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
For dynamic programming, define dp[i] to represent the least number of perfect square numbers that sum to i.
class Solution:
def numSquares(self, n: int) -> int:
dp = [0] * (n + 1)
for i in range(1, n + 1):
j, mi = 1, float('inf')
while j * j <= i:
mi = min(mi, dp[i - j * j])
j += 1
dp[i] = mi + 1
return dp[-1]class Solution {
public int numSquares(int n) {
int[] dp = new int[n + 1];
for (int i = 1; i <= n; ++i) {
int mi = Integer.MAX_VALUE;
for (int j = 1; j * j <= i; ++j) {
mi = Math.min(mi, dp[i - j * j]);
}
dp[i] = mi + 1;
}
return dp[n];
}
}class Solution {
public:
int numSquares(int n) {
vector<int> dp(n + 1);
for (int i = 1; i <= n; ++i) {
int mi = 100000;
for (int j = 1; j * j <= i; ++j) {
mi = min(mi, dp[i - j * j]);
}
dp[i] = mi + 1;
}
return dp[n];
}
};function numSquares(n: number): number {
let dp = new Array(n + 1).fill(0);
for (let i = 1; i <= n; ++i) {
let min = Infinity;
for (let j = 1; j * j <= i; ++j) {
min = Math.min(min, dp[i - j * j]);
}
dp[i] = min + 1;
}
return dp.pop();
};func numSquares(n int) int {
dp := make([]int, n+1)
for i := 1; i <= n; i++ {
mi := 100000
for j := 1; j*j <= i; j++ {
mi = min(mi, dp[i-j*j])
}
dp[i] = mi + 1
}
return dp[n]
}
func min(a, b int) int {
if a < b {
return a
}
return b
}