tester!

Comparing the Two Solutions

Greedy (Best):

Pros: O(n), O(n), optimal and fast.
Cons: Insight-based.

Recursive (Alternative):

Pros: O(2ⁿ), O(n), exhaustive.
Cons: Impractical for large n.

Greedy wins for efficiency!

Additional Examples and Edge Cases

[5]: "5".
[10, 2, 5]: "10/(2/5)".
[2, 4]: "2/4".

Greedy handles them all!

Complexity Recap

Greedy: Time O(n), Space O(n).
Recursive: Time O(2ⁿ), Space O(n).

Greedy’s the speed champ!

Key Takeaways

Greedy: Insightful optimization—learn at Python Basics!
Recursive: Full exploration.
Division: Math is fun.
Python: Greedy or recurse, your pick!

Final Thoughts: Maximize That Division!

LeetCode 553: Optimal Division in Python is a clever math challenge. Greedy string construction is your fast track, while recursive evaluation offers a thorough dive. Want more? Try LeetCode 227: Basic Calculator II or LeetCode 224: Basic Calculator. Ready to divide? Head to Solve LeetCode 553 on LeetCode and optimize that expression today!