https://leetcode.com/problems/clumsy-factorial/
X. https://leetcode.com/problems/clumsy-factorial/discuss/252247/C%2B%2BJava-Brute-Force
X. Stack
https://leetcode.com/problems/clumsy-factorial/discuss/252833/Java-Stack
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.)
Solution 2, Improve to O(1)
N * N - 3 * N = N * N - 3 * N + 2 - 2N * (N - 3) = (N - 1) * (N - 2) - 2。（factorization）N = (N - 1) * (N - 2) / (N - 3) - 2 / (N - 3) （Divide N - 3 on both side）N - (N - 1) * (N - 2) / (N - 3) = - 2 / (N - 3)- 2 / (N - 3) = 0, If N - 3 > 2.
So when
N > 5, N - (N - 1) * (N - 2) / (N - 3) = 0
Now it's
O(1) public int clumsy(int N) {
if (N == 0) return 0;
if (N == 1) return 1;
if (N == 2) return 2;
if (N == 3) return 6;
return N * (N - 1) / (N - 2) + helper(N - 3);
}
public int helper(int N) {
if (N == 0) return 0;
if (N == 1) return 1;
if (N == 2) return 1;
if (N == 3) return 1;
if (N == 4) return -2;
if (N == 5) return 0;
return helper((N - 2) % 4 + 2);
}
Solution 3, Better Format
Now make a summary.
N = 0, return 0
N = 1, return 1
N = 2, return 2
N = 3, return 6
N = 4, return 7
N = 5 + 4K, return N + 2
N = 6 + 4K, return N + 2
N = 7 + 4K, return N - 1
N = 8 + 4K, return N + 1
N = 0, return 0
N = 1, return 1
N = 2, return 2
N = 3, return 6
N = 4, return 7
N = 5 + 4K, return N + 2
N = 6 + 4K, return N + 2
N = 7 + 4K, return N - 1
N = 8 + 4K, return N + 1
public int clumsy(int N) {
if (N == 1) return 1;
if (N == 2) return 2;
if (N == 3) return 6;
if (N == 4) return 7;
if (N % 4 == 1) return N + 2;
if (N % 4 == 2) return N + 2;
if (N % 4 == 3) return N - 1;
return N + 1;
}
As defined in the description of problem,
Additionally, the division that we use is floor division such that 10 * 9 / 8 equals 11.
We can easily observe below:
5 * 4 / 3 = 6
6 * 5 / 4 = 7
10 * 9 / 8 = 11
...
...
...
so we can get this formula:
i * (i-1) / (i-2) = i+1 when i >= 5
we can simplify our computation as below:
i * (i-1) / (i-2) + (i-3) - (i-4) * (i-5) / (i-6) + (i-7) - (i-8) * .... + rest elments
= (i+1) + "(i-3)" - "(i-4) * (i-5) / (i-6)" + "(i-7)" - "(i-8) * " .... + rest elments
= (i+1) + "(i-3) - (i-3)" + "(i-7) - (i-7)" + .... + rest elments
= (i+1) + rest elments
we can call each 4 numbers a
chunk, so from N // 4 we can know how many chunks there are, then the rest 0, 1, 2 and 3 elements will influence our final result.- when
0element left: final result is(i+1) + ... + 5 - (4*3/2) + 1, which isi+1 - when
1element left: final result is(i+1) + ... + 6 - (5*4/3) + 2 - 1, which isi+2 - when
2element left: final result is(i+1) + ... + 7 - (6*5/4) + 3 - 2 * 1, which isi+2 - when
3element left: final result is(i+1) + ... + 8 - (7*6/5) + 4 - 3 * 2 / 1, which isi-1
After consider the corner case, we can arrive at the solution:
def clumsy(self, N: int) -> int:
if N <= 2:
return N
if N <= 4:
return N + 3
if (N - 4) % 4 == 0:
return N + 1
elif (N - 4) % 4 <= 2:
return N + 2
else:
return N - 1
Note that x-(x-1)*(x-2)/(x-3)=0 for x>5.
for example, clumsy(22)=22*21/20+(19-18*17/16)+(15-14*13/12)+(11-10*9/8)+(7-6*5/4)+3-2*1,
in which 19-18*17/16=15-14*13/12=11-10*9/8=7-6*5/4=0
And for the same reason, N*(N-1)/(N-2)=N+1. (N>5)
for example, clumsy(22)=22*21/20+(19-18*17/16)+(15-14*13/12)+(11-10*9/8)+(7-6*5/4)+3-2*1,
in which 19-18*17/16=15-14*13/12=11-10*9/8=7-6*5/4=0
And for the same reason, N*(N-1)/(N-2)=N+1. (N>5)
int clumsy(int N) {
int a[5]={-1, 1, 2, 6, 7};
int b[4]={1, 2, 2, -1};
return N<5?a[N]:N+b[N%4];
}public int clumsy(int N) {
if (N == 1) {return 1;}
if (N == 2) {return 2;}
if (N == 3) {return 6;}
if (N == 4) {return 7;}
int res = N + 1;
int remain = N - 3;
if (remain % 4 == 0) {res -= 2;}
if (remain % 4 == 2 || (remain % 4 == 3)) {res += 1;}
return res;
}
given a * (a - 1) / (a - 2) always equal to (a + 1) for a > 4
we find the whole thing becomes
N * (N - 1) / (N - 2) +
(N - 3) -
(N-4) * (N - 5) / (N - 6) + .......
(N - 3) -
(N-4) * (N - 5) / (N - 6) + .......
where (N-4) * (N - 5) / (N - 6) = (N - 3)
so the whole thing becomes N * (N - 1) / (N - 2) + Some EdgeCases
X. https://leetcode.com/problems/clumsy-factorial/discuss/252247/C%2B%2BJava-Brute-Force
public int clumsy(int N) {
if (N == 0) return 0;
if (N == 1) return 1;
if (N == 2) return 2;
if (N == 3) return 6;
return N * (N - 1) / (N - 2) + helper(N - 3);
}
public int helper(int N) {
if (N == 0) return 0;
if (N == 1) return 1;
if (N == 2) return 1;
if (N == 3) return 1;
return N - (N - 1) * (N - 2) / (N - 3) + helper(N - 4);
}
https://leetcode.com/problems/clumsy-factorial/discuss/252833/Java-Stack
public int clumsy(int N) {
Stack<Integer> stack = new Stack<>();
char[] op = new char[]{ '*', '/', '+', '-' };
stack.push(N--);
int index = 0;
while (N > 0) {
if (op[index] == '*') {
stack.push(stack.pop() * N--);
} else if (op[index] == '/') {
stack.push(stack.pop() / N--);
} else if (op[index] == '+') {
stack.push(N--);
} else if (op[index] == '-') {
stack.push(-1 * (N--));
}
index = (index + 1) % 4;
}
int sum = 0;
while (!stack.isEmpty()) {
sum += stack.pop();
}
return sum;
}
