Problem Description

You need to divide a circle into n equal slices using the minimum number of cuts.

A valid cut in a circle can be one of two types:

1. A cut that passes through the center and touches two points on the edge of the circle (this creates 2 slices with one cut)
2. A cut that passes through the center and touches only one point on the edge of the circle (this creates 1 slice with one cut)

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

The key insight is that when making cuts through the center:

• If you need an even number of slices, you can make cuts that go completely across the circle (touching two edge points), where each cut creates 2 slices
• If you need an odd number of slices, you cannot use only full diameter cuts, so you need n individual cuts from the center to the edge

For example:

• To get 1 slice (the whole circle): 0 cuts needed
• To get 2 slices: 1 cut across the diameter
• To get 3 slices: 3 cuts from center to edge
• To get 4 slices: 2 cuts across the diameter
• To get 5 slices: 5 cuts from center to edge
• To get 6 slices: 3 cuts across the diameter

The solution follows the pattern:

• When n = 1: return 0 (no cuts needed)
• When n is odd and n > 1: return n (need n individual cuts)
• When n is even: return n/2 (can use diameter cuts)

Intuition

Let's think about how cuts work when they pass through the center of a circle.

When you make a cut that goes completely across the circle (through the center and touching two edge points), you're essentially drawing a diameter. This single cut divides the circle into 2 equal parts. If you make another diameter cut at a different angle, you now have 4 equal parts with just 2 cuts. This pattern continues - with k diameter cuts, you can create 2k equal slices.

This is why even numbers are efficient - to get n slices where n is even, you only need n/2 cuts by using full diameter cuts.

But what happens when we need an odd number of slices? Let's say we want 3 equal slices. If we make one diameter cut, we get 2 slices. Now we need to split one of these halves into two parts while leaving the other half intact. But any cut through the center that enters one half must exit somewhere - if it exits through the opposite edge, it would cut both halves. The only way to cut just one slice is to make a cut from the center to the edge and stop there (a radius cut).

For odd numbers, we can't leverage the efficiency of diameter cuts that create 2 slices at once. We need to make n individual radius cuts from the center, each creating exactly one slice.

The special case is when n = 1, which means we want the whole circle as one slice - this requires 0 cuts since the circle is already a single piece.

This leads us to the formula:

• If n = 1: return 0
• If n is odd and n > 1: return n
• If n is even: return n/2 or n >> 1 (using bit shift for division by 2)

Learn more about Math patterns.

Solution Approach

The implementation uses a simple conditional expression to handle the three distinct cases we identified:

Case 1: When n = 1

• The circle is already a single slice, so no cuts are needed
• Return 0

Case 2: When n is odd and greater than 1

• We cannot use diameter cuts efficiently since each diameter cut creates 2 slices
• We need exactly n radius cuts from the center to the edge
• Return n

Case 3: When n is even

• We can use diameter cuts where each cut creates 2 slices
• We need n/2 diameter cuts
• Return n/2

The solution combines these cases into a single expression:

def numberOfCuts(self, n: int) -> int:
    return n if (n > 1 and n & 1) else n >> 1

Implementation Details:

1. Checking if n is odd: The expression n & 1 uses bitwise AND operation to check if n is odd. This works because:

   • Odd numbers have their least significant bit set to 1
   • n & 1 returns 1 (truthy) for odd numbers, 0 (falsy) for even numbers

2. The conditional logic: n if (n > 1 and n & 1) else n >> 1

   • If n > 1 AND n is odd: return n
   • Otherwise: return n >> 1 (which is n/2 using right bit shift)

3. Handling the special case: When n = 1:

   • The condition (n > 1 and n & 1) evaluates to False
   • So we return n >> 1 which is 1 >> 1 = 0

The mathematical formula for the answer is:

This solution runs in O(1) time complexity and O(1) space complexity, as it only performs a constant number of operations regardless of the input size.

Example Walkthrough

Let's walk through the solution with a few examples to understand how the algorithm works.

