Perfect Squares 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 <= 104
Perfect Squares Solutions
✅Time: O(log n)
✅Space: O(n log n)
C++
class Solution {
public:
int numSquares(int n) {
vector<int> dp(n + 1, n); // 1^2 x n
dp[0] = 0; // no way
dp[1] = 1; // 1^2
for (int i = 2; i <= n; ++i)
for (int j = 1; j * j <= i; ++j)
dp[i] = min(dp[i], dp[i - j * j] + 1);
return dp[n];
}
};
Java
class Solution {
public int numSquares(int n) {
int[] dp = new int[n + 1];
Arrays.fill(dp, n); // 1^2 x n
dp[0] = 0; // no way
dp[1] = 1; // 1^2
for (int i = 2; i <= n; ++i)
for (int j = 1; j * j <= i; ++j)
dp[i] = Math.min(dp[i], dp[i - j * j] + 1);
return dp[n];
}
}
Python
class Solution:
def numSquares(self, n: int) -> int:
dp = [n] * (n + 1)
dp[0] = 0
dp[1] = 1
for i in range(2, n + 1):
j = 1
while j * j <= i:
dp[i] = min(dp[i], dp[i - j * j] + 1)
j += 1
return dp[n]
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