Search in Rotated Sorted Array in DSA Python - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of search in rotated sorted array

What is the output of the following code that searches for a target in a rotated sorted array?

DSA Python

def search(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

nums = [6,7,8,1,2,3,4,5]
target = 3
print(search(nums, target))

A -1

B 3

C 5

D 2

💡 Hint

Trace the binary search steps considering the rotated sorted array halves.

The function uses binary search adapted for rotated sorted arrays. The target 3 is at index 5 in the array.

❓ Predict Output

intermediate

Find index of target not present

What does the function return when the target is not in the rotated sorted array?

DSA Python

def search(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

nums = [4,5,6,7,0,1,2]
target = 3
print(search(nums, target))

A 0

B -1

C 6

D 3

💡 Hint

If the target is not found, the function returns -1.

The target 3 is not in the array, so the function returns -1.

🔧 Debug

advanced

Identify the error in this rotated array search code

What error does this code produce when searching for target 2?

DSA Python

def search(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

nums = [3,4,5,1,2]
target = 2
print(search(nums, target))

A -1

B 3

C IndexError

D Infinite loop

💡 Hint

Check the loop condition and how it affects the search range.

The loop condition uses < instead of <=, so it may miss the correct index and return -1.

❓ Predict Output

advanced

Output of search with duplicates in rotated array

What is the output of the code searching for target 2 in a rotated array with duplicates?

DSA Python

def search(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

nums = [2,3,4,2,2,2]
target = 2
print(search(nums, target))

A 5

B 3

C -1

D 0

💡 Hint

The function returns the first found index of the target.

The target 2 appears multiple times; the function returns the first found index 0.

🧠 Conceptual

expert

Why binary search works on rotated sorted arrays

Which statement best explains why binary search can be adapted to work on rotated sorted arrays?

A Because at least one half of the array is always sorted, allowing us to decide which half to search next.

B Because the array is fully sorted, so normal binary search applies without changes.

C Because the rotation does not affect the order of elements, so linear search is faster.

D Because the array elements are all unique, so we can use hashing instead.

💡 Hint

Think about how the array is split and what properties remain after rotation.

In a rotated sorted array, one half is always sorted, which helps decide where to continue searching.

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