Sort Integers by The Power Value

Medium

Asked by:

11 views

Topics:

ArraysSortingDynamic Programming

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 k^th 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

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the possible ranges for the integers in the input array and for the power value calculation resulting from those integers?
What is the specified range [lo, hi] for calculating the power value, and what happens if lo > hi or either is negative?
In case of ties (multiple integers with the same power value), what should be the sorting order among them?
Should the returned array contain the sorted numbers themselves or their original indices in the input range [lo, hi]?
If k is greater than the number of integers within the range [lo, hi], what should the function return?

Brute Force Solution

Approach

The brute force method for this problem involves figuring out a special value (called the power value) for each number in a given range. Then, we sort all the numbers based on these power values. If two numbers have the same power value, we sort them by their original value.

Here's how the algorithm would work step-by-step:

For each number in the given range, calculate its power value.
To calculate the power value of a number, repeatedly apply a rule. If the number is even, divide it by 2. If the number is odd, multiply it by 3 and add 1. Count how many steps it takes to reach the number 1.
Store each number along with its calculated power value.
After calculating the power value for all numbers in the range, sort them based on their power values. Numbers with smaller power values should come first.
If two numbers have the same power value, sort them based on their original value. Smaller numbers should come first.
The problem asks for the number at a specific position after this sorting. Return the number at that position.

Code Implementation

def sort_by_power(low, high, kth_element):
    power_values = []

    for number in range(low, high + 1):
        power = calculate_power(number)
        power_values.append((number, power))

    # Sort based on power value, then original value
    power_values.sort(key=lambda item: (item[1], item[0]))

    return power_values[kth_element - 1][0]

def calculate_power(number):
    power_count = 0
    current_number = number

    # Simulate the power calculation process
    while current_number != 1:
        if current_number % 2 == 0:
            current_number //= 2
        else:
            current_number = 3 * current_number + 1
        power_count += 1

    return power_count

Big(O) Analysis

Time Complexity

O(n log n) – We iterate through n numbers in the given range to calculate the power value for each number. Calculating the power value for a single number involves a sequence of operations that, in the worst case, may not be easily bound. However, the dominant cost comes from sorting the n numbers based on their power values. Using an efficient sorting algorithm such as merge sort or quicksort yields a time complexity of O(n log n). Since the power value calculation is performed n times, and sorting is O(n log n), the overall complexity is dominated by the sorting step. Thus, the time complexity is O(n log n).

Space Complexity

O(N) – The algorithm stores each number in the given range along with its calculated power value. This requires creating a data structure, such as a list or an array, to hold these number-power value pairs. Therefore, the space needed is directly proportional to the number of elements in the range, which we can denote as N. The auxiliary space complexity is thus O(N).

Optimal Solution

Approach

The trick to solving this problem efficiently is to remember the 'power value' for numbers we've already calculated. We'll use this 'memory' to avoid repeating calculations and speed things up considerably.

Here's how the algorithm would work step-by-step:

Create a way to store the power value for each number we calculate.
Write a simple rule to calculate the power value of a number. Keep applying this rule until the number becomes 1.
As you calculate the power value for a number, store it in your 'memory'.
When you need to find the power value of another number, first check if you've already calculated and stored it. If you have, use the stored value directly instead of recalculating.
Once you have the power values for all the numbers in the given range, sort the numbers based on their power values. If two numbers have the same power value, sort them by their original value.
Return the number at the requested position in the sorted list.

Code Implementation

def sort_integers_by_power_value(low, high, k):
    power_value_memory = {}

    def calculate_power_value(number):
        if number in power_value_memory:
            return power_value_memory[number]
        
        steps = 0
        current_number = number
        
        while current_number != 1:
            if current_number % 2 == 0:
                current_number = current_number // 2
            else:
                current_number = 3 * current_number + 1
            steps += 1

            if current_number in power_value_memory:
                steps += power_value_memory[current_number]
                break

        power_value_memory[number] = steps
        return steps

    numbers_with_power = []
    for number in range(low, high + 1):
        power = calculate_power_value(number)
        numbers_with_power.append((number, power))

    # Sort based on power value, then original value.
    numbers_with_power.sort(key=lambda item: (item[1], item[0]))

    # Return the k-th smallest number based on power value.
    return numbers_with_power[k - 1][0]

Big(O) Analysis

Time Complexity

O(n * log(n)) – Calculating the power value for each number from lo to hi involves repeatedly applying the transformation rule until the number reaches 1. In the worst case, this can take logarithmic time relative to the initial number. We memoize these power values, which avoids recomputation. We then sort n numbers (hi - lo + 1) based on their power values, which takes O(n * log(n)) time using a standard sorting algorithm. Thus, the overall time complexity is dominated by the sorting step, yielding O(n * log(n)).

Space Complexity

O(N) – The algorithm uses a memory (likely a hash map or array) to store the calculated power values for each number within the given range [lo, hi], where N is the number of integers in that range (hi - lo + 1). This memoization technique avoids redundant calculations by storing intermediate results. The size of this memory scales linearly with the number of integers in the range. Thus, the auxiliary space complexity is O(N).

Edge Cases

Case	How to Handle
k is zero or negative	Return an empty array or throw an exception, as the problem specifies a positive k.
lo is greater than hi	Return an empty array since the range is invalid.
lo and hi are equal	Calculate power of the single number and return it in a list if k is 1, otherwise return an empty list.
Integer overflow when calculating the power value	Use a long data type to store the power value to avoid integer overflow or handle overflow cases to avoid errors in sorting or comparisons.
Large range (hi - lo) with small k	Optimize the caching strategy to only store power values for numbers within the range to avoid excessive memory usage.
Duplicate power values within the range	Ensure the sorting algorithm is stable to maintain the original order of numbers with the same power value.
Extreme boundary values of lo and hi within the integer range	The solution should handle extreme value boundaries correctly without causing errors from calculation, such as overflows.
Memory constraints for caching power values within a very large range	Implement an LRU cache or a similar eviction strategy to limit the memory usage for caching power values, or calculate the power on demand.

k is zero or negative

How to Handle:

Return an empty array or throw an exception, as the problem specifies a positive k.

lo is greater than hi

How to Handle:

Return an empty array since the range is invalid.

lo and hi are equal

How to Handle:

Calculate power of the single number and return it in a list if k is 1, otherwise return an empty list.

Integer overflow when calculating the power value

How to Handle:

Use a long data type to store the power value to avoid integer overflow or handle overflow cases to avoid errors in sorting or comparisons.

Large range (hi - lo) with small k

How to Handle:

Optimize the caching strategy to only store power values for numbers within the range to avoid excessive memory usage.

Duplicate power values within the range

How to Handle:

Ensure the sorting algorithm is stable to maintain the original order of numbers with the same power value.

Extreme boundary values of lo and hi within the integer range

How to Handle:

The solution should handle extreme value boundaries correctly without causing errors from calculation, such as overflows.

Memory constraints for caching power values within a very large range

How to Handle:

Implement an LRU cache or a similar eviction strategy to limit the memory usage for caching power values, or calculate the power on demand.