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Sample Practice Problems on Complexity Analysis of Algorithms

Last Updated : 04 Apr, 2024
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Prerequisite: Asymptotic Analysis, Worst, Average and Best Cases, Asymptotic Notations, Analysis of loops.

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(√n).
Time Complexity: O(√n).

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)

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.
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