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Sort Integers by The Power Value

Difficulty: Medium

Problem Description

Given three integers lo, hi, and k, sort all integers in the interval [lo, hi] by their power value in ascending order. 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. Return the kth integer in the sorted order.

Key Insights

  • The transformation of integers follows the Collatz conjecture rules.
  • The power value determines how many steps it takes to reach 1.
  • We can use a sorting algorithm based on power values, using tuples to maintain order for integers with the same power value.
  • The constraints allow us to use straightforward computation without performance concerns.

Space and Time Complexity

Time Complexity: O(n log n), where n is the number of integers in the range [lo, hi] due to sorting. Space Complexity: O(n), for storing the power values and the list of integers.

Solution

To solve the problem, we will iterate through each integer in the range [lo, hi], calculate its power value, and store these values in a list of tuples where each tuple contains the power value and the integer itself. After that, we will sort this list based on the power value and the integer value. Finally, we will return the kth integer from the sorted list.

Code Solutions

def get_power_value(x):
    steps = 0
    while x != 1:
        if x % 2 == 0:
            x //= 2
        else:
            x = 3 * x + 1
        steps += 1
    return steps

def getKth(lo, hi, k):
    power_values = []
    for i in range(lo, hi + 1):
        power = get_power_value(i)
        power_values.append((power, i))
    # Sort by power value first, then by the integer value
    power_values.sort()
    return power_values[k - 1][1]
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