# Sample Practice Problems on Complexity Analysis of Algorithms

Last Updated : 04 Apr, 2024

### Problem 1: Find the complexity of the 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
he solution to the recurrence relation here gives a different result.

### Problem 3: Find the complexity of the below program:

C++ ```void function(int n) { if (n==1) return; for (int i=1; i<=n; i++) { for (int j=1; j<=n; j++) { cout << "*"; break; } cout << endl; } } ``` Java ```/*package whatever //do not write package name here */ static void function(int n) { if (n==1) return; for (int i=1; i<=n; i++) { for (int j=1; j<=n; j++) { System.out.print("*"); break; } System.out.println(); } } // The code is contributed by Nidhi goel. ``` Python ```def funct(n): if (n==1): return for i in range(1, n+1): for j in range(1, n + 1): print("*", end = "") break print() # The code is contributed by Nidhi goel. ``` C# ```/*package whatever //do not write package name here */ public static void function(int n) { if (n==1) return; for (int i=1; i<=n; i++) { for (int j=1; j<=n; j++) { Console.Write("*"); break; } Console.WriteLine(); } } // The code is contributed by Nidhi goel. ``` JavaScript ```function funct(n) { if (n==1) return; for (let i=1; i<=n; i++) { for (let j=1; j<=n; j++) { console.log("*"); break; } console.log(); } } // The code is contributed by Nidhi goel. ```

Solution: Consider the comments in the following function.

C++ ```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; } } } ``` Java ```public class Main { public static void main(String[] args) { int n = 5; // You can change the value of n as needed printPattern(n); } static void printPattern(int n) { for (int i = 1; i <= n; i++) { // Print a single '*' on each line System.out.println("*"); } } } ``` Python ```def my_function(n): if n == 1: return for i in range(1, n+1): # Inner loop executes only one # time due to break statement. for j in range(1, n+1): print("*", end="") break my_function(5) # Example: calling the function with n=5 #this code is contributed by Monu Yadav. ``` C# ```using System; class Program { static void PrintPattern(int n) { if (n == 1) return; for (int i = 1; i <= n; i++) { // Inner loop executes only once // due to the break statement. #pragma warning disable CS0162 for (int j = 1; j <= n; j++) { Console.Write("*"); // The following break statement exits the inner loop // after printing a single '*' character. // If you want the inner loop to iterate through the entire range, // you can remove or comment out the break statement. break; } #pragma warning restore CS0162 } } static void Main() { int n = 5; // You can change the value of 'n' as needed PrintPattern(n); } } ``` JavaScript ```function printStars(n) { if (n === 1) { return; // If n is 1, exit the function } for (let i = 1; i <= n; i++) { // Outer loop to iterate from 1 to n // Inner loop executes only once due to the break statement for (let j = 1; j <= n; j++) { console.log("*"); // Print a star break; // Break the inner loop after printing one star } } } // Example usage: printStars(5); // Call the function with n = 5 ```

Time Complexity: 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:

C++ ```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 contributed by rutvik_56. ``` Python ```def function1(n): count = 0 for i in range(n // 2, n + 1): for j in range(1, n + 1, 2 * j): for k in range(1, n + 1, k * 2): count += 1 ``` 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. ``` JavaScript ```<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.

C++ ```#include <iostream> using namespace std; 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++; } } } cout << "Count: " << count << endl; } int main() { // Example usage function(10); // Call function with an example value of n return 0; } ``` Java ```public class Main { public static 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++; } } } System.out.println("Count: " + count); } public static void main(String[] args) { // Example usage function(10); // Call function with an example value of n } } //This code is contributed by Adarsh ``` Python ```def function(n): count = 0 # Outer loop executes n/2 times for i in range(n // 2, n + 1): # Middle loop executes O(Log n) times j = 1 while j <= n: # Inner loop also executes O(Log n) times k = 1 while k <= n: count += 1 k *= 2 j *= 2 print("Count:", count) # Example usage function(10) ``` JavaScript ```function functionMain(n) { let count = 0; for (let i = Math.floor(n / 2); i <= n; i++) { // Executes O(Log n) times for (let j = 1; j <= n; j *= 2) { // Executes O(Log n) times for (let k = 1; k <= n; k *= 2) { count++; } } } console.log("Count: " + count); } functionMain(10); // Example usage ```

Time Complexity: O(n log2n).

### Problem 5: Find the complexity of the below program:

C++ ```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++; } ``` Java ```static 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++; } // This code is contributed by Pushpesh Raj. ``` Python3 ```def function(n): # Initialize count to 0 count = 0 # Outer loop starts from n/2 and goes up to n for i in range(n//2, n+1): # Middle loop starts from 1 and goes up to n/2 for j in range(1, n//2+1): # Inner loop starts from 1 and doubles at each step for k in range(1, n+1, 2): # Increment count at each iteration count += 1 ``` JavaScript ```function myFunction(n) { let count = 0; // Outer loop runs from n/2 to n for (let i = n / 2; i <= n; i++) { // Middle loop runs from 1 to n - n/2 for (let j = 1; j + n / 2 <= n; j++) { // Inner loop runs from 1 to n, doubling k in each iteration for (let k = 1; k <= n; k = k * 2) { count++; } } } return count; } ```

Solution: Consider the comments in the following function.

C++ ```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++; } // The code is contributed by Nidhi goel. ``` Java ```static 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++; } // This code is contributed by Aman Kumar ``` Python ```def function(n): count = 0 # outer loop executes n/2 times for i in range(n//2, n+1): # middle loop executes n/2 times for j in range((1, n//2 + 1): # inner loop executes logn times for k in range(1, n+1, 2): count++ # The code is contributed by Nidhi goel. ``` C# ```using System; public static 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++; } // The code is contributed by Nidhi goel. ``` JavaScript ```function function(n) { let count = 0; // outer loop executes n/2 times for (let i= Math.floor(n/2); i<=n; i++) // middle loop executes n/2 times for (let j=1; j+n/2<=n; j = j++) // inner loop executes logn times for (let k=1; k<=n; k = k * 2) count++; } // The code is contributed by Nidhi goel. ```

Time Complexity: O(n2logn).

### Problem 6: Find the complexity of the below program:

C++ ```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(âˆšn).
Time Complexity: O(âˆšn).

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

C++ ```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("*"); } } ``` Java ```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("*"); } } ``` JavaScript ```function myFunction(n) { for (let i = 0; i < n; i++) { for (let j = i; j < i * i; j++) { if (j % i === 0) { // Check if j is divisible by i for (let k = 0; k < j; k++) { process.stdout.write("*"); // Print '*' to the console } } } } } ```

Solution: Consider the comments in the following function.

C++ ```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: O(n5)

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