LeetCode 553 - Optimal Division

The problem gives us an array of positive integers, and the array elements are combined using division operations in order from left to right. For example, if the input is: the default expression becomes: which evaluates as: because division is left associative.

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LeetCode Problem 553

Difficulty: 🟡 Medium
Topics: Array, Math, Dynamic Programming

Solution

Problem Understanding

The problem gives us an array of positive integers, and the array elements are combined using division operations in order from left to right.

For example, if the input is:

[2, 3, 4]

the default expression becomes:

2/3/4

which evaluates as:

(2/3)/4

because division is left associative.

The key part of the problem is that we are allowed to insert parentheses anywhere we want in order to maximize the final numerical result. We are not changing the order of the numbers, only the grouping of operations.

The output is not the numerical value itself. Instead, we must return the expression string that produces the maximum value.

The problem also states that the returned expression should not contain redundant parentheses. This means parentheses should only appear where they actually change evaluation order.

The constraints are very small:

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

Since the array length is at most 10, even exponential or dynamic programming solutions are technically feasible. However, the mathematical structure of division allows us to derive a much simpler optimal solution.

Several important edge cases immediately stand out:

  • A single number, where no division exists at all.
  • Two numbers, where parentheses are unnecessary because there is only one possible expression.
  • Three or more numbers, where parentheses can significantly change the result.
  • Avoiding redundant parentheses such as "1000/((100/10)/2)", because the inner grouping already determines the order.

Another important guarantee is that there is only one optimal expression. This means we do not need to worry about ties.

Approaches

Brute Force Approach

A brute force solution would try every possible way to parenthesize the division expression.

This is similar to the classic "Different Ways to Add Parentheses" problem. For every subarray, we could recursively compute:

  • The maximum value obtainable from that subexpression
  • The minimum value obtainable from that subexpression
  • The corresponding expression strings

For every split point, we would combine the left and right results using division.

This works because every valid parenthesization is explored, so the optimal answer must eventually be found.

However, the number of possible parenthesizations grows exponentially. The number of binary expression trees corresponds to Catalan numbers, which become large quickly. Even though n <= 10 makes this technically manageable, it is far more complicated than necessary.

Key Insight

The critical observation is based on how division behaves mathematically.

Suppose we have:

a / b / c / d

We want to maximize the result.

Division by a smaller number produces a larger result. Therefore, after the first number, we want the denominator to become as small as possible.

The best strategy is:

a / (b / c / d / ...)

Why?

Because:

b / c / d

evaluates to a relatively small number, and dividing a by a smaller denominator maximizes the final value.

For example:

1000 / (100 / 10 / 2)
= 1000 / ((100 / 10) / 2)
= 1000 / 5
= 200

Any other parenthesization produces a smaller denominator reduction effect.

This leads to a remarkably simple optimal solution:

  • If there is only one number, return it directly.
  • If there are two numbers, return "a/b".
  • If there are three or more numbers, place all numbers after the first inside one pair of parentheses.

Approach Comparison

Approach Time Complexity Space Complexity Notes
Brute Force O(Catalan(n)) O(Catalan(n)) Explores all valid parenthesizations recursively
Optimal O(n) O(n) Uses mathematical observation about minimizing the denominator

Algorithm Walkthrough

  1. Check whether the array contains only one number.

If the array length is 1, there is no division operation at all. The answer is simply the number itself converted to a string. 2. Check whether the array contains exactly two numbers.

With only two numbers, there is only one possible expression:

a/b

Parentheses cannot change anything, so we return the straightforward division string. 3. Handle the general case where the array has three or more numbers.

The optimal structure is always:

nums[0] / (nums[1] / nums[2] / ... / nums[n-1])

This minimizes the denominator and therefore maximizes the total result. 4. Construct the expression string.

Convert each number to a string and join them using '/'.

The numerator is simply:

str(nums[0])

The denominator portion is:

"/".join(...)

wrapped inside parentheses. 5. Return the final expression.

Why it works

The expression:

a / x

becomes larger when x becomes smaller.

All numbers are positive integers greater than 1. Therefore, to maximize the overall result, we must minimize the denominator.

For any sequence:

b / c / d / ...

the smallest possible value occurs when evaluation proceeds left to right:

((b / c) / d) / ...

Wrapping the entire suffix in parentheses forces exactly this evaluation order. As a result, the denominator becomes minimal, and the overall expression becomes maximal.

Python Solution

from typing import List

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]}"

        denominator = "/".join(str(num) for num in nums[1:])

        return f"{nums[0]}/({denominator})"

The implementation follows the mathematical observation directly.

First, we compute the array length because the behavior depends entirely on how many numbers exist.

If there is only one number, we return it immediately as a string. No operators or parentheses are needed.

If there are two numbers, we simply return the standard division format. Adding parentheses would be redundant because there is only one operation.

