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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]
    }
}

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