Binary Search as Divide and Conquer in DSA C - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the main idea behind the divide and conquer approach in binary search?

Divide the problem into smaller parts by splitting the array into halves, then conquer by searching in the half where the target might be.

beginner

In binary search, what condition decides which half of the array to search next?

Compare the target value with the middle element; if target is smaller, search left half; if larger, search right half.

beginner

Why must the array be sorted for binary search to work?

Because binary search relies on order to decide which half to search next; without sorting, this decision is not possible.

intermediate

What is the time complexity of binary search and why?

O(log n) because each step halves the search space, quickly reducing the problem size.

intermediate

How does binary search demonstrate the divide and conquer strategy?

It divides the array into two halves and conquers by recursively searching the half that may contain the target until found or empty.

What is the first step in binary search on a sorted array?

A Find the middle element

B Sort the array

C Search the entire array linearly

D Swap elements

Binary search starts by finding the middle element to divide the array.

If the target is greater than the middle element, which part of the array do we search next?

A Right half

B Left half

C Both halves

D None

If target > middle element, search the right half because the array is sorted.

What happens if the target equals the middle element in binary search?

A Search left half

B Search right half

C Return the middle element's index

D Continue searching both halves

If target equals middle element, binary search returns its index immediately.

What is the worst-case time complexity of binary search?

A O(n)

B O(log n)

C O(n log n)

D O(1)

Binary search halves the search space each step, leading to O(log n) complexity.

Why can't binary search be used on an unsorted array?

A Because it needs duplicates

B Because it requires linear search

C Because it only works on arrays of size 1

D Because it relies on sorted order to decide which half to search

Binary search depends on sorted order to eliminate half the array each step.

Explain how binary search uses the divide and conquer approach to find a target in a sorted array.

Think about how the array is split and how the search space shrinks.

Describe why binary search is more efficient than linear search for large sorted arrays.

Focus on how quickly the search space reduces.

Practice

1. What is the main idea behind Binary Search as a divide and conquer algorithm?

easy

A. It searches every element one by one until the target is found.

B. It divides the search range in half each time to find the target efficiently.

C. It sorts the array before searching for the target.

D. It uses recursion to add all elements in the array.

Solution

1. Step 1: Understand divide and conquer in binary search

   Binary search splits the array into two halves and decides which half to search next.

2. Step 2: Identify the efficiency of halving the search space

   By halving the search space each time, it quickly narrows down the target location.

3. Final Answer:

   It divides the search range in half each time to find the target efficiently. -> Option B

4. Quick Check:

   Divide and conquer = halving search space [OK]

Hint: Binary search halves the search space each step [OK]

Common Mistakes:

• Thinking binary search checks every element
• Confusing binary search with sorting
• Believing binary search sums elements

2. Which of the following is the correct syntax to calculate the middle index in binary search in C?

easy

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

B. int mid = (low * high) / 2;

C. int mid = low + (high - low) / 2;

D. int mid = (low - high) / 2;

Solution

1. Step 1: Understand safe middle index calculation

   Using low + (high - low) / 2 prevents integer overflow when low and high are large.

2. Step 2: Compare with other options

   int mid = (low + high) / 2; can overflow if low + high exceeds integer limit; options B and C are incorrect formulas.

3. Final Answer:

   int mid = low + (high - low) / 2; -> Option C

4. Quick Check:

   Safe mid calculation = low + (high - low)/2 [OK]

Hint: Use low + (high - low)/2 to avoid overflow [OK]

Common Mistakes:

• Using (low + high)/2 causing overflow
• Subtracting incorrectly in mid calculation
• Multiplying low and high instead of adding

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

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

medium

A. 2

B. 3

C. 1

D. 4

Solution

1. Step 1: Trace the binary search steps

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

2. Step 2: Print the mid index

   Since arr[mid] == target, the code prints 2 and breaks the loop.

3. Final Answer:

   2 -> Option A

4. Quick Check:

   Target found at index 2 [OK]

Hint: Check mid index value matches target first [OK]

Common Mistakes:

• Not updating low and high correctly
• Confusing index with value
• Off-by-one errors in mid calculation

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

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

medium

A. Change while condition to low <= high

B. Change mid calculation to (low + high) / 2

C. Remove else if and use only if statements

D. Return 0 instead of -1 when not found

Solution

1. Step 1: Analyze the loop condition

   The loop uses while (low < high), which misses checking when low equals high.

2. Step 2: Correct the loop condition

   Changing to while (low <= high) ensures the last element is checked.

3. Final Answer:

   Change while condition to low <= high -> Option A

4. Quick Check:

   Loop must include low == high to check all elements [OK]

Hint: Use while(low <= high) to include all elements [OK]

Common Mistakes:

• Using < instead of <= in loop condition
• Incorrect mid calculation causing overflow
• Returning wrong value when target not found

5. Given a sorted array with possible duplicate elements, how can you modify binary search to find the first occurrence of a target value?

hard

A. Use linear search instead of binary search.

B. Return the first found index immediately without further search.

C. Search only the right half after finding the target.

D. After finding target, continue searching left half to find earlier occurrence.

Solution

1. Step 1: Understand duplicates in binary search

   Standard binary search returns any occurrence, not necessarily the first.

2. Step 2: Modify search to find first occurrence

   After finding target at mid, continue searching left half to check for earlier occurrences.

3. Final Answer:

   After finding target, continue searching left half to find earlier occurrence. -> Option D

4. Quick Check:

   Find first occurrence by searching left after match [OK]

Hint: Search left half after match to find first occurrence [OK]

Common Mistakes:

• Stopping at first found index ignoring duplicates
• Searching right half instead of left after match
• Switching to linear search unnecessarily