# Perfect Squares - Amazon Top Interview Questions

### Problem Statement :

```Write a program that determines the smallest number of square numbers that sum up to n.

Constraints

1 ≤ n ≤ 100,000

Example 1

Input
n = 4

Output
1

Explanation
4 is already the square of 2.

Example 2

Input
n = 17

Output
2

Explanation
16 + 1

Example 3

Input
n = 18

Output
2

Explanation
9 + 9```

### Solution :

```                        ```Solution in C++ :

int solve(int n) {
int m = 1;
while (n % 2 == 0) {
if (n % 4 == 0) {
n /= 4;
} else {
n /= 2;
m *= 2;
}
}
// now n is odd.
bool check_4p3 = false;

// prime factorize with O(sqrt(n))
// you can reduce time by any other prime factorize method.
for (int i = 3; i <= n / i; i += 2) {
if (n % i == 0) {
int state = 0;
while (n % i == 0) {
n /= i;
state ^= 1;
}
if (state) {
m *= i;
if (i % 4 == 3) {
check_4p3 = true;
}
}
}
}
if (n > 1) {
m *= n;
if (n % 4 == 3) check_4p3 = true;
}
// now original n = m * (some square number)
// case 1 : n is already square number.
if (m == 1) return 1;
// case 2 : there is no prime factor s.t. p%4 == 3
if (!check_4p3) return 2;
// case 3 : m % 8 is not 7. I don't know why!
if (m % 8 != 7) return 3;
// case 4 : else. by lagrange four square theorem
return 4;
}```
```

```                        ```Solution in Java :

import java.lang.Math;

class Solution {
public int solve(int n) {
if (isSquare(n)) {
return 1;
}
for (int i = (int) Math.floor(Math.sqrt(n)); (int) Math.pow(i, 2) >= n / 2; i--) {
if (isSquare(n - (int) Math.pow(i, 2))) {
return 2;
}
}
// Legendre's three-square theorem
return isLegendres(n) ? 4 : 3;
}

private boolean isSquare(int n) {
double root = Math.sqrt(n);
return root == Math.floor(root);
}
private boolean isLegendres(int n) {
for (int a = 1; a <= n - 7; a *= 4) {
if (n % a == 0 && (((n / a) - 7) % 8 == 0)) {
return true;
}
}
return false;
}
}```
```

```                        ```Solution in Python :

class Solution:
def solve(self, n):
squares = []
for i in range(1, int(n ** 0.5) + 1):
squares.append(i ** 2)

squares = squares[::-1]  # start from the largest one would reduce the time
q = deque([0])  # cur value
visited = set()
step = 0
while q:
for _ in range(len(q)):
node = q.popleft()
if node == n:
return step

if node > n:
continue

if node in visited:
continue
visited.add(node)

for s in squares:
q.append(node + s)

step += 1```
```

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