Q.Write a program that takes as input a list of 10 integers and a key value and applies binary search to find whether the key is present in the list or not. If the key is present it should display the position of the key in the list otherwise it should print an appropriate message. Run the program for at least 3 different key values and note the results.

Shape of Question

This is a code task with an embedded conceptual error: the question asks for binary search but tags the concept as linear search. Binary search requires a sorted list; the question does not specify that the input is sorted. I will implement binary search as requested (since that is what the question explicitly asks for), note the precondition, and also show the runs.

Solution

Binary search works only on a sorted list. It repeatedly divides the search interval in half: compare the key with the middle element, then search the left or right half accordingly. If the list is unsorted, binary search may fail to find an element that is actually present.

The program below implements binary search and runs it on three different keys.

def binary_search(lst, key):
    """
    Performs binary search on a sorted list.
    Returns the index (0-based) if found, otherwise -1.
    """
    left = 0
    right = len(lst) - 1
    
    while left <= right:
        mid = (left + right) // 2
        
        if lst[mid] == key:
            return mid  # Key found at index mid
        elif lst[mid] < key:
            left = mid + 1  # Search in the right half
        else:
            right = mid - 1  # Search in the left half
    
    return -1  # Key not found

# Input: a sorted list of 10 integers
numbers = [5, 12, 18, 23, 29, 34, 41, 50, 67, 78]

# Test with three different keys
keys = [29, 40, 5]

for key in keys:
    position = binary_search(numbers, key)
    
    if position != -1:
        print(f"Key {key}: Found at position {position} (0-based index) / position {position + 1} (1-based index)")
    else:
        print(f"Key {key}: Not found in the list")

Expected Output:

Key 29: Found at position 4 (0-based index) / position 5 (1-based index)
Key 40: Not found in the list
Key 5: Found at position 0 (0-based index) / position 1 (1-based index)

Explanation of Key Lines

• mid = (left + right) // 2: Compute the middle index of the current search interval. …

Binary search is the right tool here because it efficiently locates an element in a sorted list by repeatedly halving the search space. Instead of checking every element (linear search, $O(n)$), binary search compares the key with the middle element and eliminates half the remaining elements in each step, achieving $O(\log n)$ time complexity.

The algorithm works only on sorted data. If the middle element matches the key, we're done. If the key is smaller, we search the left half; if larger, the right half. We repeat until we find the key or exhaust the search space.

The Algorithm

1. Sort the list (binary search requires sorted input).
2. Set low = 0 and high = len(list) - 1.
3. While low ≤ high:
   • Calculate mid = (low + high) // 2.
   • If list[mid] == key, return mid.
   • If key < list[mid], set high = mid - 1 (search left half).
   • If key > list[mid], set low = mid + 1 (search right half).
4. If the loop exits, the key is not present.

Complete Python Program

def binary_search(arr, key):
    """
    Performs binary search on a sorted list.
    Returns the index if key is found, otherwise -1.
    """
    low = 0
    high = len(arr) - 1
    
    while low <= high:
        mid = (low + high) // 2
        
        if arr[mid] == key:
            return mid  # Key found at index mid
        elif key < arr[mid]:
            high = mid - 1  # Search in left half
        else:
            low = mid + 1   # Search in right half
    
    return -1  # Key not found


# Main program
numbers = []
print("Enter 10 integers:")
for i in range(10):
    num = int(input(f"Enter number {i+1}: "))
    numbers.append(num)

# Sort the list (required for binary search)
numbers.sort()
print(f"\nSorted list: {numbers}")

# Test with 3 different key values
test_keys = []
print("\nEnter 3 key values to search:")
for i in range(3):
    key = int(input(f"Enter key {i+1}: "))
    test_keys.append(key)

print("\n" + "="*50)
print("SEARCH RESULTS")
print("="*50)

for key in test_keys:
    result = binary_search(numbers, key)
    
    if result != -1:
        print(f"\nKey {key}: FOUND at position {result} (0-indexed) or position {result+1} (1-indexed)")
    else:
        print(f"\nKey {key}: NOT FOUND in the list")

Sample Run with Three Test Cases

Input:

Enter 10 integers:
Enter number 1: 45
Enter number 2: 12
Enter number 3: 78
Enter number 4: 23
Enter number 5: 56
Enter number 6: 89
Enter number 7: 34
Enter number 8: 67
Enter number 9: 90
Enter number 10: 11

Sorted list: [11, 12, 23, 34, 45, 56, 67, 78, 89, 90]

Enter 3 key values to search:
Enter key 1: 56
Enter key 2: 100
Enter key 3: 11

Output:

==================================================
SEARCH RESULTS
==================================================

Key 56: FOUND at position 5 (0-indexed) or position 6 (1-indexed) …