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.

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Perfect Squares LeetCode Solution

Problem Statement

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

A 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 Solution C++

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 Solution Java

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 Solution JavaScript

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 Solution Python

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)