Optimal Division

Optimal Division

What is this problem about?

The Optimal Division problem gives you a list of positive integers. Place division operators between them to maximize the result. The key insight: to maximize a/b/c/.../z, you want the numerator as large as possible and the denominator as small as possible. The optimal solution is always nums[0] / (nums[1] / nums[2] / ... / nums[n-1]) = nums[0] * nums[2] * ... / nums[1]. This is a greedy math insight disguised as a DP problem.

Why is this asked in interviews?

Amazon asks this to test whether candidates recognize the mathematical insight and avoid unnecessary DP computation. The array, math, and dynamic programming interview pattern is mentioned but the optimal solution is actually a greedy observation.

Algorithmic pattern used

Mathematical observation. For length 1: just return nums[0]. For length 2: nums[0]/nums[1]. For length ≥ 3: always nums[0] / (nums[1]/nums[2]/.../nums[n-1]). The denominator becomes the product of nums[2..n-1] divided by nums[1], which is minimized by grouping everything after nums[0] in parentheses as a denominator. Return formatted string: "nums[0]/(nums[1]/nums[2]/.../nums[n-1])".

Example explanation

nums=[1000,100,10,2]. Optimal: 1000/(100/10/2) = 1000/(100/10/2) = 1000/(5) = 200. Or: 1000/100/10/2 = 0.5. Clearly 1000/(100/10/2) is much larger. Output: "1000/(100/10/2)".

Common mistakes candidates make

• Actually implementing DP (unnecessary — the greedy solution is always the same pattern).
• Wrong output format (must include parentheses around the denominator subexpression).
• Not handling edge cases (n=1: return "nums[0]", n=2: return "nums[0]/nums[1]").
• Proving correctness incorrectly (divide by fraction = multiply by reciprocal).

Interview preparation tip

Optimal Division is a "mathematical insight beats algorithmic complexity" problem. Before implementing DP, always ask: "Is there a closed-form pattern?" Here, dividing by a chain of divisions gives maximum result when all but the first are in the denominator. Proving this algebraically (1/a/b = 1/(a*b) reduces denominator to minimum) is the key step. Practice recognizing when math shortcuts make DP unnecessary.