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leetcode 1387. Sort Integers by The Power Value (python)

2022-01-31 15:54:48 Wang Daya

「 This is my participation 11 The fourth of the yuegengwen challenge 16 God , Check out the activity details :2021 One last more challenge

describe

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 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.
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Example 2:

Input: lo = 1, hi = 1, k = 1
Output: 1
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Example 3:

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.
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Example 4:

Input: lo = 10, hi = 20, k = 5
Output: 13
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Example 5:

Input: lo = 1, hi = 1000, k = 777
Output: 570
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Note:

1 <= lo <= hi <= 1000
1 <= k <= hi - lo + 1
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analysis

According to the meaning , Is to give a calculation of a number power Methods , That is to put a number x Turn into 1 The number of times to go through the following operations :

  • x If it's even, then x/=2
  • x If it's odd, then x = 3*x+1

Then the range is given [lo, hi] and k , Ask us to rank the interval in ascending power order [lo, hi] Sort all integers in , If two or more integers have the same power value , Sort them in ascending order . Returns the values sorted by power [lo, hi] Within the scope of k It's an integer . The title ensures that all certificates can be obtained from x Turn into 1 .

The simplest way is to follow the meaning of the question , Write a to get a number power The value of the function getPower . Then all integers are sorted by power Value permutation , Finally, return to No k It's just an integer .

answer

class Solution(object):
    def getKth(self, lo, hi, k):
        """
        :type lo: int
        :type hi: int
        :type k: int
        :rtype: int
        """
        def getPower(n):
            result = 0
            while n!=1:
                if n%2==0:
                    n = n//2
                else:
                    n = n * 3 + 1
                result += 1
            return result
        d = []
        for i in range(lo,hi+1):
            d.append([i,getPower(i)])
        d = sorted(d, key=lambda x: x[1])
        return d[k-1][0]
            
        	      
		
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Running results

Runtime: 708 ms, faster than 35.66% of Python online submissions for Sort Integers by The Power Value.
Memory Usage: 13.6 MB, less than 86.82% of Python online submissions for Sort Integers by The Power Value.
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analysis

The above solution is from [lo, hi] There are many pairs of numbers in the process of calculating one by one power The process of value is repeated , You can use a dictionary d Direct records already know power The number of , In this way, a lot of calculation time can be saved in the later calculation process .

answer

class Solution(object):
    def getKth(self, lo, hi, k):
        """
        :type lo: int
        :type hi: int
        :type k: int
        :rtype: int
        """
        d = {1: 1}
        def getPower(n):
            if n not in d:
                if n % 2 == 0:
                    d[n] = 1 + getPower(n / 2)
                else:
                    d[n] = 1 + getPower(3*n + 1)
            return d[n]
        return sorted((getPower(i), i) for i in range(lo, hi+1))[k-1][1]
            
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Running results

Runtime: 164 ms, faster than 95.35% of Python online submissions for Sort Integers by The Power Value.
Memory Usage: 42.7 MB, less than 11.63% of Python online submissions for Sort Integers by The Power Value.
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Original link :leetcode.com/problems/so…

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author[Wang Daya],Please bring the original link to reprint, thank you.
https://en.pythonmana.com/2022/01/202201311554469020.html

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