For arrays with three or more numbers, we build the denominator by joining all elements after the first with '/'. This creates a string like:

100/10/2

We then wrap this entire denominator inside parentheses and prepend the first number:

1000/(100/10/2)

The implementation is concise because the mathematical proof eliminates the need for recursion or dynamic programming.

Go Solution

package main

import (
	"strconv"
	"strings"
)

func optimalDivision(nums []int) string {
	n := len(nums)

	if n == 1 {
		return strconv.Itoa(nums[0])
	}

	if n == 2 {
		return strconv.Itoa(nums[0]) + "/" + strconv.Itoa(nums[1])
	}

	parts := make([]string, 0, n-1)

	for _, num := range nums[1:] {
		parts = append(parts, strconv.Itoa(num))
	}

	denominator := strings.Join(parts, "/")

	return strconv.Itoa(nums[0]) + "/(" + denominator + ")"
}

The Go solution mirrors the Python logic closely.

Since Go does not provide Python-style string interpolation, we use strconv.Itoa() to convert integers into strings.

We use a slice of strings to collect the denominator components, then combine them with strings.Join().

There are no overflow concerns because we never actually compute the numerical result. We only construct the expression string.

Worked Examples

Example 1

Input:

nums = [1000, 100, 10, 2]

Step-by-step execution

Step Action Result
1 Length is 4 Use general case
2 First number becomes numerator "1000"
3 Join remaining numbers "100/10/2"
4 Wrap denominator in parentheses "(100/10/2)"
5 Construct final expression "1000/(100/10/2)"

Evaluation

1000 / ((100 / 10) / 2)
= 1000 / (10 / 2)
= 1000 / 5
= 200

Example 2

Input:

nums = [2, 3, 4]

Step-by-step execution

Step Action Result
1 Length is 3 Use general case
2 Numerator "2"
3 Join remaining numbers "3/4"
4 Add parentheses "(3/4)"
5 Final expression "2/(3/4)"

Evaluation

2 / (3 / 4)
= 2 * (4 / 3)
= 8 / 3
≈ 2.667

Complexity Analysis

Measure Complexity Explanation
Time O(n) Each number is processed once while building the output string
Space O(n) The output string and intermediate joined strings require linear space

The algorithm is linear because we only traverse the array once to build the denominator string. No recursion, dynamic programming table, or repeated computation is needed.

The space complexity is also linear because the resulting expression itself contains all numbers and operators.

Test Cases

from typing import List

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]}"

        denominator = "/".join(str(num) for num in nums[1:])

        return f"{nums[0]}/({denominator})"

solution = Solution()

assert solution.optimalDivision([1000, 100, 10, 2]) == "1000/(100/10/2)"  # provided example
assert solution.optimalDivision([2, 3, 4]) == "2/(3/4)"  # provided example

assert solution.optimalDivision([5]) == "5"  # single element
assert solution.optimalDivision([8, 2]) == "8/2"  # two elements

assert solution.optimalDivision([6, 2, 3]) == "6/(2/3)"  # smallest nontrivial case
assert solution.optimalDivision([10, 5, 2, 1]) == "10/(5/2/1)"  # multiple divisions

assert solution.optimalDivision([100, 10, 10, 10, 10]) == "100/(10/10/10/10)"  # longer chain
assert solution.optimalDivision([2, 2, 2, 2, 2, 2]) == "2/(2/2/2/2/2)"  # repeated values

Test Summary

Test Why
[1000,100,10,2] Validates official example
[2,3,4] Validates minimal parenthesized case
[5] Tests single-element edge case
[8,2] Tests two-element edge case
[6,2,3] Smallest case requiring parentheses
[10,5,2,1] Tests longer denominator chain
[100,10,10,10,10] Verifies formatting with many elements
[2,2,2,2,2,2] Ensures repeated values are handled correctly

Edge Cases

A very important edge case is when the input contains only one number. In this scenario, there are no division operators at all, so adding parentheses makes no sense. A buggy implementation might accidentally add unnecessary formatting like "5()" or attempt to join nonexistent denominator elements. The implementation handles this cleanly by returning the number immediately.

Another critical edge case occurs when the array contains exactly two numbers. Some implementations incorrectly add parentheses and produce expressions like "8/(2)". While mathematically correct, the problem explicitly forbids redundant parentheses. The solution avoids this by treating the two-element case separately.

A more subtle edge case is ensuring the denominator includes all remaining numbers inside one pair of parentheses. A naive implementation might only parenthesize part of the suffix, such as:

1000/(100/10)/2

This does not minimize the denominator and therefore does not maximize the result. The implementation correctly groups the entire suffix after the first number.

Repeated values are another useful edge case because they make it harder to visually verify correctness. For example:

[2,2,2,2]

The algorithm still works because the mathematical principle depends on operation structure, not distinct values.