Solution Approach

The solution implements a binary search using Python's bisect_left function with a custom key function to find the n-th magical number.

Step 1: Calculate the Least Common Multiple (LCM)

c = lcm(a, b)

We need the LCM to apply the inclusion-exclusion principle correctly. Python's [math](/problems/math-basics) module provides the lcm function directly.

Step 2: Define the Search Range

r = (a + b) * n

This sets an upper bound for our binary search. The n-th magical number will definitely be less than (a + b) * n, giving us a safe search range from 0 to r.

Step 3: Binary Search with Custom Key Function

bisect_left(range(r), x=n, key=lambda x: x // a + x // b - x // c)

This is the core of the solution:

• range(r) creates our search space from 0 to r-1
• The key function lambda x: x // a + x // b - x // c counts how many magical numbers exist up to and including x
  • x // a: count of numbers divisible by a
  • x // b: count of numbers divisible by b
  • x // c: count of numbers divisible by both (to remove double-counting)
• bisect_left finds the leftmost position where the count equals n

Step 4: Apply Modulo

% mod

Since the answer can be very large, we return it modulo 10^9 + 7 as required.

The beauty of this approach is that bisect_left automatically handles finding the exact n-th magical number by searching for the smallest value x where the count of magical numbers up to x is at least n.

Example Walkthrough

Let's find the 4th magical number where a = 2 and b = 3.

Step 1: Calculate LCM

• lcm(2, 3) = 6 (since 2 and 3 are coprime)

Step 2: Set up search range

• Upper bound: (2 + 3) * 4 = 20
• We'll search from 0 to 20

Step 3: Binary search process

Let's trace through how bisect_left works with our counting function:

• The key function for any value x is: x // 2 + x // 3 - x // 6

Let's calculate counts for several values:

• x = 1: 1//2 + 1//3 - 1//6 = 0 + 0 - 0 = 0 magical numbers
• x = 2: 2//2 + 2//3 - 2//6 = 1 + 0 - 0 = 1 magical number (just 2)
• x = 3: 3//2 + 3//3 - 3//6 = 1 + 1 - 0 = 2 magical numbers (2, 3)
• x = 4: 4//2 + 4//3 - 4//6 = 2 + 1 - 0 = 3 magical numbers (2, 3, 4)
• x = 5: 5//2 + 5//3 - 5//6 = 2 + 1 - 0 = 3 magical numbers (2, 3, 4)
• x = 6: 6//2 + 6//3 - 6//6 = 3 + 2 - 1 = 4 magical numbers (2, 3, 4, 6)

The binary search looks for where the count first reaches 4:

• Start with range [0, 20]
• Middle values are tested, narrowing down the range
• Eventually finds that at x = 6, we have exactly 4 magical numbers

Step 4: Return result

• The 4th magical number is 6
• Apply modulo: 6 % (10^9 + 7) = 6

Verification: The magical numbers in order are 2, 3, 4, 6... and indeed the 4th one is 6.

The key insight is that we never generated the actual sequence - we just counted how many magical numbers exist up to each point and used binary search to find where that count equals n.

Time and Space Complexity

Time Complexity: O(log(n * max(a, b)))

The algorithm uses binary search via bisect_left on a range from 0 to r = (a + b) * n. The binary search performs O(log(r)) iterations, where r = (a + b) * n. Since a and b are positive integers, this simplifies to O(log(n * (a + b))), which can be expressed as O(log(n * max(a, b))) in the worst case.

For each iteration of the binary search, the key function lambda x: x // a + x // b - x // c is evaluated. This function performs:

• Three integer division operations: O(1)
• Two addition/subtraction operations: O(1)
• The lcm(a, b) calculation is done once before the binary search: O(log(min(a, b))) using the GCD algorithm

Therefore, the overall time complexity is dominated by the binary search: O(log(n * max(a, b))).

Space Complexity: O(1)

The algorithm uses only a constant amount of extra space:

• Variables mod, c, and r each require O(1) space
• The bisect_left function uses O(1) auxiliary space for binary search
• The range object in Python 3 is a lazy iterator that doesn't create the full list, using O(1) space
• The lambda function doesn't create any additional data structures

Thus, the total space complexity is O(1).

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

Common Pitfalls

1. Using Wrong Binary Search Template Variant

The most critical mistake is using a different binary search variant that doesn't match the standard template. The template-compliant approach uses first_true_index to track the answer, while left <= right, and right = mid - 1 when feasible.

Wrong approach (alternative variant):

left, right = 1, n * min(a, b)
while left < right:
    mid = (left + right) // 2
    if count(mid) >= n:
        right = mid  # WRONG: should be right = mid - 1
    else:
        left = mid + 1
return left  # WRONG: should track first_true_index

Correct template-compliant approach:

left, right = 1, n * min(a, b)
first_true_index = -1

while left <= right:
    mid = (left + right) // 2
    if count(mid) >= n:
        first_true_index = mid
        right = mid - 1
    else:
        left = mid + 1

return first_true_index % MOD

2. Incorrect LCM Calculation Leading to Integer Overflow

The naive formula lcm = (a * b) // gcd(a, b) can cause integer overflow during the multiplication step, especially in languages like Java/C++.

Solution:

# Option 1: Use built-in function (Python 3.9+)
lcm_value = math.lcm(a, b)

# Option 2: Divide first to prevent overflow
gcd_value = math.gcd(a, b)
lcm_value = (a // gcd_value) * b

3. Forgetting the Modulo Operation

The problem explicitly asks for the result modulo 10^9 + 7, but it's easy to forget this step since the binary search itself doesn't require it.

Solution: Always apply the modulo operation before returning:

MOD = 10**9 + 7
return first_true_index % MOD

4. Edge Cases with first_true_index

When using the template, always initialize first_true_index = -1 and handle the case where no valid answer is found (though for this problem, an answer is guaranteed to exist).