Solution: Big (O) of Nested Loop with Multiplication
This review provides a detailed analysis of the different ways to solve the Big (O) of nested loop with multiplication.
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Given code
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class NestedLoop {public static void main(String[] args) {int n = 10; // O(time complexity of the called function)int sum = 0; //O(1)double pie = 3.14; //O(1)int var = 1;while(var < n) { // O(log n)System.out.println("Pie: " + pie); // O(log n)for (int j = 0; j < var; j++) { // 2nsum++; // (2n-1)}var *= 2; // O(log n)} //end of while loopSystem.out.println("Sum: " + sum); //O(1)} //end of main} //end of class
Explanation
The answer is . Have a look at the slides below for an in-depth explanation of the answer.
Time Complexity
The above slides give a detailed, step-by-step analysis of the code. Here, we provide a more summarized version.
The outer loop here runs ...
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