Examples with their complexity analysis

1. Linear search algorithm:

C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

// Linearly search x in arr[].
// If x is present then return the index,
// otherwise return -1
int search(int arr[], int n, int x)
{
    int i;
    for (i = 0; i < n; i++) {
        if (arr[i] == x)
            return i;
    }
    return -1;
}

// Driver's Code
int main()
{
    int arr[] = { 1, 10, 30, 15 };
    int x = 30;
    int n = sizeof(arr) / sizeof(arr[0]);

    // Function call
    cout << x << " is present at index "
         << search(arr, n, x);

    return 0;
}
C
// C implementation of the approach
#include <stdio.h>

// Linearly search x in arr[].
// If x is present then return the index,
// otherwise return -1
int search(int arr[], int n, int x)
{
    int i;
    for (i = 0; i < n; i++) {
        if (arr[i] == x)
            return i;
    }
    return -1;
}

/* Driver's code*/
int main()
{
    int arr[] = { 1, 10, 30, 15 };
    int x = 30;
    int n = sizeof(arr) / sizeof(arr[0]);

    // Function call
    printf("%d is present at index %d", x,
           search(arr, n, x));

    getchar();
    return 0;
}
Java
// Java implementation of the approach

public class GFG {

    // Linearly search x in arr[].  If x is present then
    // return the index, otherwise return -1
    static int search(int arr[], int n, int x)
    {
        int i;
        for (i = 0; i < n; i++) {
            if (arr[i] == x) {
                return i;
            }
        }
        return -1;
    }

    /* Driver's code*/
    public static void main(String[] args)
    {
        int arr[] = { 1, 10, 30, 15 };
        int x = 30;
        int n = arr.length;

        // Function call
        System.out.printf("%d is present at index %d", x,
                          search(arr, n, x));
    }
}
C#
// C# implementation of the approach
using System;
public class GFG {

    // Linearly search x in arr[].  If x is present then
    // return the index, otherwise return -1
    static int search(int[] arr, int n, int x)
    {
        int i;
        for (i = 0; i < n; i++) {
            if (arr[i] == x) {
                return i;
            }
        }
        return -1;
    }

    /* Driver's code*/
    public static void Main()
    {
        int[] arr = { 1, 10, 30, 15 };
        int x = 30;
        int n = arr.Length;

        // Function call
        Console.WriteLine(x + " is present at index "
                          + search(arr, n, x));
    }
}
Javascript
// javascript implementation of the approach

     // Linearly search x in arr. If x is present then
    // return the index, otherwise return -1
    function search(arr , n , x) {
        var i;
        for (i = 0; i < n; i++) {
            if (arr[i] == x) {
                return i;
            }
        }
        return -1;
    }

    /* Driver program to test above functions */
    
        var arr = [ 1, 10, 30, 15 ];
        var x = 30;
        var n = arr.length;
        document.write(x+" is present at index "+ search(arr, n, x));
PHP
<?php
// PHP implementation of the approach

// Linearly search x in arr[]. If x 
// is present then return the index,
// otherwise return -1
function search($arr, $n, $x)
{
    for ($i = 0; $i < $n; $i++)
    {
    if ($arr[$i] == $x)
        return $i;
    }
    return -1;
}

// Driver's Code
$arr = array(1, 10, 30, 15);
$x = 30;
$n = sizeof($arr);

// Function call
echo $x . " is present at index ". 
             search($arr, $n, $x);
Python3
# Python 3 implementation of the approach

# Linearly search x in arr[]. If x is present
# then return the index, otherwise return -1


def search(arr, x):
    for index, value in enumerate(arr):
        if value == x:
            return index
    return -1


# Driver's Code
if __name__ == '__main__':
    arr = [1, 10, 30, 15]
    x = 30

    # Function call
    print(x, "is present at index",
          search(arr, x))

Output
30 is present at index 2


Time Complexity Analysis: (In Big-O notation)

  • Best Case: O(1), This will take place if the element to be searched is on the first index of the given list. So, the number of comparisons, in this case, is 1.
  • Average Case: O(n), This will take place if the element to be searched is on the middle index of the given list.
  • Worst Case: O(n), This will take place if:
    • The element to be searched is on the last index
    • The element to be searched is not present on the list

2. In this example, we will take an array of length (n) and deals with the following cases :

  • If (n) is even then our output will be 0
  • If (n) is odd then our output will be the sum of the elements of the array.

