Leetcode Perfect Squares problem solution

YASH PAL,

In this Leetcode Perfect Squares problem solution, we have 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.

Leetcode Perfect Squares problem solution

Problem solution in Python.

class Solution(object):
    def numSquares(self, n):
        from math import ceil, sqrt
        dp = [0, 1, 2, 3]
        if n<=3:
            return dp[n]
        else:
            for i in range(4, n + 1):
                dp.append(i)
                for x in range(1, int(ceil(sqrt(i))) + 1):
                    temp = x*x
                    if temp>i:
                        break
                    else:
                        dp[i] = min(dp[i], 1 + dp[i - temp])
        return dp[n]

Problem solution in Java.

class Solution {
    public int numSquares(int n) {
        int sqrtN = (int) Math.sqrt(n);
        int[] squares = new int [sqrtN];
        for (int i = 1; i <= sqrtN; i++) {
            squares[i - 1] = i * i;
        }
        int[] dp = new int [n + 1];
        for (int i = 1; i <= n; i++) {
            dp[i] = n + 1;
            for (int j = 0; j < squares.length; j++) {
                if (i < squares[j]) {
                    continue;
                }
                if (squares[j] == i) {
                    dp[i] = 1;
                    continue;
                } else {
                    dp[i] = Math.min(dp[i], 1 + dp[i - squares[j]]);
                }
            }
        }
        return dp[n];
    }
}

Problem solution in C++.

class Solution {
public:
    int numSquares(int n)
    {
        if (n==0) return 0;
        vector<int> per;
        int f[n+5]={0};
        f[1]=1;
        for(int i=2;i<=n;i++)
        {
            int minV=INT_MAX;
            int j=1;
            while (j*j<=i)
            {
                if (j*j==i)
                {
                    minV=1;break;
                }
                minV=min(minV,f[i-j*j]+1);
                j++;
            }
            f[i]=minV;
        }
        return f[n];
    }
};

Problem solution in C.

void solve(int n, int * ans, int a){
    if (a > *ans) return;
    if (n == 0) {
        if (a < *ans) *ans = a;
        return; 
    }
    
    int z = 1;
    
    while (z++) { 
        // Find the biggest perfect square that is less than n.
        if (z*z > n) {
            z = z-1;
            break;
        }
    }

    while (z > 0) {
        // Start with the biggest perfect square less than n and
        // continue trying every perfect square less than that.
        solve(n - z*z, ans, a+1);
        if (a+2 >= *ans) break;
        z = z-1;
    } 

    return; 
}   

int numSquares(int n) {
    int ans = INT_MAX;
    solve(n, &ans, 0);
    return ans;
}

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