Worst, Average and Best Case Analysis of Algorithms (original) (raw)

Last Updated : 29 Jan, 2026

In the previous post, we discussed how Asymptotic analysis overcomes the problems of the naive way of analyzing algorithms. Now let us learn about What is Worst, Average, and Best cases of an algorithm:

**1. Worst Case Analysis (Mostly used)

In the worst-case analysis, we calculate the upper bound on the running time of an algorithm. We must know the case that causes a maximum number of operations to be executed.
For Linear Search, the worst case happens when the element to be searched (x) is not present in the array. When x is not present, the search()function compares it with all the elements of arr[] one by one.
This is the most commonly used analysis of algorithms (We will be discussing below why). Most of the time we consider the case that causes maximum operations.

**2. Best Case Analysis (Very Rarely used)

In the best-case analysis, we calculate the lower bound on the running time of an algorithm. We must know the case that causes a minimum number of operations to be executed.
For linear search, the best case occurs when x is present at the first location. The number of operations in the best case is constant (not dependent on n). So the order of growth of time taken in terms of input size is constant.

**3. Average Case Analysis (Rarely used)

In average case analysis, we take all possible inputs and calculate the computing time for all of the inputs. Sum all the calculated values and divide the sum by the total number of inputs.
Assuming all cases (including x not being in the array) are equally likely in linear search, the average case is computed by summing all cases and dividing by n+1n+1n+1, to account for the possibility that the element is absent.

Why is Worst Case Analysis Mostly Used?

**Average Case: The average case analysis is not easy to do in most practical cases and it is rarely done. In the average case analysis, we need to consider every input, its frequency and time taken by it which may not be possible in many scenarios

**Best Case: Best-case analysis is of limited value, as knowing a lower bound gives little insight when the algorithm’s worst-case running time can be extremely large.

**Worst Case: This is easier than average case and gives an upper bound which is useful information to analyze software products.

Interesting information about asymptotic notations:

A) For some algorithms, all the cases (worst, best, average) are asymptotically the same. i.e., there are no worst and best cases.

**Example: Merge Sort does order of n log(n) operations in all cases.

B) Where as most of the other sorting algorithms have worst and best cases.

**Example 1: In the typical implementation of Quick Sort (where pivot is chosen as a corner element), the worst occurs when the input array is already sorted and the best occurs when the pivot elements always divide the array into two halves.

**Example 2: For insertion sort, the worst case occurs when the array is reverse sorted and the best case occurs when the array is sorted in the same order as output.

Examples with their complexity analysis:

**1. Linear search algorithm:

C++ `

#include #include using namespace std;

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

int main() { vector arr = {1, 10, 30, 15}; int x = 30; cout << search(arr, x); return 0; }

C

#include <stdio.h> #include <stdbool.h>

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

int main() { int arr[] = {1, 10, 30, 15}; int size = sizeof(arr) / sizeof(arr[0]); int x = 30; printf("%d", search(arr, size, x)); return 0; }

Java

import java.util.Arrays;

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

public static void main(String[] args) {
    int[] arr = {1, 10, 30, 15};
    int x = 30;
    System.out.println(search(arr, x));
}

}

Python

Linearly search target in arr.

If target is present, return the index;

otherwise, return -1

def search(arr, x): for i in range(len(arr)): if arr[i] == x: return i return -1

if name == 'main': arr = [1, 10, 30, 15] x = 30 print(search(arr, x))

C#

using System;

public class Program {

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

public static void Main() {
    int[] arr = {1, 10, 30, 15};
    int target = 30;
    Console.WriteLine(Search(arr, target));
}

}

JavaScript

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

const arr = [1, 10, 30, 15]; const x = 30; console.log(search(arr, x));

**Best Case: Constant Time irrespective of input size.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: Linear Time, This will take place if the element to be searched is at the middle index (in an average search) of the given list.
**Worst Case: The element to be searched is not present in the list

