LeetCode Problem 3857

Difficulty: 🟡 Medium
Topics: Math, Dynamic Programming

Solution

LeetCode 3857 - Minimum Cost to Split into Ones

Problem Understanding

We are given a single integer n. Starting with this integer, we repeatedly perform split operations until every resulting piece is equal to 1.

A split operation takes some integer x and divides it into two positive integers a and b such that:

$$a + b = x$$

The cost of performing this split is:

$$a \times b$$

Our goal is to choose the sequence of splits that minimizes the total accumulated cost.

The final state must consist of exactly n ones, because the sum of all pieces is preserved throughout the process. Since we start with value n, eventually we must obtain:

$$1 + 1 + \cdots + 1 = n$$

The output is the minimum possible total cost required to achieve this.

The constraint is:

• 1 <= n <= 500

This is a relatively small bound, which strongly suggests that a dynamic programming solution is feasible. Even an $O(n^2)$ algorithm would be completely acceptable since $500^2 = 250000$.

An important observation is that every intermediate number can be split in many different ways, and a greedy choice is not obviously correct. The cost paid now affects the sizes of future subproblems, so we must consider all possible partitions.

Several edge cases are worth noting:

• n = 1 requires no operations, so the answer is 0.
• Small values such as n = 2 and n = 3 help verify the recurrence.
• Larger values may have many possible split trees, making brute force impractical.
• The problem guarantees n is positive, so we never need to handle invalid inputs or zero.

Approaches

Brute Force

A natural recursive approach is to consider every possible first split.

For a number x, we can split it into:

$$(1, x-1), (2, x-2), \ldots, (x-1, 1)$$

If we choose a split (a, b), we immediately pay a * b, then recursively solve the two resulting subproblems.

This leads to the recurrence:

$$f(x) = \min_{1 \le a < x} \left( a(x-a) + f(a) + f(x-a) \right)$$

The recursion is correct because every valid splitting process begins with exactly one first split.

However, without memoization, the same values are recomputed many times. The number of recursive states grows exponentially, making the approach far too slow.

Dynamic Programming

The key observation is that the optimal cost for a number depends only on the optimal costs of smaller numbers.

The brute force recurrence already reveals overlapping subproblems:

$$dp[x] = \min_{1 \le a < x} \left( a(x-a) + dp[a] + dp[x-a] \right)$$

Since every subproblem size is smaller than x, we can compute answers in increasing order from 1 to n.

The constraint n <= 500 makes an $O(n^2)$ dynamic programming solution very efficient.

Approach Comparison

Approach	Time Complexity	Space Complexity	Notes
Brute Force	Exponential	O(n) recursion depth	Tries every split tree repeatedly
Optimal Dynamic Programming	O(n²)	O(n)	Computes each state once and tests all splits

Algorithm Walkthrough

1. Create a DP array dp of length n + 1.
2. Define dp[x] as the minimum cost required to split integer x completely into ones.
3. Initialize the base case:

• dp[1] = 0

No operation is needed because the number is already a one. 4. Process values from 2 through n. 5. For each value x, initialize its answer to infinity. 6. Try every possible first split:

• Let a range from 1 to x - 1.
• Let b = x - a.

7. The total cost of choosing this split is:

$$a \times b + dp[a] + dp[b]$$

The first term is the cost of the current split. The remaining terms are the optimal costs for finishing the two resulting pieces. 8. Update:

$$dp[x] = \min(dp[x], a \times b + dp[a] + dp[b])$$ 9. After considering all possible splits, dp[x] stores the optimal answer for value x. 10. Return dp[n].

Why it works

Any valid splitting process for a number x must begin with exactly one split into two positive integers a and b. Once that first split is chosen, the remaining work is independent: we must optimally split a into ones and optimally split b into ones. Therefore every valid solution has cost:

$$a \times b + dp[a] + dp[b]$$

Taking the minimum over all possible first splits guarantees that dp[x] equals the minimum achievable cost. Since all smaller states are computed before larger ones, the dynamic programming table correctly evaluates the recurrence for every value up to n.

Python Solution

class Solution:
    def minCost(self, n: int) -> int:
        dp = [0] * (n + 1)

        for x in range(2, n + 1):
            best = float("inf")

            for a in range(1, x):
                b = x - a
                best = min(best, a * b + dp[a] + dp[b])

            dp[x] = best

        return dp[n]

The implementation directly follows the recurrence.

The array dp stores the optimal answer for every value from 1 to n. Since dp[1] is already zero, no explicit initialization is required beyond creating the array.

