Leetcode 2481: Minimum Cuts to Divide a Circle
You are given a circle, and you need to divide it into ’n’ equal slices using the minimum number of straight cuts. A valid cut can either be through the center of the circle touching two points on the edge or touching just one point and the center.
Problem
Approach
Steps
Complexity
Input: The input is a single integer, n, which represents the number of slices the circle should be divided into.
Example: n = 6
Constraints:
• 1 <= n <= 100
Output: Return the minimum number of cuts needed to divide the circle into 'n' equal slices.
Example: Output: 3
Constraints:
• The output should be a single integer representing the minimum cuts needed.
Goal: The goal is to determine the fewest number of cuts needed to divide a circle into exactly 'n' equal parts using valid cuts.
Steps:
• If n is 1, no cuts are required.
• For even numbers, cuts can be made in half, and for odd numbers, each cut should be made to increase the number of equal slices incrementally.
Goal: You are guaranteed that 1 <= n <= 100.
Steps:
• 1 <= n <= 100
Assumptions:
• The circle is perfectly symmetric, and the goal is to divide it into equal slices.
• Input: Example 2: n = 5
• Explanation: Since 5 is an odd number, 5 cuts are needed to divide the circle into 5 equal parts, as no other number of cuts can evenly divide it.
Approach: The solution involves calculating the minimum number of cuts based on whether the number of slices is odd or even.
Observations:
• If n is even, a smaller number of cuts can be used by cutting through the center of the circle symmetrically.
• If n is odd, each cut must be added incrementally to get the required number of slices.
• For even values of n, the answer is n/2. For odd values of n, the answer is simply n.
Steps:
• Check if n is 1. If true, return 0 cuts.
• If n is even, return n / 2 as the minimum cuts.
• If n is odd, return n as the minimum cuts.
Empty Inputs:
• n will always be greater than or equal to 1, so no empty inputs are possible.
Large Inputs:
• The largest value for n is 100, which is manageable in terms of time complexity.
Special Values:
• If n = 1, no cuts are needed.
Constraints:
• Ensure that n always falls within the bounds of 1 to 100.
int numberOfCuts(int n) {
if (n == 1) return 0;
return n % 2 ? n : n / 2;
}
1 : Function Declaration
int numberOfCuts(int n) {
This is the declaration of the function `numberOfCuts`, which takes an integer `n` as input and returns the number of cuts (an integer) needed to divide `n` into pieces.
2 : Base Case
if (n == 1) return 0;
Checks if `n` is 1. If it is, no cuts are needed, so the function returns 0.
3 : Return Statement
return n % 2 ? n : n / 2;
If `n` is odd, the function returns `n` itself (since no further division is possible). If `n` is even, it returns `n / 2`, indicating that the number of cuts required is half of `n`.
Best Case: O(1)
Average Case: O(1)
Worst Case: O(1)
Description: The time complexity is O(1) since we are simply performing a constant-time check on whether n is odd or even.
Best Case: O(1)
Worst Case: O(1)
Description: The space complexity is O(1) because we only use a constant amount of space for the computation.
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