Example 1: n = 4 (even number)

• We need to divide the circle into 4 equal slices
• Since 4 is even, we can use diameter cuts (cuts that go completely across the circle)
• Each diameter cut creates 2 slices
• First cut: Creates 2 slices (180° each)
• Second cut: Intersects the first cut at the center, creating 4 slices (90° each)
• Total cuts needed: 4/2 = 2 cuts
• Using our formula: n & 1 evaluates to 4 & 1 = 0 (false), so we return n >> 1 = 4 >> 1 = 2

Example 2: n = 3 (odd number)

• We need to divide the circle into 3 equal slices
• Since 3 is odd, we cannot use only diameter cuts (each creates 2 slices)
• We must make 3 radius cuts from the center to the edge
• First cut: From center to edge at 0°
• Second cut: From center to edge at 120°
• Third cut: From center to edge at 240°
• This creates 3 equal slices of 120° each
• Total cuts needed: 3 cuts
• Using our formula: n > 1 and n & 1 evaluates to 3 > 1 and 3 & 1 = true and 1 = true, so we return n = 3

Example 3: n = 1 (special case)

• We need the whole circle as a single slice
• The circle is already one complete slice
• No cuts are needed
• Total cuts needed: 0 cuts
• Using our formula: n > 1 and n & 1 evaluates to 1 > 1 and 1 & 1 = false and 1 = false, so we return n >> 1 = 1 >> 1 = 0

The algorithm efficiently determines the minimum cuts by recognizing that even numbers can leverage diameter cuts (2 slices per cut), while odd numbers require individual radius cuts (1 slice per cut).

Solution Implementation

Time and Space Complexity

The time complexity is O(1) because the function performs a constant number of operations regardless of the input size. It only involves:

• A single comparison operation (n > 1)
• A bitwise AND operation (n & 1) to check if n is odd
• A conditional return with either the value n or a right shift operation (n >> 1)

All these operations execute in constant time.

The space complexity is O(1) because the function uses only a fixed amount of extra space. No additional data structures are created, and the space usage doesn't depend on the input value n. The function only works with the input parameter and returns a single integer value.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Forgetting the Special Case of n = 1

A common mistake is to only consider whether n is odd or even without handling n = 1 separately:

# Incorrect implementation
def numberOfCuts(self, n: int) -> int:
    if n % 2 == 1:  # Odd number
        return n
    else:  # Even number
        return n // 2

This would incorrectly return 1 for n = 1, when the correct answer is 0 (no cuts needed for a whole circle).

Solution: Always check for n > 1 when handling odd numbers:

def numberOfCuts(self, n: int) -> int:
    if n > 1 and n % 2 == 1:
        return n
    else:
        return n // 2

2. Misunderstanding the Problem Statement

Some might interpret the problem as requiring cuts that don't necessarily pass through the center, leading to incorrect logic like trying to minimize cuts by making parallel lines or other geometric arrangements.

Solution: Remember that all valid cuts must pass through the center of the circle. This constraint is what makes the problem deterministic with a clear pattern.

3. Integer Division Issues in Other Languages

When implementing in languages like Python 2 or C++, using / for division might cause issues:

# Potential issue in Python 2
return n / 2  # Returns float in Python 3, integer in Python 2

Solution: Use integer division explicitly:

return n // 2  # Integer division in Python 3
# OR use bit shift for better performance
return n >> 1  # Right shift by 1 bit equals division by 2

4. Over-complicating the Logic

Some solutions might use unnecessary nested conditions or separate handling for multiple cases:

# Overly complex
def numberOfCuts(self, n: int) -> int:
    if n == 1:
        return 0
    elif n == 2:
        return 1
    elif n % 2 == 0:
        return n // 2
    else:
        return n

Solution: Simplify to the essential logic that covers all cases:

def numberOfCuts(self, n: int) -> int:
    return n if (n > 1 and n & 1) else n >> 1

5. Not Considering Edge Cases in Testing

Developers might test with typical values like 2, 3, 4 but forget to verify:

• n = 1 (should return 0)
• Large odd numbers
• Large even numbers

Solution: Always test boundary cases and ensure your solution handles:

• Minimum valid input (n = 1)
• Small odd and even numbers
• Large values to verify no overflow issues