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Algorithms for Coding Interviews in C++

Solution: Nested Loop with Multiplication (Pro)

Explore how to determine the time complexity of a nested loop where the inner loop doubles its iterator. Understand the use of logarithmic analysis and how to simplify expressions to Big O notation to evaluate algorithm efficiency.

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Solution #

C++
int main(){
  int n =10;    // you can change the value of n
  int sum = 0;
  int j = 1;
  float pie = 3.14;
  
  for (int i = 0; i < n; i++){
    cout << pie << endl;
    while (j < i){
      sum+=1;
      j*=2;  
    }
  }
  cout << sum << endl;
}

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 O(log2(n))O(log_2(n)) times for all the iterations of the outer loop.

Note: Inner while loop will run at most once for each iteration of the outer loop.

Statement Number of Executions
int n = 10;
1
int sum = 0;
1
int j = 1;
1
float pie = 3.14;
1
int i=0;
11
i<n;
n+1n+1
i++
...