LeetCode 1006: Clumsy Factorial Solution

Master LeetCode problem 1006 (Clumsy Factorial), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

1006. Clumsy Factorial

Medium

Math Stack Simulation

Problem Explanation

Explanation

To solve this problem, we can simulate the process of calculating the clumsy factorial by following the given order of operations. We can use a stack to keep track of the intermediate results as we iterate over the numbers from n to 1. We will perform the operations in the order of multiply, divide, add, and subtract.

Here is the step-by-step algorithm:

Initialize a stack to store the intermediate results.
Iterate from n to 1 and perform the following operations:
- If the stack is empty, push the current number onto the stack.
- If the stack is not empty, check the current position in the cycle (multiply, divide, add, subtract) and perform the corresponding operation with the current number.
- Update the current position in the cycle.
After iterating over all numbers, sum up the remaining elements in the stack to get the final result.

Time complexity: O(n) Space complexity: O(n)

Solution Code

class Solution {
    public int clumsy(int n) {
        Deque<Integer> stack = new ArrayDeque<>();
        stack.push(n);
        n--;

        int index = 0; // 0 for multiply, 1 for divide, 2 for add, 3 for subtract

        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);
            }
            index++;
            n--;
        }

        int result = 0;
        while (!stack.isEmpty()) {
            result += stack.pop();
        }

        return result;
    }
}

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