Optimal Division - Video Solutions

Problem Overview: You get an array nums representing a chain of divisions: nums[0] / nums[1] / nums[2] / ... / nums[n-1]. You can insert parentheses anywhere to change evaluation order. The goal is to return a string expression that produces the maximum possible value.

Approach 1: Brute Force Parenthesization with Recursion/DP (Time: O(n^3), Space: O(n^2))

This problem resembles the classic matrix-chain multiplication structure. For every subarray [i, j], compute the minimum and maximum value obtainable by placing parentheses in different ways. Try every split point k where the expression becomes (i..k) / (k+1..j). Because division is not associative, both min and max values must be tracked for each interval. Dynamic programming reduces repeated work by storing results in a table. This guarantees correctness but requires evaluating many partitions and storing intermediate expressions, which makes it heavier than needed for this specific problem.

Approach 2: Greedy Observation (Time: O(n), Space: O(1))

The key insight comes from the behavior of division. For a / b / c / d, the result equals ((a / b) / c) / d by default. However, placing parentheses as a / (b / c / d) makes the denominator smaller because you divide b by multiple numbers. A smaller denominator increases the overall value. Therefore, to maximize the result, keep the first number outside and group all remaining numbers inside a single denominator.

If the array length is 1, return the number. If it is 2, simply return a/b. When the array has three or more elements, construct the expression nums[0] / (nums[1] / nums[2] / ... / nums[n-1]). This structure guarantees the largest value without exploring all parenthesis placements. Implementation is straightforward: iterate through the array and build the string while inserting one opening parenthesis after the first division and a closing parenthesis at the end.

This greedy rule works because every additional division inside the denominator shrinks it, which increases the overall quotient. The solution relies on simple math properties rather than heavy dynamic programming.

Recommended for interviews: Interviewers expect the greedy observation. Starting with the brute-force or DP reasoning shows you understand the search space of parenthesization. Recognizing that all numbers after the first should be grouped into a single denominator demonstrates the optimization insight and leads to the O(n) solution.