Minimum Cost to Split into Ones - Solution & Explanation

Problem Overview: You are given an integer n. In one operation you split a number into two positive integers whose sum equals the original number. Each split incurs a cost equal to the value being split. The goal is to keep splitting until every piece becomes 1, while minimizing the total cost.

Approach 1: Dynamic Programming Simulation (O(n^2) time, O(n) space)

A straightforward way to reason about the problem is to treat it as a dynamic programming task. Let dp[x] represent the minimum cost to split the value x into all ones. For every x, try every possible split x = a + b. The total cost becomes x + dp[a] + dp[b] because the act of splitting x costs x. Iterate through all partitions and keep the minimum. This method clearly models the optimal substructure but quickly becomes inefficient since every state tries up to x-1 partitions.

Approach 2: Mathematical Insight (O(1) time, O(1) space)

The key observation comes from looking at how many splits are actually required. To convert a single number n into n ones, you must perform exactly n - 1 splits. Every split increases the number of pieces by exactly one. Starting with one piece and ending with n pieces means n - 1 operations regardless of how you partition the numbers.

Each operation costs the value being split. If you examine any valid sequence of splits, every intermediate piece ultimately contributes exactly once to the total cost before it breaks down further. Summing across the process leads to a deterministic total: the minimum total cost simplifies to n - 1. This eliminates the need for simulation, recursion, or state storage. The result can be computed directly with simple arithmetic, making it a pure math observation rather than a full dynamic programming implementation.

Recommended for interviews: Start by describing the DP formulation to show you understand the optimal substructure and recurrence. Then derive the mathematical simplification. Interviewers usually expect the constant-time solution because recognizing the invariant (exactly n-1 splits) demonstrates stronger problem-solving intuition.