Binary Search as Divide and Conquer in DSA Go - Time & Space Complexity

Time Complexity: Binary Search as Divide and Conquer

O(log n)

Understanding Time Complexity

We want to understand how fast binary search finds a number in a sorted list.

How does the time to find the number change as the list gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

func binarySearch(arr []int, target int) int {
    low, high := 0, len(arr)-1
    for low <= high {
        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
}

This code searches for a target number in a sorted list by repeatedly dividing the search range in half.

Identify Repeating Operations

• Primary operation: Checking the middle element and adjusting the search range.
• How many times: The loop runs until the search range is empty, roughly halving the range each time.

How Execution Grows With Input

Each step cuts the list size in half, so the number of steps grows slowly as the list grows.

Input Size (n)	Approx. Operations
10	About 4 steps
100	About 7 steps
1000	About 10 steps

Pattern observation: Doubling the list size adds only one extra step.

Final Time Complexity

Time Complexity: O(log n)

This means the time to find the target grows slowly, adding just a few more steps even if the list gets much bigger.

Common Mistake

[X] Wrong: "Binary search checks every element one by one like linear search."

[OK] Correct: Binary search skips half the list each time, so it does far fewer checks than linear search.

Interview Connect

Understanding binary search time helps you explain efficient searching clearly, a key skill in many coding interviews.

Self-Check

"What if the list was not sorted? How would the time complexity of searching change?"

Practice

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

easy

A. It uses recursion to add elements to the array.

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

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

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

Solution

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

   Binary search splits the search space into two halves and discards one half each time.

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

   By halving the range repeatedly, it quickly narrows down to the target element.

3. Final Answer:

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

4. Quick Check:

   Divide and conquer means halving search space [OK]

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

Common Mistakes:

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

2. Which of the following is the correct Go syntax to start a binary search loop on a sorted slice arr?

easy

A. for left := 0; left < len(arr); left++ { }

B. for i := 0; i < len(arr); i++ { }

C. while left < right { }

D. for left, right := 0, len(arr)-1; left <= right; { }

Solution

1. Step 1: Identify Go for-loop syntax for binary search

   Binary search uses two pointers, left and right, initialized to start and end indices.

2. Step 2: Confirm loop condition and initialization

   The loop continues while left <= right, using Go's for syntax with multiple variables.

3. Final Answer:

   for left, right := 0, len(arr)-1; left <= right; { } -> Option D

4. Quick Check:

   Go for-loop with two vars and condition [OK]

Hint: Binary search loop uses two pointers and condition left <= right [OK]

Common Mistakes:

• Using for-loop with single index only
• Using while syntax (not valid in Go)
• Incorrect loop condition like left < right

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

arr := []int{2, 4, 6, 8, 10}
target := 6
left, right := 0, len(arr)-1
found := false
for left <= right {
  mid := left + (right-left)/2
  if arr[mid] == target {
    found = true
    break
  } else if arr[mid] < target {
    left = mid + 1
  } else {
    right = mid - 1
  }
}
fmt.Println(found)

medium

A. false

B. true

C. 6

D. runtime error

Solution

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

   Initial left=0, right=4, mid=2 (arr[2]=6), matches target, so found=true.

2. Step 2: Confirm loop breaks and prints found

   Loop breaks immediately after finding target, so prints true.

3. Final Answer:

   true -> Option B

4. Quick Check:

   Target found at mid index 2 [OK]

Hint: Check mid element equals target first [OK]

Common Mistakes:

• Not updating left or right correctly
• Off-by-one errors in mid calculation
• Assuming output is the target value

4. Identify the bug in this binary search Go code snippet:

arr := []int{1, 3, 5, 7, 9}
target := 4
left, right := 0, len(arr)
for left <= right {
  mid := (left + right) / 2
  if arr[mid] == target {
    fmt.Println("Found")
    break
  } else if arr[mid] < target {
    left = mid + 1
  } else {
    right = mid - 1
  }
}

medium

A. right should be initialized to len(arr)-1, not len(arr)

B. mid calculation should be mid := left - right

C. left should be incremented by 2 instead of 1

D. The loop condition should be left < right

Solution

1. Step 1: Check initialization of right pointer

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

2. Step 2: Correct right initialization and explain impact

   Setting right to len(arr)-1 fixes out-of-range error when accessing arr[mid].

3. Final Answer:

   right should be initialized to len(arr)-1, not len(arr) -> Option A

4. Quick Check:

   Array indices go from 0 to len(arr)-1 [OK]

Hint: Initialize right to last valid index len(arr)-1 [OK]

Common Mistakes:

• Using len(arr) as index causes runtime error
• Wrong mid calculation formula
• Incorrect loop condition causing infinite loop

5. Given a sorted slice arr := []int{1, 2, 2, 2, 3, 4}, how can you modify binary search to find the first occurrence of 2?

hard

A. After finding 2, continue searching left half to find earlier occurrence

B. Return immediately when 2 is found at any position

C. Search the right half after finding 2 to find last occurrence

D. Use linear search instead of binary search

Solution

1. Step 1: Understand the problem of duplicates in binary search

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

2. Step 2: Modify search to continue left after finding target

   When target found, update result and move right pointer to mid-1 to find earlier occurrence.

3. Final Answer:

   After finding 2, continue searching left half to find earlier occurrence -> Option A

4. Quick Check:

   Find first occurrence by searching left side after match [OK]

Hint: Keep searching left half after finding target to get first occurrence [OK]

Common Mistakes:

• Stopping at first found occurrence
• Searching right half instead of left
• Using linear search unnecessarily