Python Tutorial

In many Competitive exams, we encounter problems in finding the middle number. It is also important to consider the constraints given in the problem, like minimum comparisons. Let's see different solutions to the problem.

1. Traditional Approach

In the traditional approach, using mathematical calculation, the middle element will be found. In the first step, the total of three numbers is calculated, and then the maximum and minimum values from the list are retrieved using the max and min functions. Now, the middle element is calculated by subtracting min and max from the total.

The approach is efficient in terms of the readability of the code and the complexity of computation.

Code

# Approach 1: Using Mathematical Calculation

def middle_of_three_math(num1, num2, num3):
    # Calculate the sum of all three numbers
    total = num1 + num2 + num3

    # Find the minimum and maximum values
    min_num = min(num1, num2, num3)
    max_num = max(num1, num2, num3)

    # Calculate the middle value
    middle = total - min_num - max_num

    return middle

# Test cases
print("Approach 1 Output:")
print(middle_of_three_math(5, 2, 8))  # Should return 5
print(middle_of_three_math(1, 10, 3))  # Should return 3
print(middle_of_three_math(7, 7, 7))  # Should return 7

Output:

Approach 5 Output:
5
3
7

Time Complexity : O(1)

Space Complexity: O(1)

2. Conditional Statements

If else statements are used in this approach to compare the numbers and then find the middle element. This approach is less efficient.

Code

# Approach 2: Using Conditional Statements

def middle_of_three (num1, num2, num3):
    if (num1 <= num2 <= num3) or (num3 <= num2 <= num1):
        return num2
    elif (num2 <= num1 <= num3) or (num3 <= num1 <= num2):
        return num1
    else:
        return num3

# Test cases
print("\nApproach 2 Output:")
print(middle_of_three(5, 2, 8))  # Should return 5
print(middle_of_three(1, 10, 3))  # Should return 3
print(middle_of_three(7, 7, 7))  # Should return 7

Output:

Approach 5 Output:
5
3
7

Time Complexity : O(1)

Space Complexity: O(1)

3. Sorting

In this approach, all the elements are sorted according to ascending order, and then the middle element is found. With the goal of finding the middle value, this is not efficient as the computation complexity is very low for this approach.

Code

# Approach 3: Using Sorting

def middle_of_three_sorting(num1, num2, num3):
    # Create a list and sort it
    nums = [num1, num2, num3]
    nums.sort()

    # Return the middle element
    return nums[1]

# Test cases
print("\nApproach 3 Output:")
print(middle_of_three_sorting(5, 2, 8))  # Should return 5
print(middle_of_three_sorting(1, 10, 3))  # Should return 3
print(middle_of_three_sorting(7, 7, 7))  # Should return 7

Output:

Approach 5 Output:
5
3
7

Time Complexity : O(1)

Space Complexity: O(1)

4. Bitwise Operators

Bitwise operators are used for the comparison, and then the middle element is found. These operators are not commonly used for these tasks, but for minimum comparisons, bitwise operators are a better choice. This approach is efficient but not readable.

Code

# Approach 4: Using Bitwise Operators

def middle_of_three_bitwise(num1, num2, num3):
    # Calculate the sum without the minimum and maximum values
    middle = num1 + num2 + num3 - max(num1, num2, num3) - min(num1, num2, num3)

    return middle

# Test cases
print("\nApproach 4 Output:")
print(middle_of_three_bitwise(5, 2, 8))  # Should return 5
print(middle_of_three_bitwise(1, 10, 3))  # Should return 3
print(middle_of_three_bitwise(7, 7, 7))  # Should return 7

Output:

Approach 5 Output:
5
3
7

Time Complexity : O(1)

Space Complexity: O(1)

5. Ternary Operator

Ternary operator 'x if condition else y' is used to find the middle element. This is used for a better readable and compact solution. The approach is efficient in terms of the readability of the code and the complexity of computation.

Code

# Approach 5: Using Ternary Operator
def middle_of_three_ternary(num1, num2, num3):
    # Calculate the middle value using a ternary operator
    middle = num1 if num1 <= num2 <= num3 or num3 <= num2 <= num1 else num2 if num2 <= num1 <= num3 or num3 <= num1 <= num2 else num3

    return middle

# Test cases
print("\n Approach 5 Output:")
print(middle_of_three_ternary(5, 2, 8))  # Should return 5
print(middle_of_three_ternary(1, 10, 3))  # Should return 3
print(middle_of_three_ternary(7, 7, 7))  # Should return 7

Output: