Why Binary Search and What Sorted Order Gives You in DSA C - See It Work

Why Binary Search and What Sorted Order Gives You

📖 Scenario: Imagine you have a list of book titles sorted alphabetically on a shelf. You want to find if a certain book is there quickly without checking every book one by one.

🎯 Goal: You will create a sorted array of book titles, set a target book title to find, implement binary search to check if the book is in the list, and print the result.

📋 What You'll Learn

Create a sorted array of 7 book titles in alphabetical order

Create a variable called target with the exact book title to search

Write a function called binary_search that takes the array, its size, and the target title, and returns the index if found or -1 if not

Print the index returned by binary_search

💡 Why This Matters

🌍 Real World

Binary search is used in many apps and systems to quickly find data like names, files, or records when the data is sorted.

💼 Career

Understanding binary search and sorted data is essential for software developers to write efficient search functions and optimize performance.

1

Create a sorted array of book titles

Create a sorted array of 7 book titles called books with these exact titles in this order: "Algorithms", "Data Structures", "Design Patterns", "Operating Systems", "Programming", "Software Engineering", "Systems Programming".

DSA C

char *books[7] = {"Algorithms", "Data Structures", "Design Patterns", "Operating Systems", "Programming", "Software Engineering", "Systems Programming"};
// Your code here

Hint

Use an array of strings with the exact titles in alphabetical order.

2

Set the target book title to search

Create a variable called target and set it to the exact string "Programming".

DSA C

char *books[7] = {"Algorithms", "Data Structures", "Design Patterns", "Operating Systems", "Programming", "Software Engineering", "Systems Programming"};
char *target = "Programming";
// Your code here

Hint

Use a string pointer variable named target with the exact value.

3

Write the binary search function

Write a function called binary_search that takes char *arr[], int size, and char *target. Use a loop with variables low, high, and mid to find target in arr using strcmp. Return the index if found, or -1 if not found.

DSA C

char *books[7] = {"Algorithms", "Data Structures", "Design Patterns", "Operating Systems", "Programming", "Software Engineering", "Systems Programming"};
char *target = "Programming";

int binary_search(char *arr[], int size, char *target) {
    // Your code here
}

Hint

Use strcmp to compare strings and adjust low and high accordingly.

4

Print the result of binary search

Call binary_search with books, 7, and target. Store the result in index. Print index using printf with format "%d\n".

DSA C

char *books[7] = {"Algorithms", "Data Structures", "Design Patterns", "Operating Systems", "Programming", "Software Engineering", "Systems Programming"};
char *target = "Programming";

#include 
#include 

int binary_search(char *arr[], int size, char *target) {
    int low = 0;
    int high = size - 1;
    while (low <= high) {
        int mid = (low + high) / 2;
        int cmp = strcmp(arr[mid], target);
        if (cmp == 0) {
            return mid;
        } else if (cmp < 0) {
            low = mid + 1;
        } else {
            high = mid - 1;
        }
    }
    return -1;
}

int main() {
    // Your code here
}

Hint

Call binary_search and print the returned index.

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