Binary Search Iterative Approach in DSA Python - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the middle index in the binary search loop.

DSA Python

while low <= high:
    mid = (low [1] high) // 2
    if arr[mid] == target:
        return mid
    elif arr[mid] < target:
        low = mid + 1
    else:
        high = mid - 1
return -1

Drag options to blanks, or click blank then click option'

A-

B+

C*

D/

Attempts:

3 left

💡 Hint

Common Mistakes

Using '-' instead of '+' causes wrong mid calculation.

Using '*' or '/' leads to incorrect index.

To find the middle index, we add low and high and divide by 2. So the operator between low and high is '+'.

2fill in blank

medium

Complete the code to update the low index when the middle element is less than the target.

DSA Python

if arr[mid] < target:
    low = mid [1] 1

Drag options to blanks, or click blank then click option'

A//

B-

C+

D*

Attempts:

3 left

💡 Hint

Common Mistakes

Using '-' decreases low, causing infinite loop.

Using '*' or '//' is invalid for index update.

When the middle element is less than the target, we move low to mid + 1 to search the right half.

3fill in blank

hard

Fix the error in the condition to continue the binary search loop.

DSA Python

while low [1] high:
    mid = (low + high) // 2
    if arr[mid] == target:
        return mid
    elif arr[mid] < target:
        low = mid + 1
    else:
        high = mid - 1
return -1

Drag options to blanks, or click blank then click option'

A<=

B>

C<

D>=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<' misses checking when low equals high.

Using '>' or '>=' causes the loop to never run or run incorrectly.

The loop should continue while low is less than or equal to high to cover all elements.

4fill in blank

hard

Fill both blanks to correctly update the high index when the middle element is greater than the target.

DSA Python

if arr[mid] > target:
    high = mid [1] [2] 1

Drag options to blanks, or click blank then click option'

A//

B+

C*

D-

Attempts:

3 left

💡 Hint

Common Mistakes

Using '+' instead of '-' moves high in wrong direction.

Using '*' or '//' is invalid for index update.

When middle element is greater than target, high moves to mid - 1 to search left half.

5fill in blank

hard

Fill all three blanks to complete the iterative binary search function.

DSA Python

def binary_search(arr, target):
    low, high = 0, len(arr) [1] 1
    while low [2] high:
        mid = (low + high) // 2
        if arr[mid] == target:
            return mid
        elif arr[mid] < target:
            low = mid + 1
        else:
            high = mid [3] 1
    return -1

Drag options to blanks, or click blank then click option'

A-

B<=

C<

D+

Attempts:

3 left

💡 Hint

Common Mistakes

Setting high to len(arr) causes index error.

Using '<' instead of '<=' misses last element.

Updating high with '+' causes wrong search direction.

We set high to last index by subtracting 1 from length. Loop runs while low <= high. High updates to mid - 1 when middle element is greater than target.

Practice

1. What is the main purpose of the mid variable in the iterative binary search algorithm?

easy

A. To store the target value we are searching for

B. To keep track of the number of iterations

C. To find the middle index of the current search range

D. To hold the entire list being searched

Solution

1. Step 1: Understand the role of mid in binary search

   The mid variable calculates the middle index between the current low and high indices to split the search range.

2. Step 2: Explain why mid is important

   By checking the element at mid, the algorithm decides whether to search the left or right half next.

3. Final Answer:

   To find the middle index of the current search range -> Option C

4. Quick Check:

   mid = middle index [OK]

Hint: Remember binary search splits list by middle index [OK]

Common Mistakes:

• Confusing mid with the target value
• Using mid as a counter instead of index
• Not updating mid inside the loop

2. Which of the following is the correct way to calculate the middle index mid in an iterative binary search in Python?

easy

A. mid = (low + high) // 2

B. mid = low - high // 2

C. mid = low + high

D. mid = (low + high) / 2

Solution

1. Step 1: Identify correct integer division in Python

   In Python, // performs integer division, which is needed for an index.

2. Step 2: Explain why / is incorrect

   / returns a float, which cannot be used as a list index.

3. Final Answer:

   mid = (low + high) // 2 -> Option A

4. Quick Check:

   Use integer division for index [OK]

Hint: Use // for integer division in Python [OK]

Common Mistakes:

• Using / which returns float
• Adding low and high without division
• Incorrect operator precedence without parentheses

3. What will be the output of the following code snippet?

def binary_search(arr, target):
    low, high = 0, len(arr) - 1
    while low <= high:
        mid = (low + high) // 2
        if arr[mid] == target:
            return mid
        elif arr[mid] < target:
            low = mid + 1
        else:
            high = mid - 1
    return -1

arr = [2, 4, 6, 8, 10]
print(binary_search(arr, 6))

medium

A. 1

B. 3

C. -1

D. 2

Solution

1. Step 1: Trace the binary search for target 6

   Initial low=0, high=4, mid=2. arr[2]=6 matches target.

2. Step 2: Return the index where target is found

   Since arr[mid] == target, function returns 2 immediately.

3. Final Answer:

   2 -> Option D

4. Quick Check:

   Index of 6 in array is 2 [OK]

Hint: Check mid index value first to find target fast [OK]

Common Mistakes:

• Confusing index with value
• Returning value instead of index
• Off-by-one errors in mid calculation

4. Identify the error in the following iterative binary search code:

def binary_search(arr, target):
    low = 0
    high = len(arr)
    while low <= high:
        mid = (low + high) // 2
        if arr[mid] == target:
            return mid
        elif arr[mid] < target:
            low = mid + 1
        else:
            high = mid - 1
    return -1

medium

A. high should be initialized as len(arr) - 1

B. low should start from 1 instead of 0

C. mid calculation should use float division

D. The loop condition should be while low < high

Solution

1. Step 1: Check initialization of high

   High is set to len(arr), which is out of valid index range (0 to len(arr)-1).

2. Step 2: Explain correct initialization

   High should be len(arr) - 1 to avoid index out of range errors when accessing arr[mid].

3. Final Answer:

   high should be initialized as len(arr) - 1 -> Option A

4. Quick Check:

   High = len(arr) - 1 for valid index [OK]

Hint: Array indices go from 0 to length-1 [OK]

Common Mistakes:

• Setting high to len(arr) causes index error
• Changing low start incorrectly
• Using float division for mid

5. Given a sorted list arr = [1, 3, 5, 7, 9, 11, 13], how many iterations will the iterative binary search take to find the target 7?

medium

A. 3

B. 1

C. 2

D. 4

Solution

1. Step 1: Trace iterations to find 7

   Iteration 1: low=0, high=6, mid=3, arr[3]=7 matches target.

2. Step 2: Count iterations

   Since target found at first mid check, total iterations = 1.

3. Step 3: Reconsider counting loop passes

   Loop runs once, but mid calculation and comparison count as one iteration.

4. Final Answer:

   1 -> Option B

5. Quick Check:

   Target found at first mid check [OK]

Hint: Count each mid calculation as one iteration [OK]

Common Mistakes:

• Counting iterations incorrectly
• Assuming always 3 iterations for 7 elements
• Confusing iterations with index value