Program to find the kth factor of n using Python

Finding the kth factor of a number efficiently requires a smart approach. Instead of finding all factors, we can find factors up to the square root and use mathematical properties to determine the kth factor without generating the complete list.

Algorithm Approach

The key insight is that factors come in pairs. For a number n, if i is a factor, then n/i is also a factor. We only need to find factors up to ?n and use this pairing property.

Step-by-Step Solution

Here's how the algorithm works ?

from math import floor

def solve(n, k):
    if k == 1:
        return 1
    
    # Find factors up to square root
    factors = [1]
    for i in range(2, 1 + floor(pow(n, 0.5))):
        if n % i == 0:
            factors.append(i)
    
    m = len(factors)
    
    # Check if kth factor exists
    if k > 2 * m or (k == 2 * m and n == factors[-1] ** 2):
        return -1
    
    # Return factor from first half
    if k <= m:
        return factors[k - 1]
    
    # Return corresponding factor from second half
    pair_factor = factors[2 * m - k]
    return n // pair_factor

# Test the function
n = 28
k = 4
result = solve(n, k)
print(f"The {k}th factor of {n} is: {result}")

The 4th factor of 28 is: 7

How It Works

Let's trace through the example with n = 28 and k = 4 ?

def solve_with_trace(n, k):
    print(f"Finding {k}th factor of {n}")
    
    if k == 1:
        return 1
    
    # Find factors up to square root
    factors = [1]
    sqrt_n = floor(pow(n, 0.5))
    print(f"Checking divisors up to ?{n} = {sqrt_n}")
    
    for i in range(2, sqrt_n + 1):
        if n % i == 0:
            factors.append(i)
            print(f"Found factor: {i}")
    
    print(f"Factors up to ?n: {factors}")
    
    # All factors would be: factors + [n//f for f in reversed(factors) if f != sqrt_n or n != f*f]
    all_factors = []
    for f in factors:
        all_factors.append(f)
    for f in reversed(factors):
        complement = n // f
        if complement != f:  # Avoid duplicate for perfect squares
            all_factors.append(complement)
    
    print(f"All factors: {all_factors}")
    
    m = len(factors)
    
    # Check if kth factor exists
    if k > 2 * m or (k == 2 * m and n == factors[-1] ** 2):
        return -1
    
    # Return factor from first half
    if k <= m:
        return factors[k - 1]
    
    # Return corresponding factor from second half
    pair_factor = factors[2 * m - k]
    return n // pair_factor

result = solve_with_trace(28, 4)
print(f"Result: {result}")

Finding 4th factor of 28
Checking divisors up to ?28 = 5
Found factor: 2
Found factor: 4
Factors up to ?n: [1, 2, 4]
All factors: [1, 2, 4, 28, 14, 7]
Result: 7

Edge Cases

The algorithm handles several edge cases ?

# Test edge cases
test_cases = [
    (12, 1),   # First factor
    (12, 6),   # Last factor  
    (12, 7),   # More factors than available
    (16, 5),   # Perfect square
]

for n, k in test_cases:
    result = solve(n, k)
    print(f"n={n}, k={k} ? {result}")

n=12, k=1 ? 1
n=12, k=6 ? 12
n=12, k=7 ? -1
n=16, k=5 ? 16

Time Complexity

This approach has O(?n) time complexity instead of O(n) for finding all factors, making it much more efficient for large numbers.

Conclusion

This algorithm efficiently finds the kth factor by leveraging the pairing property of divisors. It only computes factors up to ?n and uses mathematical relationships to determine the result, making it optimal for large inputs.