<p>Given an integer <code>n</code>, return <em>the least number of perfect square numbers that sum to</em> <code>n</code>.</p>

<p>A <strong>perfect square</strong> is an integer that is the square of an integer; in other words, it is the product of some integer with itself. For example, <code>1</code>, <code>4</code>, <code>9</code>, and <code>16</code> are perfect squares while <code>3</code> and <code>11</code> are not.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> n = 12
<strong>Output:</strong> 3
<strong>Explanation:</strong> 12 = 4 + 4 + 4.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> n = 13
<strong>Output:</strong> 2
<strong>Explanation:</strong> 13 = 4 + 9.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n &lt;= 10<sup>4</sup></code></li>
</ul>

'''
- Greedy approach can be whatever the number is given,, consider the Nearest Left-Sided Perfect Square and Subtract it and further do the process for the other resulting guys
- Greedy approach will give non-optimal answers
- So the remaining way is to do for all possibilities
- Okay,, so for a given number check with all the perfect squares which are less than or equal to that number
'''

class Solution:
    def numSquares(self, n: int) -> int:
        memo = [-1] * (n+1)
        def recur(n):
            if n <= 0:
                return 0
            
            if memo[n] != -1:
                return memo[n]

            # For each recursive call the number will change,, so the Possible Denominations will change,, so for each number we need to find the minimum count
            min_count = float('inf')

            # Generating the Perfect Squares which will come before the number
            for i in range(1, n+1):
                if i * i > n:
                    break
                
                # Else if that perfect square is less than n
                resulting_num = n - (i * i)
                res = 1 + recur(resulting_num)
                # In above line Adding 1,, denotes 1 perfect square is considered

                min_count = min(min_count , res)
            
            memo[n] = min_count
            return memo[n]
                
        return recur(n)
# ------------------------------------
'''
from functools import lru_cache
class Solution:
    def numSquares(self, n: int) -> int:
        @lru_cache(None)
        def recur(n):
            if n <= 0:
                return 0
            
            # For each recursive call the number will change,, so the Possible Denominations will change,, so for each number we need to find the minimum count
            min_count = float('inf')

            # Generating the Perfect Squares which will come before the number
            for i in range(1, n+1):
                if i * i > n:
                    break
                
                # Else if that perfect square is less than n
                resulting_num = n - (i * i)
                res = 1 + recur(resulting_num)
                # In above line Adding 1,, denotes 1 perfect square is considered

                min_count = min(min_count , res)
            
            return min_count
                
        return recur(n)
'''
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