hard Nth Magical Number

Problem Statement

Given the three integers n, a, and b, return the n^th magical number modulo + 7 as it can be very large.

A positive integer is called magical if it is divisible by either a or b.

Examples

• Example 1:

  • Input: n = 3, a = 2, b = 3
  • Expected Output: 4
  • Justification: The first three magical numbers divisible by 2 or 3 are 2, 3, and 4. The third magical number is 4.

• Example 2:

  • Input: n = 5, a = 3, b = 4
  • Expected Output: 9
  • Justification: The sequence of magical numbers that are divisible by 3 or 4 begins as 3, 4, 6, 8, 9. The fifth one in this sequence is 9.

• Example 3:

  • Input: n = 20, a = 3, b = 5
  • Expected Output: 42
  • Justification: The 20^th magical number for a = 30, and b = 5 is 42.

Solution

To solve this problem, we leverage the concept of binary search along with some mathematical principles to efficiently find the nth "magical number." A magical number, in this context, is defined as any positive integer that is divisible by either of two given numbers, A or B. The key to solving this problem lies in understanding how these numbers distribute over a range and utilizing the least common multiple (LCM) of A and B to avoid redundancy. The solution involves a binary search strategy, focusing on narrowing down the range within which the nth magical number lies by counting the number of magical numbers up to a certain point and adjusting the search bounds accordingly. This method significantly reduces the number of checks needed compared to a linear search, especially for large values of N, A, and B.

The algorithm's effectiveness is rooted in its use of mathematical properties to simplify the problem. By calculating the LCM of A and B, we can accurately determine the intervals at which magical numbers occur and leverage this to estimate the position of the nth magical number within a bounded search space. Combining this with the efficiency of binary search, we can quickly converge on the exact value of the nth magical number. This approach minimizes computational overhead and ensures that the solution can handle large input values without performance degradation, making it an optimal strategy for solving the given problem.

Step-by-Step Algorithm

1. Calculate the Least Common Multiple (LCM) of A and B using the Greatest Common Divisor (GCD) method. This is crucial for accurately identifying the distribution of magical numbers.
2. Initialize the search bounds: Set the lower bound (lo) to 0 and the upper bound (hi) to N * min(A, B). This range guarantees that the nth magical number lies within.
3. Perform binary search: Repeat the following steps until lo is less than hi:
   • Calculate the mid-point (mi) as lo + (hi - lo) / 2.
   • Count the total number of magical numbers less than or equal to mi by adding the numbers divisible by A, those divisible by B, and subtracting those divisible by both (to avoid double counting).
   • If the count is less than N, adjust lo to mi + 1 to search the upper half. Otherwise, adjust hi to mi to search the lower half.
4. Return the result: Once the loop concludes, lo will be the nth magical number. Return lo % MOD to ensure the result is within the integer limit.

Algorithm Walkthrough

Let's consider the Input: n = 20, a = 3, b = 5

1. Calculate LCM and Initialize:

   • The LCM of 3 and 5 is found using the gcd function: LCM(3, 5) = 15.
   • Set MOD = 1_000_000_007.
   • Initialize search bounds: lo = 0, hi = 60 (since n * min(a, b) = 20 * 3 = 60).

2. Binary Search Iterations:

   • Iteration 1:
     • Mid = (0 + 60) / 2 = 30.
     • Count = 30 / 3 + 30 / 5 - 30 / 15 = 10 + 6 - 2 = 14. Since 14 < 20, lo becomes 31.
   • Iteration 2:
     • New bounds are lo = 31, hi = 60.
     • Mid = (31 + 60) / 2 = 45.5 (using integer division, Mid = 45).
     • Count = 45 / 3 + 45 / 5 - 45 / 15 = 15 + 9 - 3 = 21. Since 21 >= 20, hi becomes 45.
   • Iteration 3:
     • New bounds are lo = 31, hi = 45.
     • Mid = (31 + 45) / 2 = 38.
     • Count = 38 / 3 + 38 / 5 - 38 / 15 = 12 + 7 - 2 = 17. Since 17 < 20, lo becomes 39.
   • Iteration 4:
     • New bounds are lo = 39, hi = 45.
     • Mid = (39 + 45) / 2 = 42.
     • Count = 42 / 3 + 42 / 5 - 42 / 15 = 14 + 8 - 2 = 20. Since 20 == 20, and we're looking for the least value of lo that meets this condition, hi becomes 42.
   • Iteration 5:
     • New bounds are lo = 39, hi = 42.
     • Mid = (39 + 42) / 2 = 40.5 (using integer division, Mid = 40).
     • Count = 40 / 3 + 40 / 5 - 40 / 15 = 13 + 8 - 2 = 19. Since 19 < 20, lo becomes 41.
   • Iteration 6:
     • New bounds are lo = 41, hi = 42.
     • Mid = (41 + 42) / 2 = 41.5 (using integer division, Mid = 41).
     • Count = 41 / 3 + 41 / 5 - 41 / 15 = 13 + 8 - 2 = 19 (same as last time, since Mid didn't change effectively). Since 19 < 20, lo becomes 42.
   • Iteration 7:
     • Now lo = 42 and hi = 42, so the loop terminates.

3. Final Result:

   • The loop has exited with lo = 42, which is the value we consider to be the 20th magical number given a = 3 and b = 5.

4. Modulus Operation:

   • Apply the modulus operation: 42 % MOD = 42.

Code

public class Solution {

  // Method to find the nth magical number
  public int nthMagicalNumber(int N, int A, int B) {
    int MOD = 1_000_000_007; // Define the modulus value to ensure the result fits within the integer limits
    int L = (A / gcd(A, B)) * B; // Calculate the least common multiple (LCM) of A and B using the gcd

    long lo = 0; // Set the lower bound of our search
    long hi = (long) N * Math.min(A, B); // Set the upper bound of our search based on the minimum of A and B
    // Use binary search to find the nth magical number
    while (lo < hi) {
      long mi = lo + (hi - lo) / 2; // Calculate the middle point of our current search space
      // If there are not enough magical numbers below or equal to mi...
      if (mi / A + mi / B - mi / L < N) lo = mi + 1; // ...then we need to search the higher half
      else hi = mi; // ...else search the lower half
    }

    return (int) (lo % MOD); // Return the nth magical number modulo MOD
  }

  // Helper method to compute the greatest common divisor (gcd) of x and y
  public int gcd(int x, int y) {
    if (x == 0) return y; // Base case: if x is 0, gcd is y
    return gcd(y % x, x); // Recursive call: gcd of y % x and x
  }

  // Main method to test the nthMagicalNumber method with example inputs
  public static void main(String[] args) {
    Solution solution = new Solution();

    // Example 1: Input: N = 3, A = 2, B = 3
    System.out.println(solution.nthMagicalNumber(3, 2, 3));
    // Expected Output: 2

    // Example 2: Input: N = 5, A = 3, B = 4
    System.out.println(solution.nthMagicalNumber(5, 3, 4));
    // Expected Output: 9

    // Example 3: Input: N = 20, A = 30, B = 5
    System.out.println(solution.nthMagicalNumber(20, 3, 5));
    // Expected Output: 42
  }
}

Complexity Analysis

Time Complexity

• : The main operation in this code is the binary search algorithm. The search space is between 0 and . The binary search algorithm has a logarithmic time complexity relative to the size of the search space. Therefore, the time complexity is , as each iteration of the loop halves the search space until the solution is found.

Space Complexity

• : The algorithm uses a constant amount of space regardless of the input size. Thus, the space complexity of the algorithm is , indicating that it uses a fixed amount of memory.