Why Binary Search and What Sorted Order Gives You in DSA C - Test Your Knowledge

Practice - 5 Tasks

1fill in blank

easy

Complete the code to start binary search with the correct initial left index.

DSA C

int left = [1];

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

A0

B1

C-1

Darray_length

Attempts:

3 left

💡 Hint

Common Mistakes

Starting left at 1 or -1 causes skipping the first element.

The left index in binary search starts at 0, the first element's position in the array.

2fill in blank

medium

Complete the code to calculate the middle index correctly in binary search.

DSA C

int mid = left + ([1] - left) / 2;

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

Aright

Bleft

Cmid

Darray_length

Attempts:

3 left

💡 Hint

Common Mistakes

Using mid or array_length instead of right causes wrong mid calculation.

The middle index is calculated between left and right to avoid overflow.

3fill in blank

hard

Fix the error in the condition to move the left pointer when target is greater than mid value.

DSA C

if (target > arr[[1]]) {
    left = mid + 1;
}

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

Aleft

Bright

Cmid

Dtarget

Attempts:

3 left

💡 Hint

Common Mistakes

Using left or right index instead of mid causes wrong comparisons.

We compare target with the middle element arr[mid] to decide which half to search next.

4fill in blank

hard

Fill both blanks to correctly update pointers when target is less or greater than mid value.

DSA C

if (target < arr[[1]]) {
    [2] = mid - 1;
}

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

Amid

Bleft

Cright

Dtarget

Attempts:

3 left

💡 Hint

Common Mistakes

Updating left instead of right pointer causes infinite loops.

If target is less than arr[mid], we move the right pointer to mid - 1 to search left half.

5fill in blank

hard

Fill all three blanks to complete the binary search loop condition and return statement.

DSA C

while ([1] <= [2]) {
    int mid = [1] + ([2] - [1]) / 2;
    if (arr[mid] == target) {
        return mid;
    }
    // update pointers
}
return -1;

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

Aleft

Bright

Cmid

Dtarget

Attempts:

3 left

💡 Hint

Common Mistakes

Using mid or target in loop condition causes wrong loop behavior.

The loop runs while left is less or equal to right. Mid is calculated from left and right.

Practice

1. Why is it important for a list to be sorted before applying binary search?

easy

A. Because binary search can only search in lists with odd number of elements

B. Because binary search works only on lists with duplicate elements

C. Because binary search divides the list and compares middle elements to find the target efficiently

D. Because binary search requires the list to be reversed

Solution

1. Step 1: Understand binary search working

   Binary search works by repeatedly dividing the sorted list and comparing the middle element to the target value.

2. Step 2: Importance of sorted order

   Sorted order allows us to decide which half to discard after each comparison, making search efficient.

3. Final Answer:

   Because binary search divides the list and compares middle elements to find the target efficiently -> Option C

4. Quick Check:

   Sorted list needed for binary search = Because it divides and compares middle elements [OK]

Hint: Binary search needs sorted list to halve search space fast [OK]

Common Mistakes:

• Thinking binary search works on unsorted lists
• Assuming binary search needs duplicates
• Believing binary search requires odd length lists

2. Which of the following is the correct syntax to start a binary search loop in C for an array arr of size n?

easy

A. int low = 1, high = n - 1; while (low <= high) { ... }

B. int low = 0, high = n - 1; while (low <= high) { ... }

C. int low = 0, high = n; while (low < high) { ... }

D. int low = 1, high = n; while (low < high) { ... }

Solution

1. Step 1: Identify correct index range for array

   Array indices in C start at 0 and end at n-1, so low = 0 and high = n-1.

2. Step 2: Correct loop condition for binary search

   Loop continues while low is less than or equal to high to cover all elements.

3. Final Answer:

   int low = 0, high = n - 1; while (low <= high) { ... } -> Option B

4. Quick Check:

   Array indices 0 to n-1 and loop while low <= high [OK]

Hint: Array indices start at 0; loop while low ≤ high [OK]

Common Mistakes:

• Starting low at 1 instead of 0
• Using high = n instead of n-1
• Using < instead of <= in loop condition

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

int arr[] = {2, 4, 6, 8, 10};
int low = 0, high = 4, mid;
int target = 6;
while (low <= high) {
  mid = (low + high) / 2;
  if (arr[mid] == target) {
    printf("Found at index %d", mid);
    break;
  } else if (arr[mid] < target) {
    low = mid + 1;
  } else {
    high = mid - 1;
  }
}

medium

A. Found at index 1

B. Found at index 3

C. Not found

D. Found at index 2

Solution

1. Step 1: Initialize variables and first mid calculation

   low=0, high=4, mid=(0+4)/2=2; arr[2]=6 matches target 6.

2. Step 2: Check condition and print result

   Since arr[mid] == target, print "Found at index 2" and break loop.

3. Final Answer:

   Found at index 2 -> Option D

4. Quick Check:

   Target 6 found at index 2 [OK]

Hint: Calculate mid as (low+high)/2 and compare to target [OK]

Common Mistakes:

• Calculating mid incorrectly
• Off-by-one errors in index
• Printing wrong index

4. Identify the error in this binary search code snippet and choose the correct fix:

int arr[] = {1, 3, 5, 7, 9};
int low = 0, high = 5, mid;
int target = 7;
while (low <= high) {
  mid = (low + high) / 2;
  if (arr[mid] == target) {
    printf("Found at %d", mid);
    break;
  } else if (arr[mid] < target) {
    low = mid + 1;
  } else {
    high = mid - 1;
  }
}

medium

A. Change high to 4 instead of 5

B. Change low to 1 instead of 0

C. Change mid calculation to mid = (low - high) / 2

D. No error, code is correct

Solution

1. Step 1: Check array index bounds

   Array size is 5, so highest index is 4. Setting high=5 causes out-of-bounds access.

2. Step 2: Fix high initialization

   Set high = 4 to stay within valid indices and avoid runtime errors.

3. Final Answer:

   Change high to 4 instead of 5 -> Option A

4. Quick Check:

   Array indices 0 to 4, high must be 4 [OK]

Hint: High index must be last valid index (size-1) [OK]

Common Mistakes:

• Setting high to array size instead of size-1
• Changing low incorrectly
• Incorrect mid calculation

5. Given a sorted array arr = {1, 3, 5, 7, 9, 11, 13}, you want to find the smallest element greater than or equal to 8 using binary search. Which approach correctly finds this element?

hard

A. Modify binary search to move high = mid - 1 when arr[mid] >= 8, else low = mid + 1

B. Use normal binary search to find 8 directly; if not found, return -1

C. Sort the array again before binary search

D. Use linear search instead of binary search

Solution

1. Step 1: Understand the problem

   We want the smallest element >= 8, which may not be exactly 8 in the array.

2. Step 2: Modify binary search for lower bound

   When arr[mid] >= 8, move high to mid - 1 to find smaller candidates; else move low to mid + 1.

3. Step 3: After loop, low points to smallest element >= 8

   This approach finds the correct element (9 at index 4).

4. Final Answer:

   Modify binary search to move high = mid - 1 when arr[mid] >= 8, else low = mid + 1 -> Option A

5. Quick Check:

   Lower bound binary search finds smallest element ≥ target [OK]

Hint: Adjust binary search to find lower bound by moving high on >= [OK]

Common Mistakes:

• Using normal binary search without modification
• Sorting array again unnecessarily
• Choosing linear search over binary search