Clumsy Factorial

Clumsy Factorial - Problem

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 make a clumsy factorial using the integers in decreasing order by swapping out the multiply operations for a fixed rotation of operations with 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 = 90 / 8 = 11.

Given an integer n, return the clumsy factorial of n.

Input & Output

Example 1 — Small Number

$ Input: n = 4

› Output: 7

💡 Note: clumsy(4) = 4 * 3 / 2 + 1 = 12 / 2 + 1 = 6 + 1 = 7

Example 2 — Larger Number

$ Input: n = 10

› Output: 12

💡 Note: clumsy(10) = 10 * 9 / 8 + 7 - 6 * 5 / 4 + 3 - 2 * 1 = 11 + 7 - 7 + 3 - 2 = 12

Example 3 — Minimum Case

$ Input: n = 1

› Output: 1

💡 Note: clumsy(1) = 1 (no operations needed)

Constraints

1 ≤ n ≤ 10⁴

Visualization

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Asked in

G Google 15 a Amazon 8 f Facebook 5

The key insight is handling operator precedence properly - multiplication and division must be evaluated before addition and subtraction. The stack approach elegantly handles this by immediately applying * and / operations while deferring + and - by pushing values. For optimal performance, a mathematical pattern can be recognized: result ≈ n+1 with variations based on n%4. Time: O(n) for simulation, O(1) for pattern. Space: O(n) for stack, O(1) for pattern.

Common Approaches

✓ Mathematical Pattern

⏱️ Time: O(1) Space: O(1)

After analyzing several cases, we can observe that clumsy factorial follows a pattern: for n≥5, the result is approximately n+1, with small variations. This allows us to compute the result directly without simulation.

Stack-Based Evaluation

⏱️ Time: O(n) Space: O(n)

Push numbers to stack and handle operators as we go. For multiplication and division, immediately apply to stack top. For addition and subtraction, handle them later by pushing positive/negative values.

Step by Step Calculation

⏱️ Time: O(n) Space: O(1)

Process the expression by handling multiplication and division first (left to right), then addition and subtraction. Break down into groups like (n*(n-1)/(n-2)) + (n-3) - (n-4)*(n-5)/(n-6) + (n-7) - ...

Mathematical Pattern — Algorithm Steps

Handle base cases n=1,2,3,4 directly
For n≥5, apply the mathematical pattern
Return result based on n mod 4

Visualization

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Step-by-Step Walkthrough

Base Cases

Handle n=1,2,3,4 separately

Pattern Recognition

n%4 determines the result offset

Direct Calculation

Apply formula based on n%4 value

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int solution(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 == 0) {
        return n + 1;
    } else if (n % 4 <= 2) {
        return n + 2;
    } else {
        return n - 1;
    }
}

int main() {
    int n;
    scanf("%d", &n);
    int result = solution(n);
    printf("%d\n", result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(1)

Direct calculation based on mathematical pattern, no loops

✓ Linear Growth

Space Complexity

O(1)

Only using constant extra variables

✓ Linear Space

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Input & Output

Constraints

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Common Approaches

Mathematical Pattern — Algorithm Steps

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Code -

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