# Miscellaneous Problems of Time Complexity

**Prerequisite : **Asymptotic Notations

**Time Complexity :**

Time complexity is the time needed by an algorithm expressed as a function of the size of a problem. It can also be defined as the amount of computer time it needs to run a program to completion. When we solve a problem of time complexity then this definition help the most –

“It is the number of operations an algorithm performs to complete its task with respect to the input size.”

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There are following some miscellaneous problems of time complexity which are always frequently asking in different types of quizzes.

**1. What is the time complexity of the following code –**

## C

`void` `function(` `int` `n)` `{` ` ` `int` `i = 1, s = 1;` ` ` `while` `(s < n) {` ` ` `s = s + i;` ` ` `i++;` ` ` `}` `}` |

**Solution – **

Time complexity = O(√n).

**Explanation – **

We can define the ‘S’ terms according to the relation S_{i} = S_{i-1} + i. Let *k* is the total number of iterations taken by the program

i | S |

1 | 1 |

2 | 2 |

3 | 2 + 2 |

4 | 2 + 2 + 3 |

… | … |

k | 2 + 2 + 3 + 4 + ……+ k |

When S**>=**n , then loop will stop at k^{th} iterations,

⇒ S>=n ⇒ S=n

⇒ 2 + 2 + 3 + 4 + ……+ k = n

⇒ 1 + (k * (k + 1))/2 = n

⇒ k^{2} = n

k = √n

Hence, the time complexity is O(√n).

**2. What is the time complexity of the following code : **

## C++

`void` `fun(` `int` `n)` `{` ` ` `if` `(n < 5)` ` ` `cout << ` `"GeeksforGeeks"` `;` ` ` `else` `{` ` ` `for` `(` `int` `i = 0; i < n; i++) {` ` ` `cout << i << ` `" "` `;` ` ` `}` ` ` `}` `}` |

**Solution – **

Time complexity = O(1) in best case and O(n) in worst case.

**Explanation –**

This program contains if and else condition. Hence, there are 2 possibilities of time complexity. If the value of n is less than 5, then we get only * GeeksforGeeks *as output and its time complexity will be O(1).

But, if n>=5, then for loop will execute and time complexity becomes O(n), it is considered as worst case because it takes more time.

**3. What is the time complexity of the following code : **

## C++

`void` `fun(` `int` `a, ` `int` `b)` `{` ` ` `while` `(a != b) {` ` ` `if` `(a > b)` ` ` `a = a - b;` ` ` `else` ` ` `b = b - a;` ` ` `}` `}` |

**Solution –**

Time complexity = O(1) in best case and O(max(a, b)) worst case.

**Explanation –**

If the value of a and b are the same, then while loop will not be executed. Hence, time complexity will be O(1).

But if a!=b, then the while loop will be executed. Let a=16 and b=5;

no. of iterations | a | b |

1 | 16 | 5 |

2 | 16-5=11 | 5 |

3 | 11-5=6 | 5 |

4 | 6-5=1 | 5 |

5 | 1 | 5-1=4 |

6 | 1 | 4-1=3 |

7 | 1 | 3-1=2 |

8 | 1 | 2-1=1 |

For this case, while loop executed 8 times (a/2⇒16/2⇒8).

If a=5 and b=16, then also the loop will be executed 8 times. So we can say that time complexity is O(max(a/2,b/2))⇒O(max(a, b))**,** it is considered as worst case because it takes more time.

**4. What is the time complexity of the following code : **

## C++

`void` `fun(` `int` `n)` `{` ` ` `for` `(` `int` `i=0;i*i<n;i++)` ` ` `cout<<` `"GeeksforGeeks"` `;` `}` |

**Solution – **

Time complexity = O(√n).

**Explanation –**

Let **k **be the no. of iteration of the loop.

i | i*i |

1 | 1 |

2 | 2^{2} |

3 | 3^{2} |

4 | 4^{2} |

… | … |

k | k^{2} |

**⇒** The loop will stop when i * i >=n i.e., i*i=n**⇒** i*i=n ⇒ k^{2} = n**⇒**k =√n

Hence, the time complexity is O(√n).

