Search in Rotated Sorted Array in DSA Python - Time & Space Complexity

Time Complexity: Search in Rotated Sorted Array

O(log n)

Understanding Time Complexity

We want to understand how fast we can find a number in a rotated sorted list.

How does the search time change as the list gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

def search_rotated(nums, target):
    left, right = 0, len(nums) - 1
    while left <= right:
        mid = (left + right) // 2
        if nums[mid] == target:
            return mid
        if nums[left] <= nums[mid]:
            if nums[left] <= target <= nums[mid]:
                right = mid - 1
            else:
                left = mid + 1
        else:
            if nums[mid] <= target <= nums[right]:
                left = mid + 1
            else:
                right = mid - 1
    return -1

This code searches for a target number in a rotated sorted list using a modified binary search.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The while loop that halves the search range each time.
• How many times: The loop runs until the search range is empty, roughly halving the size each time.

How Execution Grows With Input

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

Input Size (n)	Approx. Operations
10	About 4 to 5 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 search time grows slowly, adding just a few more steps even if the list gets much bigger.

Common Mistake

[X] Wrong: "Since the list is rotated, we must check every element, so it takes linear time."

[OK] Correct: The code cleverly uses the sorted parts to skip half the list each time, keeping the search fast.

Interview Connect

Understanding this search shows you can adapt classic methods to tricky situations, a skill valued in problem solving.

Self-Check

"What if the list was not rotated but still sorted? How would the time complexity change?"

Practice

1. What is the main challenge when searching for an element in a rotated sorted array compared to a normal sorted array?

easy

A. The array elements are all negative numbers.

B. The array contains duplicate elements which confuse the search.

C. The array is not fully sorted, so normal binary search cannot be directly applied.

D. The array size is unknown, so we cannot set boundaries.

Solution

1. Step 1: Understand normal binary search

   Binary search works efficiently on fully sorted arrays by dividing the search space in half each time.

2. Step 2: Identify the challenge with rotated arrays

   In a rotated sorted array, the array is sorted but rotated at some pivot, so it is not fully sorted in one direction, making normal binary search invalid without adjustments.

3. Final Answer:

   The array is not fully sorted, so normal binary search cannot be directly applied. -> Option C

4. Quick Check:

   Rotated array breaks full sorting = normal binary search fails [OK]

Hint: Rotated means partial sorting, normal binary search fails [OK]

Common Mistakes:

• Assuming array is fully sorted
• Ignoring the rotation pivot
• Thinking duplicates cause the main problem

2. Which of the following is the correct syntax to start a binary search on a rotated sorted array named arr with target in Python?

easy

A. left, right = 1, len(arr) - 1

B. left, right = 1, len(arr)

C. left, right = 0, len(arr)

D. left, right = 0, len(arr) - 1

Solution

1. Step 1: Understand array indexing in Python

   Python arrays are zero-indexed, so the first element is at index 0 and the last at len(arr) - 1.

2. Step 2: Set search boundaries correctly

   To search the entire array, left should start at 0 and right at len(arr) - 1.

3. Final Answer:

   left, right = 0, len(arr) - 1 -> Option D

4. Quick Check:

   Zero-based indexing means last index is length minus one [OK]

Hint: Start left at 0 and right at length minus one [OK]

Common Mistakes:

• Starting left at 1 instead of 0
• Setting right to length instead of length - 1
• Confusing inclusive and exclusive ranges

3. What will be the output of the following Python code?

def search_rotated(arr, target):
    left, right = 0, len(arr) - 1
    while left <= right:
        mid = (left + right) // 2
        if arr[mid] == target:
            return mid
        if arr[left] <= arr[mid]:
            if arr[left] <= target < arr[mid]:
                right = mid - 1
            else:
                left = mid + 1
        else:
            if arr[mid] < target <= arr[right]:
                left = mid + 1
            else:
                right = mid - 1
    return -1

print(search_rotated([4,5,6,7,0,1,2], 0))

medium

A. -1

B. 4

C. 0

D. 6

Solution

1. Step 1: Trace the binary search on rotated array

   Array: [4,5,6,7,0,1,2], target: 0. Initial left=0, right=6, mid=3 (value 7). Since arr[left] <= arr[mid], left half is sorted. Target 0 is not in [4..7], so move left to mid+1=4.

2. Step 2: Continue search in right half

   Now left=4, right=6, mid=5 (value 1). arr[left] <= arr[mid], left half sorted. Target 0 is less than arr[mid], so right=mid-1=4. Next mid=4 (value 0), matches target, return 4.

3. Final Answer:

   4 -> Option B

4. Quick Check:

   Target 0 found at index 4 [OK]

Hint: Check which half is sorted, then narrow search [OK]

Common Mistakes:

• Not updating left/right correctly
• Confusing comparison operators
• Returning wrong index or -1 incorrectly

4. Identify the bug in the following code snippet for searching in a rotated sorted array:

def search(arr, target):
    left, right = 0, len(arr) - 1
    while left < right:
        mid = (left + right) // 2
        if arr[mid] == target:
            return mid
        if arr[left] <= arr[mid]:
            if arr[left] <= target < arr[mid]:
                right = mid - 1
            else:
                left = mid + 1
        else:
            if arr[mid] < target <= arr[right]:
                left = mid + 1
            else:
                right = mid - 1
    return -1

medium

A. The loop condition should be while left <= right, not while left < right.

B. The mid calculation is incorrect; it should be mid = (left + right + 1) // 2.

C. The target comparison should use <= instead of < in all conditions.

D. The function should return left instead of -1 at the end.

Solution

1. Step 1: Analyze loop condition

   The loop uses while left < right, which stops when left == right, possibly missing the last element check.

2. Step 2: Correct loop condition for full search

   Using while left <= right ensures the search includes the last element, preventing missing the target at the boundary.

3. Final Answer:

   The loop condition should be while left <= right, not while left < right. -> Option A

4. Quick Check:

   Include last element by using <= in loop condition [OK]

Hint: Use <= in loop to check all elements including boundaries [OK]

Common Mistakes:

• Using < instead of <= in loop
• Miscomputing mid index
• Returning wrong value when not found

5. Given a rotated sorted array arr = [30, 40, 50, 10, 20], what is the index of the target 10 using an efficient search algorithm?

hard

A. 3

B. 2

C. 4

D. -1

Solution

1. Step 1: Identify the rotation and search boundaries

   Array: [30, 40, 50, 10, 20], target: 10. left=0, right=4, mid=2 (value 50). Left half is sorted (30 to 50).

2. Step 2: Decide which half to search

   Target 10 is not in left sorted half (30 to 50), so search right half: left=mid+1=3, right=4. mid=3 (value 10), matches target.

3. Final Answer:

   3 -> Option A

4. Quick Check:

   Target 10 found at index 3 [OK]

Hint: Check sorted half, then narrow search to find target [OK]

Common Mistakes:

• Assuming array is fully sorted
• Not updating left/right correctly
• Returning wrong index or -1 incorrectly