Question

Accept a list containing numbers. Sort the list using any sorting technique. Thereafter accept a number and display the position where that number is found. Also display suitable message, if the number is not found in the list.

Answer 1

The list is first sorted using selection sort. Then binary search finds the required number efficiently. Positions displayed are 1-based.

def selection_sort(numbers):
    for i in range(len(numbers) - 1):
        minimum_index = i

        for j in range(i + 1, len(numbers)):
            if numbers[j] < numbers[minimum_index]:
                minimum_index = j

        numbers[i], numbers[minimum_index] = (
            numbers[minimum_index], numbers[i]
        )

    return numbers


def binary_search(numbers, target):
    low = 0
    high = len(numbers) - 1

    while low <= high:
        mid = (low + high) // 2

        if numbers[mid] == target:
            return mid + 1  # 1-based position
        if numbers[mid] < target:
            low = mid + 1
        else:
            high = mid - 1

    return -1


numbers = list(map(int, input("Enter list elements: ").split()))
selection_sort(numbers)
print("Sorted list:", numbers)

target = int(input("Enter number to search: "))
position = binary_search(numbers, target)

if position == -1:
    print("Number not found in the list")
else:
    print("Number found at position", position)

Enter list elements: 24 5 18 9 30
Sorted list: [5, 9, 18, 24, 30]
Enter number to search: 18
Number found at position 3