The smallest number of perfect square numbers summing up to n

Question

The question arises:

For a positive integer n, we find the smallest number of perfect square numbers (for example, 1, 4, 9, 16, ...) that add up to n. Question link

Example

Given n = 12, return 3 because 12 = 4 + 4 + 4; if n = 13, return 2 because 13 = 4 + 9.

NOTE

The approach I made is similar to the Integer Knackack task with valid duplicates . First, I calculated all perfect squares that are less than the number n. Now that I have them, the problem is similar to the Integer Knapsack problem. I have a number n and a list of numbers. I want to select the minimum number of numbers from the list so that their sum is n. This problem has a DP solution that I used.

Out of 586 test cases, I went through 562 test cases and got a TLE in the following. The n value for this test is 3428.

Solution presented by me:

class Solution(object):
    def numSquares(self,n):
        if(n == 0):
            return 0
        if(n == 1):
            return 1
        squares = self.findSquares(n) # returns a list of perfect squares <= n
        rows = len(squares)
        cols = n + 1
        mat = []
        for i in range(rows):
            mat.append([0] * cols)

        for i in range(cols):
            mat[0][i] = i

        for i in range(rows):
            mat[i][0] = 0

        min = mat[0][cols - 1]
        for i in range(1,rows):
            for j in range(1,cols):
                if(j < squares[i]):
                    mat[i][j] = mat[i - 1][j]

                else:
                    mat[i][j] = self.min(mat[i - 1][j], (j // squares[i] + (mat[i][j % squares[i]])))
                if(j == cols - 1 and mat[i][j] < min):
                    min = mat[i][j]
        '''
        for i in range(rows):
            for j in range(cols):
                print(mat[i][j],end=" ")
            print()
            '''
        return min

    def min(self,a,b):
        if(a <= b):
            return a
        else:
            return b


    def findSquares(self,n):

        i = 1
        squares = []
        while (i * i <= n):
            squares.append(i * i)
            i = i + 1
        return squares
'''
x = Solution()

print(x.numSquares(43))
'''

Thanks in advance.

Answer 1

:

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]

:

• self.min
• / .
• 2d

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