Given an integer
n, return the least number of perfect square numbers that sum to
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,
16 are perfect squares while
11 are not.
Input: n = 12 Output: 3 Explanation: 12 = 4 + 4 + 4.
Input: n = 13 Output: 2 Explanation: 13 = 4 + 9.
1 <= n <= 104
Dynamic programming dp[i] means what is the minimum number of square numbers to add up to equal to i.
class Solution: def numSquares(self, n: int) -> int: square_nums = [i**2 for i in range(0, int(sqrt(n)) + 1)] dp = [float('inf')] * (n + 1) dp = 0 for i in range(1, n + 1): for square in square_nums: if i < square: break dp[i] = min(dp[i], dp[i - square] + 1) return dp[n]
- Time Complexity: O(N).
- Space Complexity: O(N).
where Q is the length of queries and N is the length of colors.
One Thought to “LeetCode 279. Perfect Squares”
I found this solution very popular and helpful: https://www.youtube.com/watch?v=QzU9oKjT1bo&ab_channel=EricProgramming