Find Peak Element Using Binary Search in DSA Go - Time & Space Complexity

Time Complexity: Find Peak Element Using Binary Search

O(log n)

Understanding Time Complexity

We want to understand how fast the binary search method finds a peak element in an array.

How does the number of steps change as the array gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

func findPeakElement(nums []int) int {
    left, right := 0, len(nums)-1
    for left < right {
        mid := left + (right-left)/2
        if nums[mid] < nums[mid+1] {
            left = mid + 1
        } else {
            right = mid
        }
    }
    return left
}

This code finds a peak element index by repeatedly narrowing the search range using binary search.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The for loop that halves the search range each time.
• How many times: The loop runs until the search range is one element, about log base 2 of n times.

How Execution Grows With Input

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

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

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

Final Time Complexity

Time Complexity: O(log n)

This means the steps needed grow slowly, making the search very efficient even for large arrays.

Common Mistake

[X] Wrong: "The loop runs n times because it looks at each element."

[OK] Correct: The loop does not check every element; it cuts the search space in half each time, so it runs much fewer times.

Interview Connect

Understanding this time complexity shows you can use efficient search methods, a key skill in many coding challenges and real-world problems.

Self-Check

"What if we changed the array to be unsorted without any peak guarantee? How would the time complexity change?"

Practice

1. What is a peak element in an array when using binary search to find it?

easy

A. An element that is greater than its neighbors

B. An element that is smaller than its neighbors

C. The first element of the array

D. The last element of the array

Solution

1. Step 1: Understand the definition of a peak element

   A peak element is one that is strictly greater than its neighbors on both sides (or only one side if at the boundary).

2. Step 2: Apply this to binary search context

   Binary search finds such an element by comparing middle elements with neighbors to decide which half to search next.

3. Final Answer:

   An element that is greater than its neighbors -> Option A

4. Quick Check:

   Peak element = greater than neighbors [OK]

Hint: Peak element is bigger than neighbors, not smaller [OK]

Common Mistakes:

• Thinking peak is smallest element
• Confusing peak with first or last element always
• Ignoring boundary conditions

2. Which Go code snippet correctly initializes the binary search boundaries to find a peak element in array nums?

easy

A. left, right := 1, len(nums)

B. left, right := 0, len(nums) - 1

C. left, right := 0, len(nums)

D. left, right := 1, len(nums) - 1

Solution

1. Step 1: Identify valid index range for array in Go

   Array indices in Go start at 0 and end at len(nums)-1, so boundaries must be set accordingly.

2. Step 2: Check each option for correct boundary initialization

   left, right := 0, len(nums) - 1 correctly sets left=0 and right=len(nums)-1, covering full array indices.

3. Final Answer:

   left, right := 0, len(nums) - 1 -> Option B

4. Quick Check:

   Array indices 0 to len-1 [OK]

Hint: Array indices go from 0 to length-1 in Go [OK]

Common Mistakes:

• Setting right to len(nums) (out of range)
• Starting left at 1 and missing first element
• Using negative indices

3. What will be the output index of the peak element for the array nums := []int{1, 3, 20, 4, 1, 0} using binary search peak finding?

medium

A. 3

B. 1

C. 2

D. 0

Solution

1. Step 1: Identify peak candidates in the array

   Array: [1, 3, 20, 4, 1, 0]. 20 at index 2 is greater than neighbors 3 and 4, so it's a peak.

2. Step 2: Confirm binary search finds index 2

   Binary search compares mid elements and moves left or right until it finds peak at index 2.

3. Final Answer:

   2 -> Option C

4. Quick Check:

   Peak at index 2 with value 20 [OK]

Hint: Peak is local max, here 20 at index 2 [OK]

Common Mistakes:

• Choosing first or last element incorrectly
• Confusing index with value
• Not checking both neighbors

4. Given the following Go code snippet to find a peak element, what is the bug?

func findPeak(nums []int) int {
    left, right := 0, len(nums)-1
    for left < right {
        mid := (left + right) / 2
        if nums[mid] < nums[mid+1] {
            left = mid + 1
        } else {
            right = mid - 1
        }
    }
    return left
}

medium

A. The right boundary update should be right = mid, not mid - 1

B. The left boundary update should be left = mid, not mid + 1

C. The loop condition should be left <= right

D. The function should return right instead of left

Solution

1. Step 1: Analyze the binary search boundary updates

   When nums[mid] >= nums[mid+1], peak is at mid or to the left, so right should be set to mid, not mid - 1 to avoid skipping mid.

2. Step 2: Identify impact of wrong right update

   Using right = mid - 1 skips mid, possibly missing the peak element.

3. Final Answer:

   The right boundary update should be right = mid, not mid - 1 -> Option A

4. Quick Check:

   Right boundary must include mid [OK]

Hint: Use right = mid to keep mid in search space [OK]

Common Mistakes:

• Setting right = mid - 1 and skipping mid
• Changing left incorrectly
• Wrong loop condition causing infinite loop

5. You have an array nums := []int{10, 20, 15, 2, 23, 90, 67}. Using binary search to find a peak element, which index will be returned and why?

hard

A. 1, because 20 is a peak greater than neighbors 10 and 15

B. 0, because first element is always a peak

C. 6, because last element is always a peak

D. 5, because 90 is a peak greater than neighbors 23 and 67

Solution

1. Step 1: Identify all peak candidates in the array

   Array: [10, 20, 15, 2, 23, 90, 67]. Peaks are 20 at index 1 and 90 at index 5.

2. Step 2: Understand binary search behavior

   Binary search compares mid and mid+1; it will move towards the higher peak 90 at index 5.

3. Step 3: Confirm returned index

   Binary search will return index 5 where 90 is a peak greater than neighbors 23 and 67.

4. Final Answer:

   5, because 90 is a peak greater than neighbors 23 and 67 -> Option D

5. Quick Check:

   Binary search finds highest local peak at index 5 [OK]

Hint: Binary search finds any peak, often the higher one found first [OK]

Common Mistakes:

• Assuming first or last element always peak
• Choosing smaller peak incorrectly
• Not understanding binary search direction