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

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Sum of Square Numbers LeetCode Solution

Problem Statement

Given a non-negative integer c, decide whether there’re two integers a and b such that a² + b² = c.

Example 1:
Input: c = 5
Output: true
Explanation: 1 * 1 + 2 * 2 = 5

Example 2:
Input: c = 3
Output: false

Sum of Square Numbers Leetcode Solution C++

class Solution {
public:
    bool judgeSquareSum(int c) {
         if(c==1 || c==0)return true;
        for(long int i=0;i*i<=c;i++){
           long int x=i*i;
           if (isPerfectSquare(c-x))
               return true;
        }
                   return false;        
    }
     bool isPerfectSquare( long int n) {
        if(n==1 || n==0)return true;
      long int s=0,e=n;
        
        while(s<=e){
            long int m=s+(e-s)/2;
            if(m*m==n)return true;
           else if(m*m<n)
                s=m+1;
            else e=m-1;
        }
        return false;
    }
};Code language: PHP (php)

Sum of Square Numbers Leetcode Solution Java

class Solution {
    public boolean judgeSquareSum(int c) {
        long s = 0;
        long e =  (long)Math.sqrt(c);
        while(s<=e){
            long mid = s*s + e*e;
            if(mid==(long)c){
                return true;
            }else if(mid>(long)c){
                e--;
            }else{
                s++;
            }
        }
        return false;
    }
}Code language: JavaScript (javascript)

Sum of Square Numbers Solution JavaScript

var judgeSquareSum = function(c) {
    var r = Math.floor(Math.sqrt(c));
    var l = 0;
    while(l <= r){
        var sum  = l**2 + r**2
        if( sum === c){
            return true
        }else if(sum < c){
            l++
        }else{
            r--
        }
    }
    return false
};Code language: JavaScript (javascript)

Sum of Square Numbers Solution Python

class Solution(object):
    def judgeSquareSum(self, c):
        cc = int(c ** 0.5)
        left = 0
        right = cc
        while left <= right:
            res = left ** 2 + right ** 2
            if res == c:
                return True
            elif res < c:
                left += 1
            elif res > c:
                right -= 1
        return False