LeetCode 2481 - Minimum Cuts to Divide a Circle

LeetCode Problem 2481

Difficulty: 🟢 Easy
Topics: Math, Geometry

Solution

Problem Understanding

The problem asks for the minimum number of straight cuts required to divide a circle into n equal slices. A valid cut can either pass through the center and touch two points on the circle, or touch one point on the circle and pass through the center. The input is a single integer n representing the desired number of slices, and the output is the minimum number of cuts required to achieve that.

Effectively, the problem is about determining how many lines are necessary to partition a circle symmetrically. For n = 1, no cut is needed because the circle is already a single slice. For n = 2, one straight cut through the center suffices. For other values, the problem reduces to checking whether the number of slices is even or odd, as this affects whether a single line can divide multiple slices at once or additional cuts are required.

The constraints are simple: 1 <= n <= 100. This means the input size is small, so we can solve the problem with a simple arithmetic or conditional approach without worrying about performance or memory.

Important edge cases include n = 1 (no cut), n = 2 (single cut), and n odd versus even, because odd slices require a different approach than even slices.

Approaches

The brute-force approach would try to simulate cutting the circle repeatedly, checking after each cut how many slices have been produced, and continue until reaching n. While this works for small n, it is unnecessarily complex and inefficient because the pattern of required cuts is predictable based on whether n is even or odd.

The key insight is that a circle can always be divided into even slices using n / 2 straight cuts through the center. For odd n greater than 1, we first make one cut to separate the circle and then require n cuts to reach n slices. Therefore, the solution reduces to a simple formula: 0 if n == 1, n if n is odd, and n / 2 if n is even.

Approach	Time Complexity	Space Complexity	Notes
Brute Force	O(n)	O(1)	Simulate cutting the circle slice by slice until reaching n
Optimal	O(1)	O(1)	Use parity check: 0 if n=1, n if n odd, n/2 if n even

Algorithm Walkthrough

1. Check if n equals 1. If so, the circle is already one slice and requires 0 cuts. Return 0.
2. Check if n is even. For an even number of slices, each cut through the center can divide the circle into two additional slices, so the total minimum cuts required is n / 2.
3. Handle odd n greater than 1. For odd numbers of slices, a single straight cut through the center cannot evenly split the circle. Therefore, we need as many cuts as slices, n, to ensure each slice is separated correctly.
4. Return the computed minimum cuts.

Why it works: The invariant is based on geometric properties of a circle. Any straight line through the center always divides the circle into two equal parts. For even n, we can systematically place lines at equal angles to achieve n slices with n / 2 cuts. For odd n, each line contributes at most one new slice beyond the first cut, requiring n cuts.

Python Solution

class Solution:
    def numberOfCuts(self, n: int) -> int:
        if n == 1:
            return 0
        if n % 2 == 0:
            return n // 2
        return n

Implementation Walkthrough: The Python solution follows the algorithm exactly. It first checks the special case of n = 1 and returns 0. Then it checks if n is even, returning n // 2. If neither condition is met, n must be odd and greater than 1, so it returns n.

Go Solution

func numberOfCuts(n int) int {
    if n == 1 {
        return 0
    }
    if n%2 == 0 {
        return n / 2
    }
    return n
}

Go-specific Notes: The Go implementation mirrors Python. Division with / is integer division for integers in Go, so n / 2 behaves correctly. No additional handling for odd numbers is needed because modulo % works as expected.

Worked Examples

Example 1: n = 4

n is even, so minimum cuts = n / 2 = 2.

Example 2: n = 3

n is odd and greater than 1, so minimum cuts = n = 3.

Example 3: n = 1

n equals 1, so minimum cuts = 0.

Example	n	n Even/Odd	Min Cuts
1	4	Even	2
2	3	Odd	3
3	1	Special	0

Complexity Analysis

Measure	Complexity	Explanation
Time	O(1)	Single conditional check, constant time
Space	O(1)	No additional memory used beyond input

The reasoning is straightforward. No iteration or recursion is needed, so time and space remain constant regardless of the input n.

Test Cases

# Provided examples
assert Solution().numberOfCuts(4) == 2  # even n
assert Solution().numberOfCuts(3) == 3  # odd n
assert Solution().numberOfCuts(1) == 0  # special case

# Additional boundary and stress tests
assert Solution().numberOfCuts(2) == 1  # smallest even n
assert Solution().numberOfCuts(5) == 5  # small odd n
assert Solution().numberOfCuts(100) == 50  # largest even n
assert Solution().numberOfCuts(99) == 99  # large odd n

Test	Why
4	Even n, simple division
3	Odd n, returns n
1	Special case, no cuts
2	Smallest even n, edge of minimum input
5	Small odd n, basic odd behavior
100	Large even n, test upper limit
99	Large odd n, test upper limit

Edge Cases

n = 1: This is a unique case because no cut is required. If the algorithm incorrectly assumes a cut is always needed, it would fail here. The implementation handles it explicitly with a conditional check.

n even versus n odd: The algorithm must differentiate between even and odd slices because even slices can be achieved with fewer cuts due to symmetry. For odd numbers, each slice effectively requires a cut, which is why returning n for odd values greater than 1 is correct.

Maximum input n = 100: This tests whether the formula scales correctly. The optimal O(1) approach handles this trivially. Edge cases like n = 100 and n = 99 ensure the logic for both even and odd large numbers is correct.

This approach covers all constraints and edge cases while remaining simple and efficient.