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Formatted question description: https://leetcode.ca/all/1006.html

1006. Clumsy Factorial

Level

Medium

Description

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

Solution

Calculate the result directly. For the terms with multiplication and division, only the first term has a positive sign, and the rest terms all have negative signs. Remember this rule so that the result can be calculated correctly.

• Java
• C++
• Python

• class Solution {
      public int clumsy(int N) {
          if (N <= 3)
              return supplementary(N);
          int num = N;
          int factorial = supplementary(num);
          num = Math.max(num - 3, 0);
          factorial += num;
          num--;
          while (num > 0) {
              factorial -= supplementary(num);
              num = Math.max(num - 3, 0);
              factorial += num;
              num--;
          }
          return factorial;
      }
  
      public int supplementary(int num) {
          if (num <= 2)
              return num;
          else
              return num * (num - 1) / (num - 2);
      }
  }
  

• // OJ: https://leetcode.com/problems/clumsy-factorial/
  // Time: O(N)
  // Space: O(1)
  class Solution {
  public:
      int clumsy(int N) {
          long ans = 0;
          for (int i = N; i >= 1; i -= 4) {
              long x = (i == N ? 1 : -1) * i;
              if (i - 1 >= 1) {
                  x *= i - 1;
                  if (i - 2 >= 1) {
                      x /= i - 2;
                      if (i - 3 >= 1) {
                          x += i - 3;
                      }
                  }
              } 
              ans += x;
          }
          return ans;
      }
  };
  

• class Solution:
      def clumsy(self, N: int) -> int:
          op = 0
          s = [N]
          for i in range(N - 1, 0, -1):
              if op == 0:
                  s.append(s.pop() * i)
              elif op == 1:
                  s.append(int(s.pop() / i))
              elif op == 2:
                  s.append(i)
              else:
                  s.append(-i)
              op = (op + 1) % 4
          return sum(s)
  
  ############
  
  class Solution(object):
      def clumsy(self, N):
          """
          :type N: int
          :rtype: int
          """
          cl = ""
          ops = ["*", "/", "+", "-"]
          op = 0
          for n in range(N, 0, -1):
              if n != 1:
                  cl += str(n) + ops[op % 4]
              else:
                  cl += "1"
              op += 1
          return eval(cl)
  

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