# Perfect Squares LeetCode Solution

Here, We see Perfect Squares LeetCode Solution. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc. with different approaches.

# List of all LeetCode Solution

## Problem Statement

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.

## Perfect Squares Leetcode SolutionC++

``````class Solution {
public:
int numSquares(int n) {
if (n <= 0)
{
return 0;
}
vector<int> cntPerfectSquares(n + 1, INT_MAX);
cntPerfectSquares[0] = 0;
for (int i = 1; i <= n; i++)
{
for (int j = 1; j*j <= i; j++)
{
cntPerfectSquares[i] =
min(cntPerfectSquares[i], cntPerfectSquares[i - j*j] + 1);
}
}
return cntPerfectSquares.back();
}
};```Code language: PHP (php)```

## Perfect Squares Leetcode SolutionJava

``````class Solution {
public int numSquares(int n) {
int[] dp = new int[n + 1];
Arrays.fill(dp, Integer.MAX_VALUE);
dp[0] = 0;
int count = 1;
while (count * count <= n) {
int sq = count * count;
for (int i = sq; i <= n; i++) {
dp[i] = Math.min(dp[i - sq] + 1, dp[i]);
}
count++;
}
return dp[n];
}
}```Code language: JavaScript (javascript)```

## Perfect Squares SolutionJavaScript

``````var numSquares = function(n) {
let dp = new Array(n + 1).fill(Infinity);
dp[0] = 0;
for (let i = 1; i <= n; ++i) {
let min_val = Infinity;
for (let j = 1; j * j <= i; ++j) {
min_val = Math.min(min_val, dp[i - j * j] + 1);
}
dp[i] = min_val;
}
return dp[n];
};
```Code language: JavaScript (javascript)```

## Perfect Squares SolutionPython

``````class Solution(object):
def numSquares(self, n):
dp = [float('inf')] * (n + 1)
dp[0] = 0
count = 1
while count * count <= n:
sq = count * count
for i in range(sq, n + 1):
dp[i] = min(dp[i - sq] + 1, dp[i])
count += 1
return dp[n]```Code language: HTML, XML (xml)```
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