Below is the implementation of the given problem:

C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

int getSum(int arr[], int n)
{
    if (n % 2 == 0) // (n) is even
    {
        return 0;
    }
    int sum = 0;
    for (int i = 0; i < n; i++) {
        sum += arr[i];
    }
    return sum; //  (n) is odd
}

// Driver's Code
int main()
{
    // Declaring two array one of length odd and other of
    // length even;
    int arr[4] = { 1, 2, 3, 4 };
    int a[5] = { 1, 2, 3, 4, 5 };

    // Function call
    cout << getSum(arr, 4)
         << endl; // print 0 because (n) is even
    cout << getSum(a, 5)
         << endl; // print sum because (n) is odd
}
// This code is contributed by Suruchi Kumari
Java
// Java implementation of the approach

public class GFG {
    static int getsum(int arr[], int n)
    {
        if (n % 2 == 0) // if (n) is even
        {
            return 0;
        }
        int sum = 0;
        for (int i = 0; i < n; i++) {
            sum += arr[i];
        }
        return sum; // if (n) is odd
    }

    /* Driver's code*/
    public static void main(String[] args)
    {
        int arr1[]
            = { 1, 2, 3,
                4 }; // Declaring an array of even length
        int n1 = arr1.length;

        int arr2[]
            = { 1, 2, 3, 4,
                5 }; // Declaring an array of odd length
        int n2 = arr2.length;

        // Function call
        System.out.println(getsum(
            arr1, n1)); // print 0 because (n) is even
        System.out.println(getsum(
            arr2,
            n2)); // print sum of array because (n) is odd
    }
} // This code is contributed by Syed Maruf Ali (Sdmrf)
C#
using System;

public class Gfg
{
  static int getSum(int[] arr, int n)
  {
    if (n % 2 == 0) // (n) is even
    {
      return 0;
    }
    int sum = 0;
    for (int i = 0; i < n; i++) {
      sum += arr[i];
    }
    return sum; //  (n) is odd
  }

  // Driver's Code
  public static void Main(string[] args)
  {
    // Declaring two array one of length odd and other of
    // length even;
    int[] arr = { 1, 2, 3, 4 };
    int[] a = { 1, 2, 3, 4, 5 };

    // Function call
    Console.Write(getSum(arr, 4)+"\n"); // print 0 because (n) is even
    Console.Write(getSum(a, 5)); // print sum because (n) is odd
  }
}

// This code is contributed by poojaagarwal2.
Javascript
// Javascript implementation of the approach
      function getSum(arr, n) {
        if (n % 2 == 0) {
          // (n) is even
          return 0;
        }
        var sum = 0;
        for (var i = 0; i < n; i++) {
          sum += arr[i];
        }
        return sum; //  (n) is odd
      }

      // Driver's Code

      // Declaring two array one of length odd and other of
      // length even;
      var arr = [1, 2, 3, 4];
      var a = [1, 2, 3, 4, 5];

      // Function call
      console.log(getSum(arr, 4)); //print 0 because (n) is even
      console.log(getSum(a, 5)); // print sum because (n) is odd
      
      // This code is contributed by satwiksuman.
PHP
<?php
// PHP implementation of the approach

function getSum($arr, $n) {
    if ($n % 2 == 0) {
        return 0;
    }

    $sum = 0;
    for ($i = 0; $i < $n; $i++) {
        $sum += $arr[$i];
    }
    return $sum;
}

