Analysis of Algorithms | Set 5 (Practice Problems) - GeeksforGeeks

31 Mar, 2022
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 the 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 - 1]
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:  
CPP
function(int n){    if (n==1)       return;    for (int i=1; i<=n; i++)    {        for (int j=1; j<=n; j++)        {            printf("*");            break;        }    }}
Solution: Consider the comments in the following function. 
CPP
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;        }    }}
Time Complexity of the above function O(n). Even though the inner loop is bounded by n, but due to the break statement, it is executing only once. Problem 4: Find the complexity of the below program: 
CPP
Java
C#
Javascript
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++;}
static 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++;} // This code is contributed by rutvik_56.
static 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++;} // This code is contributed by pratham76.
<script>function function1(n){    var count = 0;    for (i = n / 2; i <= n; i++)        for (j = 1; j <= n; j = 2 * j)            for (k = 1; k <= n; k = k * 2)                count++;} // This code is contributed by umadevi9616</script>
Solution: Consider the comments in the following function. 
CPP
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++;}
Time Complexity of the above function O(n log2n). Problem 5: Find the complexity of the below program: 
CPP
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++;}
Solution: Consider the comments in the following function.  
CPP
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++;}
Time Complexity of the above function O(n2logn). Problem 6: Find the complexity of the below program:  
CPP
void function(int n){    int i = 1, s =1;    while (s <= n)    {        i++;        s += i;        printf("*");    }}
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 the complexity of the below program:  
CPP
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("*");            }}
Solution: Consider the comments in the following function.  
CPP
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("*");            }}
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 write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above. 
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