# 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 <= 104

## Solutions

For dynamic programming, define dp[i] to represent the least number of perfect square numbers that sum to i.

• class Solution {
public int numSquares(int n) {
int m = (int) Math.sqrt(n);
int[] f = new int[n + 1];
Arrays.fill(f, 1 << 30);
f[0] = 0;
for (int i = 1; i <= m; ++i) {
for (int j = i * i; j <= n; ++j) {
f[j] = Math.min(f[j], f[j - i * i] + 1);
}
}
return f[n];
}
}

• class Solution {
public:
int numSquares(int n) {
int m = sqrt(n);
int f[n + 1];
memset(f, 0x3f, sizeof(f));
f[0] = 0;
for (int i = 1; i <= m; ++i) {
for (int j = i * i; j <= n; ++j) {
f[j] = min(f[j], f[j - i * i] + 1);
}
}
return f[n];
}
};

• class Solution:
def numSquares(self, n: int) -> int:
m = int(sqrt(n))
f = [0] + [inf] * n
for i in range(1, m + 1):
for j in range(i * i, n + 1):
f[j] = min(f[j], f[j - i * i] + 1)
return f[n]


• func numSquares(n int) int {
m := int(math.Sqrt(float64(n)))
f := make([]int, n+1)
for i := range f {
f[i] = 1 << 30
}
f[0] = 0
for i := 1; i <= m; i++ {
for j := i * i; j <= n; j++ {
f[j] = min(f[j], f[j-i*i]+1)
}
}
return f[n]
}

• function numSquares(n: number): number {
const m = Math.floor(Math.sqrt(n));
const f: number[] = Array(n + 1).fill(1 << 30);
f[0] = 0;
for (let i = 1; i <= m; ++i) {
for (let j = i * i; j <= n; ++j) {
f[j] = Math.min(f[j], f[j - i * i] + 1);
}
}
return f[n];
}


• impl Solution {
pub fn num_squares(n: i32) -> i32 {
let (row, col) = ((n as f32).sqrt().floor() as usize, n as usize);
let mut dp = vec![vec![i32::MAX; col + 1]; row + 1];
dp[0][0] = 0;
for i in 1..=row {
for j in 0..=col {
dp[i][j] = dp[i - 1][j];
if j >= i * i {
dp[i][j] = std::cmp::min(dp[i][j], dp[i][j - i * i] + 1);
}
}
}
dp[row][col]
}
}