// Driver's Code
$arr1 = [1, 2, 3, 4];  // Declaring an array of even length
$n1 = count($arr1);
$arr2 = [1, 2, 3, 4, 5];  // Declaring an array of odd length
$n2 = count($arr2);

// Function calls
echo getSum($arr1, $n1) . "\n";  // print 0 because (n) is even
echo getSum($arr2, $n2) . "\n";  // print sum of array because (n) is odd
?>
Python3
# Python 3 implementation of the approach


def getsum(arr, n):
    if n % 2 == 0:  # if (n) is even
        return 0

    Sum = 0
    for i in range(n):
        Sum += arr[i]
    return Sum  # if (n) is odd


# Driver's Code
if __name__ == '__main__':
  arr1 = [1, 2, 3, 4]  # Declaring an array of even length
  n1 = len(arr1)
  arr2 = [1, 2, 3, 4, 5]  # Declaring an array of odd length
  n2 = len(arr2)

# Function call
print(getsum(arr1, n1))  # print 0 because (n) is even

print(getsum(arr2, n2))  # print sum of array because (n) is odd

# This code is contributed by Syed Maruf Ali

Output
0
15


Time Complexity Analysis:

  • Best Case: The order of growth will be constant because in the best case we are assuming that (n) is even.
  • Average Case: In this case, we will assume that even and odd are equally likely, therefore Order of growth will be linear
  • Worst Case: The order of growth will be linear because in this case, we are assuming that (n) is always odd.

For more details, please refer: Design and Analysis of Algorithms.

Worst, Average, and Best Case Analysis of Algorithms is a technique used to analyze the performance of algorithms under different conditions. Here are some advantages, disadvantages, important points, and reference books related to this analysis technique:

Advantages:

  1. This technique allows developers to understand the performance of algorithms under different scenarios, which can help in making informed decisions about which algorithm to use for a specific task.
  2. Worst case analysis provides a guarantee on the upper bound of the running time of an algorithm, which can help in designing reliable and efficient algorithms.
  3. Average case analysis provides a more realistic estimate of the running time of an algorithm, which can be useful in real-world scenarios.


Disadvantages:

  1. This technique can be time-consuming and requires a good understanding of the algorithm being analyzed.
  2. Worst case analysis does not provide any information about the typical running time of an algorithm, which can be a disadvantage in real-world scenarios.
  3. Average case analysis requires knowledge of the probability distribution of input data, which may not always be available.


Important points:

  1. The worst case analysis of an algorithm provides an upper bound on the running time of the algorithm for any input size.
  2. The average case analysis of an algorithm provides an estimate of the running time of the algorithm for a random input.
  3. The best case analysis of an algorithm provides a lower bound on the running time of the algorithm for any input size.
  4. The big O notation is commonly used to express the worst case running time of an algorithm.
  5. Different algorithms may have different best, average, and worst case running times.


Reference books:

  1. “Introduction to Algorithms” by Thomas H. Cormen, Charles E. Leiserson, Ronald L. Rivest, and Clifford Stein is a comprehensive guide to algorithm analysis, including worst, average, and best case analysis.
  2. “Algorithm Design” by Jon Kleinberg and Éva Tardos provides a modern approach to algorithm design, with a focus on worst case analysis.
  3. “The Art of Computer Programming” by Donald Knuth is a classic text on algorithms and programming, which includes a detailed discussion of worst case analysis.
  4. “Algorithms Unlocked” by Thomas H. Cormen provides a gentle introduction to algorithm analysis, including worst, average, and best case analysis.


Worst, Average and Best Case Analysis of Algorithms

In the previous post, we discussed how Asymptotic analysis overcomes the problems of the naive way of analyzing algorithms. But let’s take an overview of the asymptotic notation and learn about What is Worst, Average, and Best cases of an algorithm:

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