**2. Special Array Sum:

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. C++ `

#include #include using namespace std;

int getSum(const vector &arr1) { int n = arr1.size(); if (n % 2 == 0) // (n) is even return 0;

int sum = 0;
for (int i = 0; i < n; i++)
{
    sum += arr1[i];
}
return sum; // (n) is odd

}

int main() {

// Declaring two vectors, one with an odd length
// and the other with an even length
vector<int> arr1 = {1, 2, 3, 4};
vector<int> arr2 = {1, 2, 3, 4, 5};
cout << getSum(arr1) << endl;
cout << getSum(arr2) << endl;
return 0;

}

C

#include <stdio.h> #include <stdlib.h>

int getSum(const int* arr1, int n) { if (n % 2 == 0) // (n) is even return 0;

int sum = 0;
for (int i = 0; i < n; i++) {
    sum += arr1[i];
}
return sum; // (n) is odd

}

int main() { // Declaring two arrays, one with an odd length // and the other with an even length int arr1[] = {1, 2, 3, 4}; int arr2[] = {1, 2, 3, 4, 5}; printf("%d\n", getSum(arr1, 4)); printf("%d\n", getSum(arr2, 5)); return 0; }

Java

import java.util.Arrays;

public class Main { public static int getSum(int[] arr1) { int n = arr1.length; if (n % 2 == 0) // (n) is even return 0;

    int sum = 0;
    for (int i = 0; i < n; i++) {
        sum += arr1[i];
    }
    return sum; // (n) is odd
}

public static void main(String[] args) {
    // Declaring two arrays, one with an odd length
    // and the other with an even length
    int[] arr1 = {1, 2, 3, 4};
    int[] arr2 = {1, 2, 3, 4, 5};
    System.out.println(getSum(arr1));
    System.out.println(getSum(arr2));
}

}

Python

def getSum(arr1): n = len(arr1) if n % 2 == 0: # (n) is even return 0

sum = 0
for i in range(n):
    sum += arr1[i]
return sum  # (n) is odd

if name == 'main': # Declaring two lists, one with an odd length # and the other with an even length arr1 = [1, 2, 3, 4] arr2 = [1, 2, 3, 4, 5] print(getSum(arr1)) print(getSum(arr2))

C#

using System;

class Program { static int getSum(int[] arr1) { int n = arr1.Length; if (n % 2 == 0) // (n) is even return 0;

    int sum = 0;
    for (int i = 0; i < n; i++) {
        sum += arr1[i];
    }
    return sum; // (n) is odd
}

static void Main() {
    // Declaring two arrays, one with an odd length
    // and the other with an even length
    int[] arr1 = {1, 2, 3, 4};
    int[] arr2 = {1, 2, 3, 4, 5};
    Console.WriteLine(getSum(arr1));
    Console.WriteLine(getSum(arr2));
}

}

JavaScript

function getSum(arr1) { const n = arr1.length; if (n % 2 === 0) // (n) is even return 0;

let sum = 0;
for (let i = 0; i < n; i++) {
    sum += arr1[i];
}
return sum; // (n) is odd

}

// Declaring two arrays, one with an odd length // and the other with an even length const arr1 = [1, 2, 3, 4]; const arr2 = [1, 2, 3, 4, 5]; console.log(getSum(arr1)); console.log(getSum(arr2));

PHP

n=count(n = count(n=count(arr1); if ($n % 2 == 0) // (n) is even return 0; $sum = 0; for ($i = 0; i<i < i<n; $i++) { sum+=sum += sum+=arr1[$i]; } return $sum; // (n) is odd } // Declaring two arrays, one with an odd length // and the other with an even length $arr1 = [1, 2, 3, 4]; $arr2 = [1, 2, 3, 4, 5]; echo getSum($arr1) . "\n"; echo getSum($arr2) . "\n"; ?>

**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 willassume that even and odd are equally likely, thereforeOrder 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.