Introduction

In this blog post, we will explore how to solve the LeetCode problem "1387" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

The power of an integer x is defined as the number of steps needed to transform x into 1 using the following steps: if x is even then x = x / 2 if x is odd then x = 3 * x + 1 For example, the power of x = 3 is 7 because 3 needs 7 steps to become 1 (3 --> 10 --> 5 --> 16 --> 8 --> 4 --> 2 --> 1). Given three integers lo, hi and k. The task is to sort all integers in the interval [lo, hi] by the power value in ascending order, if two or more integers have the same power value sort them by ascending order. Return the kth integer in the range [lo, hi] sorted by the power value. Notice that for any integer x (lo <= x <= hi) it is guaranteed that x will transform into 1 using these steps and that the power of x is will fit in a 32-bit signed integer. Example 1: Input: lo = 12, hi = 15, k = 2 Output: 13 Explanation: The power of 12 is 9 (12 --> 6 --> 3 --> 10 --> 5 --> 16 --> 8 --> 4 --> 2 --> 1) The power of 13 is 9 The power of 14 is 17 The power of 15 is 17 The interval sorted by the power value [12,13,14,15]. For k = 2 answer is the second element which is 13. Notice that 12 and 13 have the same power value and we sorted them in ascending order. Same for 14 and 15. Example 2: Input: lo = 7, hi = 11, k = 4 Output: 7 Explanation: The power array corresponding to the interval [7, 8, 9, 10, 11] is [16, 3, 19, 6, 14]. The interval sorted by power is [8, 10, 11, 7, 9]. The fourth number in the sorted array is 7. Constraints: 1 <= lo <= hi <= 1000 1 <= k <= hi - lo + 1

Explanation

• Calculate Power and Store: Calculate the power of each number in the given range [lo, hi] and store the number along with its power in a list of tuples.
  • Sort: Sort the list of tuples based on power (ascending) and then by the number itself (ascending) in case of ties.
  • Return Kth Element: Return the number at the kth position (k-1 index) in the sorted list.

• Runtime Complexity: O(N log N), where N is the size of the range (hi - lo + 1). This is due to the sorting step. The power calculation is effectively constant time per number due to the constraints.
• Storage Complexity: O(N) to store the numbers and their corresponding powers.

Code

    def getKth(lo: int, hi: int, k: int) -> int:
    """
    Given three integers lo, hi and k. Consider the integers in the interval [lo, hi] sorted by the power value in ascending order,
    if two or more integers have the same power value sort them by ascending order.
    Return the kth integer in the range [lo, hi] sorted by the power value.
    """

    def calculate_power(x: int) -> int:
        """Calculates the power of a number x."""
        power = 0
        while x != 1:
            if x % 2 == 0:
                x //= 2
            else:
                x = 3 * x + 1
            power += 1
        return power

    power_values = []
    for num in range(lo, hi + 1):
        power_values.append((num, calculate_power(num)))

    power_values.sort(key=lambda item: (item[1], item[0]))

    return power_values[k - 1][0]