Solution: Nested Loop with Multiplication (Pro)
This review provides a detailed analysis of the different ways to solve the Nested Loop with Multiplication (Pro) challenge.
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int main(){int n =10; // O(1)int sum = 0; // O(1)int j = 1; // O(1)float pie = 3.14; // O(1)for (int i = 0; i < n; i++){ // O(n)cout << pie << endl; // O(n)while (j < i){ // O(log2(n))sum+=1; // O(log2(n))j*=2; // O(log2(n))}}cout << sum << endl; // O(1)}
The outer loop in the main function has n iterations as it iterates on i from 0 to n-1. If the condition j < i is true, the inner loop is entered. However, immediately, j is doubled. Note that j is not reset to 1 in the code since the j is immediately doubled, therefore inner while loop will run ...