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# Perfect Squares LeetCode Solution – Queslers

## Problem – Perfect Squares

Given an integer `n`, return the least number of perfect square numbers that sum to `n`.

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`

### Perfect Squares LeetCode Solution in C++

``````class Solution
{
public:
int numSquares(int n)
{
if (n <= 0)
{
return 0;
}

// cntPerfectSquares[i] = the least number of perfect square numbers
// which sum to i. Note that cntPerfectSquares is 0.
vector<int> cntPerfectSquares(n + 1, INT_MAX);
cntPerfectSquares = 0;
for (int i = 1; i <= n; i++)
{
// For each i, it must be the sum of some number (i - j*j) and
// a perfect square number (j*j).
for (int j = 1; j*j <= i; j++)
{
cntPerfectSquares[i] =
min(cntPerfectSquares[i], cntPerfectSquares[i - j*j] + 1);
}
}

return cntPerfectSquares.back();
}
};
``````

### Perfect Squares LeetCode Solution in Java

``````public int numSquares(int n) {
int[] dp = new int[n + 1];
Arrays.fill(dp, Integer.MAX_VALUE);
dp = 0;
for(int i = 1; i <= n; ++i) {
int min = Integer.MAX_VALUE;
int j = 1;
while(i - j*j >= 0) {
min = Math.min(min, dp[i - j*j] + 1);
++j;
}
dp[i] = min;
}
return dp[n];
}
``````

### Perfect Squares LeetCode Solution in Python

``````def numSquares(self, n):
if n < 2:
return n
lst = []
i = 1
while i * i <= n:
lst.append( i * i )
i += 1
cnt = 0
toCheck = {n}
while toCheck:
cnt += 1
temp = set()
for x in toCheck:
for y in lst:
if x == y:
return cnt
if x < y:
break
toCheck = temp

return cnt
``````
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