Why Binary Search and What Sorted Order Gives You in DSA C - Challenge Your Understanding

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Binary Search on Sorted Array

What is the output of the following C code that performs binary search on a sorted array?

DSA C

int binarySearch(int arr[], int size, int target) {
    int left = 0, right = size - 1;
    while (left <= right) {
        int mid = left + (right - left) / 2;
        if (arr[mid] == target) return mid;
        else if (arr[mid] < target) left = mid + 1;
        else right = mid - 1;
    }
    return -1;
}

#include <stdio.h>
int main() {
    int arr[] = {2, 4, 6, 8, 10, 12};
    int index = binarySearch(arr, 6, 8);
    printf("%d\n", index);
    return 0;
}

A 2

B 3

C -1

D 4

💡 Hint

Remember array indexing starts at 0 and the array is sorted.

The target 8 is at index 3 in the sorted array. Binary search finds it correctly and returns 3.

🧠 Conceptual

intermediate

Why Must the Array Be Sorted for Binary Search?

Why is it necessary for the array to be sorted before applying binary search?

A Because binary search can only find the first element in the array.

B Because binary search works only on arrays with unique elements.

C Because binary search requires the array to be in reverse order.

D Because binary search divides the array and compares the middle element to decide which half to search next.

💡 Hint

Think about how binary search decides which half to ignore.

Binary search relies on the array being sorted to decide whether to search the left or right half based on comparison with the middle element.

❓ Predict Output

advanced

Output of Binary Search on Unsorted Array

What is the output of the following C code that performs binary search on an unsorted array?

DSA C

int binarySearch(int arr[], int size, int target) {
    int left = 0, right = size - 1;
    while (left <= right) {
        int mid = left + (right - left) / 2;
        if (arr[mid] == target) return mid;
        else if (arr[mid] < target) left = mid + 1;
        else right = mid - 1;
    }
    return -1;
}

#include <stdio.h>
int main() {
    int arr[] = {10, 8, 2, 4, 12, 6};
    int index = binarySearch(arr, 6, 8);
    printf("%d\n", index);
    return 0;
}

A 2

B 0

C -1

D 3

💡 Hint

The array is not sorted. Think about how binary search behaves then.

Binary search assumes sorted order. On an unsorted array, it may not find the target and returns -1.

🧠 Conceptual

advanced

Effect of Sorted Order on Binary Search Efficiency

How does having a sorted array improve the efficiency of searching compared to an unsorted array?

A Sorted arrays allow binary search which reduces search time from linear to logarithmic.

B Sorted arrays require checking every element, so search time is linear.

C Sorted arrays slow down search because elements are rearranged.

D Sorted arrays make no difference; search time is always the same.

💡 Hint

Think about how many elements you can ignore at each step in binary search.

Binary search on sorted arrays cuts the search space in half each step, making it much faster than linear search on unsorted arrays.

🚀 Application

expert

Finding Insert Position Using Binary Search

Given a sorted array, which option correctly returns the index where a target should be inserted to keep the array sorted if the target is not found?

DSA C

int findInsertPosition(int arr[], int size, int target) {
    int left = 0, right = size - 1;
    while (left <= right) {
        int mid = left + (right - left) / 2;
        if (arr[mid] == target) return mid;
        else if (arr[mid] < target) left = mid + 1;
        else right = mid - 1;
    }
    return left;
}

#include <stdio.h>
int main() {
    int arr[] = {1, 3, 5, 6};
    int pos = findInsertPosition(arr, 4, 4);
    printf("%d\n", pos);
    return 0;
}

A 2

B 3

C 4

D 1

💡 Hint

If target is not found, the function returns the left pointer as insert position.

The target 4 should be inserted at index 2 to keep the array sorted. The function returns left which is 2.

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