Mary is working to find the square root of a number in C++ language using a binary search algorithm. Which of the following options can she perform to get an optimized solution?Constraints0<=x<=2^31-1Optionsint fun(int x) { if(x==1) return x; int low=0,high=x; int res=0; while(low<high){ int mid=(low+high)/2; if(mid>x/mid) high=mid; else{ res=mid; low=mid+1; } } return res;}int fun(int x) { if(x==1) return x; int low=0,high=x; int res=0; while(low<=high){ int mid=low+(high-low)/2; if(mid>x/(mid*mid) high=mid; else{ res=mid; low=mid+1; } } return res;}int fun(int x) { if(x==1) return x; int low=0,high=x; int res=0; while(low<high){ int mid=low+(high-low)/2; else if((mid*mid*mid)>x) high=mid; else{ res=mid; low=mid+1; } } return res;}int fun(int x) { int res=0; for(int i=0;i<=x;i++){ if((i*i)<=x) res=i; } return res;}

The second option is the most optimized solution for finding the square root of a number in C++ using a binary search algorithm. Here's why:

int fun(int x) {
    if(x==1) return x;
    int low=0,high=x;
    int res=0;
    while(low<=high){
        int mid=low+(high-low)/2;
        if(mid>x/mid) high=mid;
        else{
            res=mid;
            low=mid+1;
        }
    }
    return res;
}

This function uses binary search to find the square root of a number. The binary search algorithm is efficient because it reduces the search space by half at each step, leading to a time complexity of O(log n).

The function starts by initializing two pointers, low and high, to 0 and x respectively. It then enters a while loop that continues until low is greater than high.

In each iteration of the loop, it calculates the middle value mid as the average of low and high. It then checks if mid is greater than x/mid. If it is, it means that mid is too large to be the square root of x, so it sets high to mid. Otherwise, it sets res to mid and low to mid+1.

Finally, the function returns res, which is the largest integer less than or equal to the square root of x.

The other options are either incorrect or less efficient. The first and third options are incorrect because they don't correctly calculate the square root. The fourth option is less efficient because it uses a linear search, which has a time complexity of O(n).