Minimum Cuts to Divide a Circle

Minimum Cuts to Divide a Circle - Problem

Math Geometry Easy

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.

Given the integer n, return the minimum number of cuts needed to divide a circle into n equal slices.

Input & Output

Example 1 — Four Equal Slices

$ Input: n = 4

› Output: 4

💡 Note: To divide a circle into 4 equal slices, we need 4 cuts through the center. Each cut creates one slice boundary.

Example 2 — Single Slice

$ Input: n = 1

› Output: 0

💡 Note: For 1 slice (the whole circle), no cuts are needed.

Example 3 — Three Equal Slices

$ Input: n = 3

› Output: 3

💡 Note: To create 3 equal slices, we need 3 cuts, each at 120° angles from each other through the center.

Constraints

1 ≤ n ≤ 10⁶

Visualization

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Asked in

G Google 15 M Microsoft 12

The key insight is that each cut through the center creates exactly one slice boundary. For n equal slices, we need exactly n cuts, except for n=1 which needs 0 cuts. Best approach uses the direct mathematical formula. Time: O(1), Space: O(1)

Common Approaches

✓ Brute Force Analysis

⏱️ Time: O(n) Space: O(1)

Try to enumerate all possible ways to make cuts and determine which gives exactly n equal slices with minimum cuts.

Mathematical Formula

⏱️ Time: O(1) Space: O(1)

Each cut through the center creates exactly one slice. For n equal slices, we need exactly n cuts, except when n=1 (no cuts needed).

Brute Force Analysis — Algorithm Steps

Step 1: Try cutting patterns from 0 to n cuts
Step 2: For each pattern, check if it creates exactly n equal slices
Step 3: Return the minimum cuts that work

Visualization

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Step-by-Step Walkthrough

Try 1 Cut

Test if 1 cut can create n slices

Try 2 Cuts

Test if 2 cuts can create n slices

Continue

Keep testing until we find the answer

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int solution(int n) {
    // Brute force: try different combinations of cuts
    if (n == 1) {
        return 0;
    }
    
    // For n slices, we need exactly n radius cuts
    // Each radius cut creates one slice boundary
    // Try from 1 cut up to n cuts and see when we get n slices
    for (int cuts = 1; cuts <= n; cuts++) {
        // With 'cuts' radius cuts, we get 'cuts' slices
        if (cuts == n) {
            return cuts;
        }
    }
    
    return n;
}

int main() {
    int n;
    scanf("%d", &n);
    int result = solution(n);
    printf("%d\n", result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Would need to test up to n different cutting patterns

✓ Linear Growth

Space Complexity

O(1)

Only using constant extra variables

✓ Linear Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force Analysis — Algorithm Steps

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Code -

Time & Space Complexity

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