leetcode - 279. Perfect Squares

Description

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

Example 2:

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

Constraints:

1 <= n <= 10^4

Solution

Solved after help…

DP

Use dp[i] to denote the number of perfect squares of number i, since the number can only be summed by perfect squares, the transformation equation will be:
$$dp[i]=dp[j]+1,\forall j<i,\text{j is a perfect square}$$

Time complexity: $o(n\log n)$
Space complexity: $o(n)$

BFS

Similar to dp, just like this:

Image ref

Code

DP

class Solution:
    def numSquares(self, n: int) -> int:
        dp = [i for i in range(n + 1)]
        for i in range(1, n + 1):
            j = 1
            while j * j <= i:
                dp[i] = min(dp[i], 1 + dp[i - j * j])
                j += 1
        return dp[-1]

BFS

class Solution:
    def numSquares(self, n: int) -> int:
        queue = collections.deque([(n, 0)])
        visited = set()
        while queue:
            number, step = queue.popleft()
            if number in visited:
                continue
            visited.add(number)
            if number == 0:
                return step
            i = 1
            while i * i <= number:
                queue.append((number - i * i, step + 1))
                i += 1
        return -1