Problem Description

You are given an integer num. Your task is to find two integers whose product equals either num + 1 or num + 2. Among all possible pairs of integers that satisfy this condition, you need to return the pair that has the smallest absolute difference between them.

For example, if num = 8, you need to check:

• For num + 1 = 9: The factor pairs are (1, 9) and (3, 3). The absolute differences are |1-9| = 8 and |3-3| = 0.
• For num + 2 = 10: The factor pairs are (1, 10) and (2, 5). The absolute differences are |1-10| = 9 and |2-5| = 3.

Among all these pairs, (3, 3) has the smallest absolute difference of 0, so you would return [3, 3].

The key insight is that for any number, the factors that are closest to each other are those nearest to its square root. The solution efficiently finds these closest divisors by starting from √x and working downward until it finds a divisor. It then compares the results for both num + 1 and num + 2 to determine which produces the pair with the smaller absolute difference.

You can return the two integers in any order.

Intuition

To find two integers with the smallest absolute difference whose product equals a given number, we need to think about the mathematical relationship between factors. For any number x, its factors come in pairs: if i is a factor, then x/i is also a factor, and their product equals x.

The key observation is that among all factor pairs of a number, the pair closest to each other will be the one where both factors are nearest to √x. Why? Consider a number like 36:

• Factor pairs: (1, 36), (2, 18), (3, 12), (4, 9), (6, 6)
• The differences: |1-36| = 35, |2-18| = 16, |3-12| = 9, |4-9| = 5, |6-6| = 0

Notice how as we move toward √36 = 6, the difference keeps decreasing. This pattern holds for any number - the factors become more "balanced" as we approach the square root.

So our strategy becomes clear: start from √x and work downward to find the first number that divides x evenly. This will give us the factor pair with the minimum difference. We start from the square root and go down because we want the first (largest) divisor that's less than or equal to √x.

Since we need to check both num + 1 and num + 2, we apply this same logic to both values and then compare which result gives us the smaller absolute difference. This ensures we find the optimal pair across both possible target products.

Learn more about Math patterns.

Solution Approach

The solution implements a helper function f(x) that finds the two factors of x with the smallest absolute difference. Here's how the implementation works:

Helper Function f(x):

1. Start iterating from i = int(sqrt(x)) down to 1
2. For each i, check if x % i == 0 (i.e., i divides x evenly)
3. When we find the first divisor, return the pair [i, x // i]
4. Since we start from √x and move downward, the first divisor we find will give us the factor pair with minimum difference

Main Logic:

1. Call f(num + 1) to get the best factor pair for num + 1, store in variable a
2. Call f(num + 2) to get the best factor pair for num + 2, store in variable b
3. Compare the absolute differences: abs(a[0] - a[1]) vs abs(b[0] - b[1])
4. Return the pair with the smaller absolute difference

Why this works:

• By starting from √x and going down, we ensure that when we find a divisor i, the corresponding factor x // i will be greater than or equal to i
• The first valid divisor we encounter gives us the most "balanced" factor pair
• Comparing results from both num + 1 and num + 2 ensures we find the globally optimal solution

Time Complexity: O(√n) for each call to f(x), where n is the value being factored Space Complexity: O(1) as we only use a constant amount of extra space

Example Walkthrough

Let's walk through the solution with num = 8:

Step 1: Find closest divisors for num + 1 = 9

• Start from i = √9 = 3
• Check if 9 % 3 == 0? Yes!
• First divisor found: i = 3
• Return pair: [3, 9/3] = [3, 3]
• Absolute difference: |3 - 3| = 0

Step 2: Find closest divisors for num + 2 = 10

• Start from i = √10 ≈ 3.16, so i = 3
• Check if 10 % 3 == 0? No
• Decrement to i = 2
• Check if 10 % 2 == 0? Yes!
• First divisor found: i = 2
• Return pair: [2, 10/2] = [2, 5]
• Absolute difference: |2 - 5| = 3

Step 3: Compare and return the best pair

• Pair from num + 1: [3, 3] with difference 0
• Pair from num + 2: [2, 5] with difference 3
• Since 0 < 3, return [3, 3]

The key insight illustrated here: by starting from the square root and working downward, we immediately find the most balanced factor pair. For 9, we hit the perfect square case right away. For 10, we skip over 3 (not a divisor) and find 2, which gives us the closest possible factors for that number.

Solution Implementation

Time and Space Complexity

The time complexity is O(√num), and the space complexity is O(1).

Time Complexity Analysis:

• The function f(x) iterates from int(sqrt(x)) down to 1, checking if each value i divides x evenly
• In the worst case, the loop runs O(√x) iterations before finding a divisor pair
• The function f is called twice: once with num + 1 and once with num + 2
• Since num + 1 and num + 2 are approximately equal to num, each call takes O(√num) time
• Two sequential O(√num) operations result in O(√num) + O(√num) = O(√num) total time complexity

Space Complexity Analysis:

• The function uses only a constant amount of extra space for variables i, a, and b
• The returned list contains exactly 2 elements regardless of input size
• No recursive calls or data structures that grow with input size are used
• Therefore, the space complexity is O(1)

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Integer Overflow in Square Root Calculation

When dealing with very large values of num, calculating int(sqrt(target)) might lose precision due to floating-point arithmetic. For example, if target is a perfect square like 10^18, the floating-point representation of its square root might be slightly off (e.g., 999999999.9999999 instead of 1000000000), causing int() to truncate incorrectly.

Solution: Use integer-based methods or add a small epsilon for rounding:

def find_closest_divisor_pair(target: int) -> List[int]:
    # Add 0.5 for proper rounding to handle floating-point precision issues
    sqrt_val = int(sqrt(target) + 0.5)
  
    # Check if sqrt_val is actually a divisor (handles perfect squares)
    if sqrt_val * sqrt_val == target:
        return [sqrt_val, sqrt_val]
  
    # Otherwise, start from floor(sqrt(target))
    for divisor in range(int(sqrt(target)), 0, -1):
        if target % divisor == 0:
            return [divisor, target // divisor]

2. Edge Case: num = 0

When num = 0, we need to find divisors of 1 or 2. The code works correctly but it's worth noting that:

• For num + 1 = 1: The only factor pair is (1, 1)
• For num + 2 = 2: The factor pairs are (1, 2)

The algorithm handles this correctly, but developers might overlook testing this edge case.

3. Assumption About Return Order

The problem states "You can return the two integers in any order," but the current implementation always returns [smaller, larger]. While this is consistent and correct, some developers might assume they need to preserve a specific order based on which number (num+1 or num+2) was used.

Solution: Document the return order clearly or make it explicit:

# Always returns [smaller_divisor, larger_divisor] for consistency
return sorted([divisor, target // divisor])

4. Performance Issue with Very Large Numbers

For extremely large values of num, iterating from sqrt(target) down to 1 could still be slow if the number is prime or has only large prime factors. For instance, if num + 1 is prime, we'd iterate all the way down to 1.

Solution: Add an early termination optimization when the difference becomes larger than the best found so far:

def closestDivisors(self, num: int) -> List[int]:
    best_pair = None
    best_diff = float('inf')
  
    for target in [num + 1, num + 2]:
        sqrt_val = int(sqrt(target))
        for divisor in range(sqrt_val, 0, -1):
            if target % divisor == 0:
                other = target // divisor
                diff = other - divisor
                if diff < best_diff:
                    best_diff = diff
                    best_pair = [divisor, other]
                break  # Found best for this target
      
        # Early termination: if we found factors with diff 0 or 1
        if best_diff <= 1:
            break
  
    return best_pair