Analysis of Algorithms | Set 5 (Practice Problems)

We have discussed Asymptotic Analysis, Worst, Average and Best Cases , Asymptotic Notations and Analysis of loops in previous posts.

In this post, practice problems on analysis of algorithms are discussed.

Problem 1: Find the complexity of below recurrence:



         { 3T(n-1), if n>0,
T(n) =   { 1, otherwise

Solution:

Let us solve using substitution.
T(n) = 3T(n-1)
     = 3(3T(n-2)) 
     = 32T(n-2)
     = 33T(n-3)
       ...
       ...
     = 3nT(n-n)
     = 3nT(0) 
     = 3n
This clearly shows that the complexity 
of this function is O(3n).

 

Problem 2: Find the complexity of the recurrence:

        { 2T(n-1) - 1, if n>0,
T(n) =   { 1, otherwise

Solution:

 Let us try solving this function with substitution.
T(n) = 2T(n-1) - 1
     = 2(2T(n-2)-1)-1 
     = 22(T(n-2)) - 2 - 1
     = 22(2T(n-3)-1) - 2 - 1 
     = 23T(n-3) - 22 - 21 - 20
       .....
       .....
     = 2nT(n-n) - 2n-1 - 2n-2 - 2n-3
       ..... 22 - 21 - 20

     = 2n - 2n-1 - 2n-2 - 2n-3
       ..... 22 - 21 - 20
     = 2n - (2n-1) 
[Note: 2n-1 + 2n-2 + ...... +  20 = 2n ]
T(n) = 1
Time Complexity is O(1). Note that while 
the recurrence relation looks exponential
the solution to the recurrence relation 
here gives a different result.

 

Problem 3: Find the complexity of the below program:

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function(int n)
{
    if (n==1)
       return;
    for (int i=1; i<=n; i++)
    {
        for (int j=1; j<=n; j++)
        {
            printf("*");
            break;
        }
    }
}

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Solution: Consider the comments in the following function.

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function(int n)
{
    if (n==1)
       return;
    for (int i=1; i<=n; i++)
    {
        // Inner loop executes only one
        // time due to break statement.
        for (int j=1; j<=n; j++)
        {
            printf("*");
            break;
        }
    }
}

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Time Complexity of the above function O(n). Even though the inner loop is bounded by n, but due to break statement it is executing only once.

 

Problem 4: Find the complexity of the below program:


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void function(int n)
{
    int count = 0;
    for (int i=n/2; i<=n; i++)
        for (int j=1; j<=n; j = 2 * j)
            for (int k=1; k<=n; k = k * 2)
                count++;
}

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Solution: Consider the comments in the following function.

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void function(int n)
{
    int count = 0;
    for (int i=n/2; i<=n; i++)
  
        // Executes O(Log n) times
        for (int j=1; j<=n; j = 2 * j)
  
            // Executes O(Log n) times
            for (int k=1; k<=n; k = k * 2)
                count++;
}

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Time Complexity of the above function O(n log2n).

 

Problem 5: Find the complexity of the below program:

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void function(int n)
{
    int count = 0;
    for (int i=n/2; i<=n; i++)
        for (int j=1; j+n/2<=n; j = j++)
            for (int k=1; k<=n; k = k * 2)
                count++;
}

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Solution: Consider the comments in the following function.

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void function(int n)
{
    int count = 0;
  
    // outer loop executes n/2 times
    for (int i=n/2; i<=n; i++)
  
        // middle loop executes  n/2 times
        for (int j=1; j+n/2<=n; j = j++)
  
            // inner loop executes logn times
            for (int k=1; k<=n; k = k * 2)
                count++;
}

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Time Complexity of the above function O(n2logn).

 

Problem 6: Find the complexity of the below program:

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void function(int n)
{
    int i = 1, s =1;
    while (s <= n)
    {
        i++;
        s += i;
        printf("*");
    }
}

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Solution: We can define the terms ‘s’ according to relation si = si-1 + i. The value of ‘i’ increases by one for each iteration. The value contained in ‘s’ at the ith iteration is the sum of the first ‘i’ positive integers. If k is total number of iterations taken by the program, then while loop terminates if: 1 + 2 + 3 ….+ k = [k(k+1)/2] > n So k = O(√n).

Time Complexity of the above function O(√n).


 

Problem 7: Find a tight upper bound on complexity of the below program:

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void function(int n)
{
    int count = 0;
    for (int i=0; i<n; i++)
        for (int j=i; j< i*i; j++)
            if (j%i == 0)
            {
                for (int k=0; k<j; k++)
                    printf("*");
            }
}

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Solution:Consider the comments in the following function.

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void function(int n)
{
    int count = 0;
  
    // executes n times
    for (int i=0; i<n; i++)
  
        // executes O(n*n) times.
        for (int j=i; j< i*i; j++)
            if (j%i == 0)
            {
                // executes j times = O(n*n) times
                for (int k=0; k<j; k++)
                    printf("*");
            }
}

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Time Complexity of the above function O(n5).

This article is contributed by Mr. Somesh Awasthi. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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