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Miscellaneous Problems of Time Complexity

  • Difficulty Level : Medium
  • Last Updated : 06 Aug, 2021

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 Si = Si-1 + i. Let k is the total number of iterations taken by the program 

i              S
11
22
32 + 2
42 + 2 + 3
k2 + 2 + 3 + 4 + ……+ k 

 When S>=n , then loop will stop at kth iterations,
⇒ S>=n ⇒ S=n
⇒ 2 + 2 + 3 + 4 + ……+ k = n
⇒ 1 + (k * (k + 1))/2  = n
⇒  k2 = 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
1165
216-5=115
311-5=65
46-5=15
515-1=4
614-1=3
713-1=2
812-1=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 
11
222
332
442
k k2

The loop will stop when i * i >=n       i.e.,  i*i=n
i*i=n ⇒ k2 = 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(logxn).

Explanation – 
Let k be the no. of iteration of the loop.

no. of itr   i=i*x
11*x=x
2x*x=x2
3x2 *x=x3
kxk-1 *x= xk

The loop will stop when i>=n ⇒ xk = n
⇒ xk=n                        (Take log both sides)
⇒ k=logxn
⇒Hence, time complexity is O(logxn).

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(n2log2n).

Explanation – 
Time complexity of 1st for loop = O(n/2) ⇒ O(n).
Time complexity of 2nd for loop = O(n/2) ⇒ O(n).
Time complexity of 3rd for loop = O(log2n).                                     (refer question number – 5)
Hence, the time complexity of function will become O(n2log2n).

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). 2nd loop executes n/i times for each value of i. 
Its time complexity is O(∑ni=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(n2).

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(n2).

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(log2n).

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(log2n).
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(log2n).

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(log2n)+1.

Explanation –
Let k be the no. of iteration of the loop. After the kth step, the value of j is 2k. Hence, k=log2n. Here, we use ceil of log2n, because log2n may be in decimal. Since we are doing one more comparison for exiting from the loop.
So, the answer is ceil(log2n)+1.




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