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:
d p [ i ] = d p [ j ] + 1 , ∀ j < i , j is a perfect square dp[i] = dp[j] + 1, \forall j < i, \text{j is a perfect square} dp[i]=dp[j]+1,∀j<i,j is a perfect square
Time complexity: o ( n log n ) o(n\log n) o(nlogn)
Space complexity: o ( n ) o(n) 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