具有二进制搜索的函数的时间复杂度和空间复杂度

Question

I wrote the following function in Java ;

int foo(int a[], int n) {
    int num1 = 1, num2 = 1; 
    while(num1 < n) {
        if (binarySearch(a,num1,num2) >= 0) {
            return num2;
        }
        num1 = 2 * num1;
        num2 = 2 * num2;
    }
    return 0;
}

I'm trying to figure out the time complexity and space complexity of this function. I know that the time complexity of binarySearch is O(logn) and space complexity of this function is O(1) . With this information, I tried to calculate those things from of foo function. I think that the time complexity of foo is O((logn)^2) and space complexity is O(1) but I am not sure about it. What is the best method to calculate those things?

Answer 1

Let's analyze a simplified while loop without the binarySearch() call:

while(num1 < n) {
    num1 = 2 * num1;
}

How many steps does this take as a function of n ?

Once you answer this question, the key to a final solution is to realize that the complexity of a while loop multiplies by the complexity of its body. This means that the final answer is the answer to the above question time log n .