Practice Questions on Time Complexity Analysis in C++

Time complexity of any algorithm is the time taken by the algorithm to complete. It is an important metric to show the efficiency of the algorithm and for comparative analysis. We tend to reduce the time complexity of algorithm that makes it more effective.

Example 1

Find the time complexity of the following code snippets

for(i= 0 ; i < n; i++){
   cout<< i << " " ;
   i++;
}

The loop has maximum value n but the i will be incremented twice in the for loop which will make the time take half. So the time complexity is O(n/2) which is equivalent to O(n).

Example 2

Find the time complexity of the following code snippets

for(i= 0 ; i < n; i++){
   for(j = 0; j<n ;j++){
      cout<< i << " ";
   }
}

The inner loop and the outer loop both are executing n times. So for single value of i, j is looping n times, for n values of i, j will loop total n*n = n 2 times. So the time complexity is O(n 2 ).

Example 3

Find the time complexity of the following code snippets

int i = n;
while(i){
   cout << i << " ";
   i = i/2;
}

In this case, after each iteration the value of i is turned into half of its previous value. So the series will be like: . So the time complexity is O(log n).

Example 4

Find the time complexity of the following code snippets

if(i > j ){
   j>23 ? cout<<j : cout<<i;
}

There are two conditional statements in the code. Each conditional statement has time complexity = O(1), for two of them it is O(2) which is equivalent to O(1) which is constant.

Example 5

Find the time complexity of the following code snippets

for(i= 0; i < n; i++){
   for(j = 1; j < n; j = j*2){
      cout << i << " ";
   }
}

The inner loop is executing (log n) times where the outer is executing n times. So for single value of i, j is executing (log n) times, for n values of i, j will loop total n*(log n) = (n log n) times. So the time complexity is O(n log n).