For each integer x, the code evaluates every possible first split. The variable best tracks the smallest total cost encountered. The expression a * b + dp[a] + dp[b] represents the cost of making the current split plus the optimal costs of finishing the two resulting subproblems.

After all splits have been checked, dp[x] is assigned the minimum value. Once the table is filled, dp[n] is returned.

Go Solution

func minCost(n int) int {
	dp := make([]int, n+1)

	for x := 2; x <= n; x++ {
		best := 1 << 60

		for a := 1; a < x; a++ {
			b := x - a
			cost := a*b + dp[a] + dp[b]

			if cost < best {
				best = cost
			}
		}

		dp[x] = best
	}

	return dp[n]
}

The Go implementation mirrors the Python solution closely. A slice is used for the DP table. The value 1 << 60 serves as a practical infinity value. Integer overflow is not a concern because n <= 500, so all intermediate costs remain far below Go's integer limits.

Worked Examples

Example 1: n = 3

Initially:

x	dp[x]
1	0

Compute dp[2].

Split	Cost
1 + 1	1×1 + 0 + 0 = 1

Therefore:

x	dp[x]
2	1

Compute dp[3].

Split	Calculation	Cost
1 + 2	1×2 + 0 + 1	3
2 + 1	2×1 + 1 + 0	3

Minimum cost is 3.

Final table:

x	dp[x]
1	0
2	1
3	3

Answer: 3

Example 2: n = 4

Existing values:

x	dp[x]
1	0
2	1
3	3

Compute dp[4].

Split	Calculation	Cost
1 + 3	1×3 + 0 + 3	6
2 + 2	2×2 + 1 + 1	6
3 + 1	3×1 + 3 + 0	6

Minimum cost is 6.

Final table:

x	dp[x]
1	0
2	1
3	3
4	6

Answer: 6

Interestingly, every split produces the same optimal value in this case.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n²)	For each value x, all x - 1 possible splits are examined
Space	O(n)	The DP table stores one value for each integer from 1 to n

The outer loop runs from 2 through n. For each state x, the algorithm iterates through all possible first splits. Summing these iterations gives:

$$\sum_{x=2}^{n}(x-1) = \frac{n(n-1)}{2}$$

which is $O(n^2)$. The only additional memory used is the DP array of size n + 1, resulting in $O(n)$ space complexity.

Test Cases

sol = Solution()

assert sol.minCost(1) == 0      # already a single one
assert sol.minCost(2) == 1      # one split: 1+1
assert sol.minCost(3) == 3      # example 1
assert sol.minCost(4) == 6      # example 2

assert sol.minCost(5) == 10     # continues observed pattern
assert sol.minCost(6) == 15     # triangular number result
assert sol.minCost(7) == 21     # larger small case
assert sol.minCost(8) == 28     # power of two

assert sol.minCost(10) == 45    # medium-sized input
assert sol.minCost(100) == 4950 # larger input
assert sol.minCost(500) == 124750 # upper constraint boundary

Test Summary

Test	Why
n = 1	Base case, no splitting needed
n = 2	Smallest nontrivial split
n = 3	Verifies example 1
n = 4	Verifies example 2
n = 5	Checks transition beyond examples
n = 6	Tests multiple competing split choices
n = 7	Odd-sized value
n = 8	Even-sized value, many possible split trees
n = 10	Medium-sized input
n = 100	Larger DP computation
n = 500	Maximum constraint

Edge Cases

Edge Case 1: n = 1

This is the smallest possible input. Since the number is already a single one, no operations are required. A common bug is accidentally forcing at least one split and returning a positive cost. The implementation correctly sets dp[1] = 0 and never enters the transition loop.

Edge Case 2: Very Small Values

Inputs such as n = 2 and n = 3 are important because they validate the recurrence. Off by one errors in the split loop often appear here. The implementation iterates a from 1 through x - 1, ensuring every legal split is considered exactly once.

Edge Case 3: Maximum Input Size

For n = 500, there are many possible split sequences. A naive recursive solution would become prohibitively expensive because of repeated subproblem evaluation. The dynamic programming approach stores each state once and performs only $O(n^2)$ work, remaining easily within limits.

Edge Case 4: Symmetric Splits

Values such as n = 4, where (1,3) and (2,2) are both legal first splits, can expose bugs in recurrence implementation. The algorithm explicitly evaluates every possible partition and chooses the minimum, guaranteeing that no potentially optimal split is overlooked.

Edge Case 5: Odd and Even Numbers

Odd values and even values have different partition structures. For example, 7 can never be split into two equal halves, while 8 can. The recurrence does not rely on parity assumptions and examines all valid partitions, so both cases are handled uniformly and correctly.