How to Compute the Clumsy Factorials using Iterative Algorithm?

Normally, the factorial of a positive integer n is the product of all positive integers less than or equal to n. For example, factorial(10) = 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1.

We instead make a clumsy factorial: using the integers in decreasing order, we swap out the multiply operations for a fixed rotation of operations: multiply (*), divide (/), add (+) and subtract (-) in this order.

For example, clumsy(10) = 10 * 9 / 8 + 7 – 6 * 5 / 4 + 3 – 2 * 1. However, these operations are still applied using the usual order of operations of arithmetic: we do all multiplication and division steps before any addition or subtraction steps, and multiplication and division steps are processed left to right.

Additionally, the division that we use is floor division such that 10 * 9 / 8 equals 11. This guarantees the result is an integer.

Implement the clumsy function as defined above: given an integer N, it returns the clumsy factorial of N.

Example 1:
Input: 4
Output: 7
Explanation: 7 = 4 * 3 / 2 + 1

Example 2:
Input: 10
Output: 12
Explanation: 12 = 10 * 9 / 8 + 7 – 6 * 5 / 4 + 3 – 2 * 1

Note:
1 <= N <= 10000
-2^31 <= answer <= 2^31 – 1 (The answer is guaranteed to fit within a 32-bit integer.)

Clumsy Factorial Algorithms

Numbers can be proccessed in batches of four. Special cases are the four numbers from N to N-3 and the rest are stored in intermediate variable num, which is (a*b/c-d). Then the num is subtracted from the result.

The remainder items (up to 3) should be also subtracted. The following C++ implements the iterative algorithm that runs at O(N) complexity, and requires O(1) constant space.

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class Solution {
public:
    int clumsyFactorial(int N) {
        int r = 0, c = 0, num = 0;
        for (int i = N; i >= 1; -- i, ++ c) {
            switch (c % 4) {
                case 0:                    
                    num = i; break;
                case 1:
                    num *= i; break;
                case 2:
                    num /= i; break;
                case 3:
                    if (N - i <= 4) r = num + i; else r = r - num + i; break;
            }
        }
        if (N <= 3) return num;
        if (c % 4 != 0) r -= num;
        return r;
    }
};

class Solution {
public:
    int clumsyFactorial(int N) {
        int r = 0, c = 0, num = 0;
        for (int i = N; i >= 1; -- i, ++ c) {
            switch (c % 4) {
                case 0:                    
                    num = i; break;
                case 1:
                    num *= i; break;
                case 2:
                    num /= i; break;
                case 3:
                    if (N - i <= 4) r = num + i; else r = r - num + i; break;
            }
        }
        if (N <= 3) return num;
        if (c % 4 != 0) r -= num;
        return r;
    }
};

–EOF (The Ultimate Computing & Technology Blog) —