Introduction
Given a rope of length n, the objective is to cut it into at least two pieces such that the product of the lengths of all pieces is maximized.
At first glance, this problem appears similar to Dynamic Programming because we need to find an optimal partition of the rope. However, after analyzing the mathematical properties of multiplication, we can derive a much more efficient greedy solution.
This approach avoids building DP tables and achieves an optimal time complexity of O(log n) using fast exponentiation.
Problem Statement
Given a rope of length n, cut it into at least two pieces such that the product of the lengths of the resulting pieces is maximized.
Return the maximum possible product.
Example 1
Input
n = 2
Output
1
Explanation
The only possible cut is:
1 + 1
Product:
1 × 1 = 1
Example 2
Input
n = 5
Output
6
Explanation
Possible cuts:
| Cut | Product |
|---|
| 1 + 4 | 4 |
| 2 + 3 | 6 |
| 1 + 1 + 3 | 3 |
Maximum product = 6
Key Mathematical Observation
To maximize a product, larger numbers are often better when broken into smaller factors.
Consider the following:
| Number | Without Cut | Best Product After Cut |
|---|
| 2 | 2 | 1 × 1 = 1 |
| 3 | 3 | 1 × 2 = 2 |
| 4 | 4 | 2 × 2 = 4 |
| 5 | 5 | 2 × 3 = 6 |
| 6 | 6 | 3 × 3 = 9 |
Notice that once the rope length exceeds 4, splitting it generally increases the product.
For example:
5 = 2 + 3
Product:
2 × 3 = 6 > 5
Similarly:
6 = 3 + 3
Product:
3 × 3 = 9 > 6
This pattern continues for larger values.
Why Is 3 Optimal?
Suppose we have a segment of length x ≥ 5.
If we cut off a piece of length 3:
x → 3 + (x - 3)
The new product becomes:
3 × (x - 3)
This is beneficial whenever:
3(x - 3) > x
Solving:
3x - 9 > x
2x > 9
x > 4.5
Therefore, whenever a segment is larger than 4, cutting off a 3 increases the product.
This mathematical proof explains why the optimal strategy is to use as many pieces of length 3 as possible.
Mathematical Cases
After repeatedly extracting 3s, only three possible remainders can exist.
Case 1: n % 3 == 0
The rope can be divided entirely into pieces of length 3.
Example
n = 9
Optimal cut:
3 + 3 + 3
Product:
3 × 3 × 3 = 27
Formula:
3^(n/3)
Case 2: n % 3 == 1
A remainder of 1 should be avoided.
Consider:
3 + 1
Product:
3 × 1 = 3
Instead, convert it into:
2 + 2
Product:
2 × 2 = 4
Since:
4 > 3
Using two 2s is always better.
Example
n = 10
Optimal cut:
3 + 3 + 4
Product:
3 × 3 × 4 = 36
Formula:
3^(n/3 - 1) × 4
Case 3: n % 3 == 2
Simply keep one piece of length 2.
Example
n = 8
Optimal cut:
3 + 3 + 2
Product:
18
Formula:
3^(n/3) × 2
Algorithm
Handle the mandatory cut cases:
If n = 2, return 1
If n = 3, return 2
Compute:
Apply the corresponding mathematical case:
Java Solution
class Solution {
public int maxProduct(int n) {
if (n == 2) return 1;
if (n == 3) return 2;
int q = n / 3;
int r = n % 3;
if (r == 0) {
return (int) power(3, q);
}
if (r == 1) {
return (int) (power(3, q - 1) * 4);
}
return (int) (power(3, q) * 2);
}
private long power(long base, int exp) {
long result = 1;
while (exp > 0) {
if ((exp & 1) == 1) {
result *= base;
}
base *= base;
exp >>= 1;
}
return result;
}
}
Dry Run
Input
n = 10
Step 1
Compute quotient and remainder:
q = 10 / 3 = 3
r = 10 % 3 = 1
Step 2
Since the remainder is 1, use Case 2:
3^(q - 1) × 4
Substitute values:
3^(3 - 1) × 4
= 3² × 4
= 9 × 4
= 36
Output
36
Complexity Analysis
Time Complexity
O(log n)
Fast exponentiation computes powers efficiently in logarithmic time.
Auxiliary Space
O(1)
No additional data structures are required.
Interview Discussion
How to Identify This Problem?
Look for these clues:
Maximize a product.
Partition a number into smaller pieces.
Mathematical optimization appears possible.
Constraints allow a non-DP solution.
Common Mistakes
Using greedy cuts of 2 everywhere.
Forgetting the special handling for remainder 1.
Returning n itself without enforcing at least one cut.
Follow-Up Question
Can this be solved using Dynamic Programming?
Yes. A DP solution exists with O(n²) complexity, but the mathematical greedy approach is significantly more efficient and is considered the optimal solution.
Key Takeaways
The maximum product is achieved by cutting the rope into as many pieces of length 3 as possible.
A remainder of 1 should be converted into 2 + 2.
The solution is based on mathematical proof rather than dynamic programming.
Fast exponentiation reduces power calculation to O(log n) time.
The overall solution achieves O(log n) time complexity and O(1) space complexity.
Summary
The Maximum Product Cutting problem can be solved using a mathematical greedy strategy rather than dynamic programming. By proving that repeatedly cutting segments of length 3 maximizes the resulting product, the problem reduces to handling three simple remainder cases based on n % 3. This observation leads to an elegant and highly efficient solution with logarithmic time complexity and constant auxiliary space.