動機

複習換硬幣的dp isqrt是個很貴的運算

Problem

Given an integer n, return the least number of perfect square numbers that sum to n.

A perfect square is an integer that is the square of an integer; in other words, it is the product of some integer with itself. For example, 1, 4, 9, and 16 are perfect squares while 3 and 11 are not.

 

Example 1:

Input: n = 12Output: 3Explanation: 12 = 4 + 4 + 4.

Example 2:

Input: n = 13Output: 2Explanation: 13 = 4 + 9.

 

Constraints:

  • 1 <= n <= 104

Sol

isqrt是個很貴的運算，所以要先把硬幣算完，再跑dp

class Solution:
    def numSquares(self, n: int) -> int:
        cs = set([x**2 for x in range(isqrt(n),0,-1)])
        @cache
        def dp(n):
            if n in cs:
                return 1
            else:
                ret = float('inf')
                for x in cs:
                    if n-x > 0:
                        ret = min(ret, dp(n-x)+1)
                return ret
        return dp(n)