The Clumsy Factorial problem asks us to compute a modified version of the factorial for a given positive integer N. Instead of multiplying all numbers from N down to 1 as in the standard factorial, we apply a sequence of operations in a fixed pattern: multiplication (*), division (/), addition (+), and subtraction (-), repeating this cycle as we decrement through the numbers.
N, then apply * to (N-1), then / to (N-2), then + to (N-3), then - to (N-4), and so on, repeating the cycle until reaching 1.N (1 ≤ N ≤ 10,000).
At first glance, it might seem like we need to simulate the entire process step by step, applying the four operations in order as we decrement from N to 1. This approach is straightforward and closely follows the problem statement. However, as N grows large, we might look for patterns or optimizations to avoid unnecessary computation.
Unlike the standard factorial, the clumsy factorial alternates between multiplication, division, addition, and subtraction. Since division is integer division, and since addition and subtraction alternate, the result can oscillate in sign and magnitude. We notice that the first few operations (multiplication and division) have the largest impact, while later additions and subtractions adjust the result incrementally.
While brute-force simulation is acceptable for the input constraints, recognizing patterns for small values of N can help us optimize further, possibly even leading to a formulaic solution for large N.
Let's break down the solution into clear steps:
N and decrementing by 1 each time. At each step, we apply the operation in the order: *, /, +, -, and repeat.
This approach ensures we respect both the operation order and the precedence required by the problem statement, and it works efficiently even for large N.
Let's walk through the process for N = 10:
But, using the stack approach:
The final result is 12.
class Solution:
def clumsy(self, N: int) -> int:
stack = [N]
N -= 1
index = 0
while N > 0:
if index % 4 == 0:
stack[-1] = stack[-1] * N
elif index % 4 == 1:
stack[-1] = int(stack[-1] / N)
elif index % 4 == 2:
stack.append(N)
else:
stack.append(-N)
N -= 1
index += 1
return sum(stack)
class Solution {
public:
int clumsy(int N) {
vector<int> stack;
stack.push_back(N--);
int index = 0;
while (N > 0) {
if (index % 4 == 0) {
stack.back() *= N;
} else if (index % 4 == 1) {
stack.back() /= N;
} else if (index % 4 == 2) {
stack.push_back(N);
} else {
stack.push_back(-N);
}
N--;
index++;
}
int result = 0;
for (int num : stack) result += num;
return result;
}
};
class Solution {
public int clumsy(int N) {
Stack<Integer> stack = new Stack<>();
stack.push(N--);
int index = 0;
while (N > 0) {
if (index % 4 == 0) {
stack.push(stack.pop() * N);
} else if (index % 4 == 1) {
stack.push(stack.pop() / N);
} else if (index % 4 == 2) {
stack.push(N);
} else {
stack.push(-N);
}
N--;
index++;
}
int result = 0;
for (int num : stack) result += num;
return result;
}
}
var clumsy = function(N) {
let stack = [N];
N--;
let index = 0;
while (N > 0) {
if (index % 4 === 0) {
stack[stack.length - 1] = stack[stack.length - 1] * N;
} else if (index % 4 === 1) {
stack[stack.length - 1] = Math.trunc(stack[stack.length - 1] / N);
} else if (index % 4 === 2) {
stack.push(N);
} else {
stack.push(-N);
}
N--;
index++;
}
return stack.reduce((a, b) => a + b, 0);
};
Brute-force approach: If we simulate every operation one by one, we perform a constant number of operations per number from N to 1. Thus, the time complexity is O(N) and the space complexity is O(N) (for the stack).
Optimized approach: The stack-based method described above also runs in O(N) time and uses O(N) space. For very large N, further mathematical optimization is possible based on patterns, but for the problem's constraints, the stack simulation is efficient and clear.
The Clumsy Factorial problem challenges us to simulate a sequence of operations with a repeating pattern on a decreasing sequence of integers. By using a stack to handle operator precedence and the alternation of addition and subtraction, we can efficiently compute the result. The approach is both intuitive and efficient, leveraging basic data structures and control flow, and is a great example of translating a problem statement directly into code while handling subtle details like integer division and operator order.