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Data Structures for Coding Interviews in C++

Solution: Big (O) of Nested Loop with Multiplication

Explore how to analyze the time complexity of a nested loop where the inner loop iterations multiply in powers of two. Understand how to apply Big O notation with logarithmic bounds and geometric series to determine an upper time complexity bound of O(n). This lesson helps you grasp key concepts for algorithm analysis and complexity measures.

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Solution #

C++
int main() {
  int n = 10;    
  int sum = 0;   
  float pie = 3.14;  
  int var = 1;
  while (var < n){      // O(log n)
    cout << pie << endl; // O(log n)
    for (int j=0; j<var; j++)  // 8n - 12
      sum+=1;    //  (4n - 6)
    var*=2;  // O(log n)
  }
  cout<<sum; 
}

Time Complexity

The outer loop here runs log(n)log(n) times. In the first iteration of the outer loop, the body of the inner loop runs once. In the second iteration, it runs twice, and so on. The number of executions of the body of the inner loop increases in powers of 2. So, if kk is the number of iterations of the outer loop, the body of the inner loop runs a total of 1+2+4+8+⋯+2k1+2+4+8+\cdots+2^k times. This geometric series sums to 2k+1−12^{k+1}-1. The inner loop condition requires that in the last time the inner loop runs, it runs at most nn times. This requires 2k<n2^k < n ...