**5. What is the time complexity of the following code : **

## C++

`void` `fun(` `int` `n, ` `int` `x)` `{` ` ` `for` `(` `int` `i = 0; i < n; i = i * x) ` `//or for(int i = n; i >=0; i = i / x)` ` ` `cout << "GeeksforGeeks;` `}` |

**Solution –**

Time complexity = O(log_{x}n).

**Explanation – **

Let **k** be the no. of iteration of the loop.

no. of itr | i=i*x |

1 | 1*x=x |

2 | x*x=x^{2} |

3 | x^{2 }*x=x^{3} |

… | … |

k | x^{k-1} *x= x^{k} |

**⇒** The loop will stop when i>=n ⇒ x^{k} = n

⇒ x^{k}=n (Take log both sides)

⇒ k=log_{x}n

⇒Hence, time complexity is O(log_{x}n).

**6. What is the time complexity of the following code : **

## C++

`void` `fun(` `int` `n)` `{` ` ` `for` `(` `int` `i = 0; i < n / 2; i++)` ` ` `for` `(` `int` `j = 1; j + n / 2 <= n; j++)` ` ` `for` `(` `int` `k = 1; k <= n; k = k * 2)` ` ` `cout << ` `"GeeksforGeeks"` `;` `}` |

**Solution –**

Time complexity = O(n^{2}log_{2}n).

**Explanation –**

Time complexity of 1^{st} for loop = O(n/2) ⇒ O(n).

Time complexity of 2^{nd} for loop = O(n/2) ⇒ O(n).

Time complexity of 3^{rd} for loop = O(log_{2}n). (refer question number – 5)

Hence, the time complexity of function will become O(n^{2}log_{2}n).

**7. What is the time complexity of the following code : **

## C++

`void` `fun(` `int` `n)` `{` ` ` `for` `(` `int` `i = 1; i <= n; i++)` ` ` `for` `(` `int` `j = 1; j <= n; j = j + i)` ` ` `cout << ` `"GeeksforGeeks"` `;` `}` |

**Solution –** Time complexity = O(nlogn).

**Explanation – **

Time complexity of 1st for loop = O(n). 2^{nd} loop executes n/i times for each value of i.

Its time complexity is O(∑^{n}_{i=1} n/i) ⇒ O(logn).

Hence, time complexity of function = O(nlogn).

** 8. What is the time complexity of the following code : **

## C++

`void` `fun(` `int` `n)` `{` ` ` `for` `(` `int` `i = 0; i <= n / 3; i++)` ` ` `for` `(` `int` `j = 1; j <= n; j = j + 4)` ` ` `cout << ` `"GeeksforGeeks"` `;` `}` |

**Solution – **Time complexity = O(n^{2}).

**Explanation – **

Time complexity of 1st for loop = O(n/3) ⇒ O(n).

Time complexity of 2nd for loop = O(n/4) ⇒ O(n).

Hence, the time complexity of function will become O(n^{2})**.**

**9. What is the time complexity of the following code : **

## C++

`void` `fun(` `int` `n)` `{` ` ` `int` `i = 1;` ` ` `while` `(i < n) {` ` ` `int` `j = n;` ` ` `while` `(j > 0) {` ` ` `j = j / 2;` ` ` `}` ` ` `i = i * 2;` ` ` `}` `}` |

**Solution – **Time complexity = O(log^{2}n).

**Explanation – **

In each iteration , i become twice (T.C=O(logn)) and j become half (T.C=O(logn)). So, time complexity will become O(log^{2}n).

We can convert this while loop into for loop.

for( int i = 1; i < n; i = i * 2)

for( int j=n ; j > 0; j = j / 2).

Time complexity of above for loop is also O(log^{2}n).

**10. Consider the following C-code,** **what is the number of comparisons made in the execution of the loop ?**

## C++

`void` `fun(` `int` `n)` `{` ` ` `int` `j = 1;` ` ` `while` `(j <= n) {` ` ` `j = j * 2;` ` ` `}` `}` |

**Solution –**

ceil(log_{2}n)+1.

**Explanation –**

Let **k** be the no. of iteration of the loop. After the kth step, the value of j is 2^{k}. Hence, k=log_{2}n. Here, we use *ceil of log _{2}n*, because log

_{2}n may be in decimal. Since we are doing one more comparison for exiting from the loop.

So, the answer is ceil(log

_{2}n)+1.