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
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++;} |
Java
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 conributed by rutvik_56. |
C#
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. |
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 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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