Least number of perfect square numbers that sums upto n

You can simplify your solution to:

def numSquares(self,n):
    if(n == 0):
        return 0
    if(n == 1):
        return 1
    squares = self.findSquares(n)
    rows = len(squares)
    cols = n + 1
    mat = [n] * cols
    mat[0] = 0

    for s in squares:
        for j in range(s,cols):
            mat[j] = min(mat[j], 1 + mat[j - s])

    return mat[n]

This avoids using:

1. the self.min function
2. the division/modulus operation in the inner loop.
3. the 2d array

and is about twice as fast.

A bit late, but I believe this answer can help others as it did me. This below is the fastest solution possible with O(sqrt(n)) time complexity

It is based on Lagrange’s four-square theorem every natural number can be represented as the sum of four integer squares. So the answer set would be 1, 2, 3 or 4.

class Solution:
    def is_perfect(self, n):
        x = int(math.sqrt(n))
        return x * x == n

    def numSquares(self, n: int) -> int:
        if n < 4:
            return n

        if self.is_perfect(n):  # number is a perfect square
            return 1

        # the result is 4 if number = 4^k*(8*m + 7)
        while n & 3 == 0:  # while divisible by 4
            n >>= 2
        if n & 7 == 7:  # 8m+7 => last 3 digits = 111
            return 4

        x = int(math.sqrt(n))
        for i in range(1, x + 1):  # sum of 2 perfect squares
            if self.is_perfect(n - i * i):
                return 2

        return 3  # by elimination