Overflow in intermediate arithmetic
Wrong move: Temporary multiplications exceed integer bounds.
Usually fails on: Large inputs wrap around unexpectedly.
Fix: Use wider types, modular arithmetic, or rearranged operations.
Build confidence with an intuition-first walkthrough focused on math fundamentals.
A valid cut in a circle can be:
Some valid and invalid cuts are shown in the figures below.
Given the integer n, return the minimum number of cuts needed to divide a circle into n equal slices.
Example 1:
Input: n = 4 Output: 2 Explanation: The above figure shows how cutting the circle twice through the middle divides it into 4 equal slices.
Example 2:
Input: n = 3 Output: 3 Explanation: At least 3 cuts are needed to divide the circle into 3 equal slices. It can be shown that less than 3 cuts cannot result in 3 slices of equal size and shape. Also note that the first cut will not divide the circle into distinct parts.
Constraints:
1 <= n <= 100Problem summary: A valid cut in a circle can be: A cut that is represented by a straight line that touches two points on the edge of the circle and passes through its center, or A cut that is represented by a straight line that touches one point on the edge of the circle and its center. Some valid and invalid cuts are shown in the figures below. Given the integer n, return the minimum number of cuts needed to divide a circle into n equal slices.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Math
4
3
smallest-even-multiple)count-total-number-of-colored-cells)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #2481: Minimum Cuts to Divide a Circle
class Solution {
public int numberOfCuts(int n) {
return n > 1 && n % 2 == 1 ? n : n >> 1;
}
}
// Accepted solution for LeetCode #2481: Minimum Cuts to Divide a Circle
func numberOfCuts(n int) int {
if n > 1 && n%2 == 1 {
return n
}
return n >> 1
}
# Accepted solution for LeetCode #2481: Minimum Cuts to Divide a Circle
class Solution:
def numberOfCuts(self, n: int) -> int:
return n if (n > 1 and n & 1) else n >> 1
// Accepted solution for LeetCode #2481: Minimum Cuts to Divide a Circle
impl Solution {
pub fn number_of_cuts(n: i32) -> i32 {
if n > 1 && n % 2 == 1 {
return n;
}
n >> 1
}
}
// Accepted solution for LeetCode #2481: Minimum Cuts to Divide a Circle
function numberOfCuts(n: number): number {
return n > 1 && n & 1 ? n : n >> 1;
}
Use this to step through a reusable interview workflow for this problem.
Simulate the process step by step — multiply n times, check each number up to n, or iterate through all possibilities. Each step is O(1), but doing it n times gives O(n). No extra space needed since we just track running state.
Math problems often have a closed-form or O(log n) solution hidden behind an O(n) simulation. Modular arithmetic, fast exponentiation (repeated squaring), GCD (Euclidean algorithm), and number theory properties can dramatically reduce complexity.
Review these before coding to avoid predictable interview regressions.
Wrong move: Temporary multiplications exceed integer bounds.
Usually fails on: Large inputs wrap around unexpectedly.
Fix: Use wider types, modular arithmetic, or rearranged operations.