Search in Rotated Sorted Array in DSA Javascript - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is a rotated sorted array?

A rotated sorted array is an array that was originally sorted but then rotated (shifted) at some pivot point. For example, [4,5,6,7,0,1,2] is a rotated version of [0,1,2,4,5,6,7].

beginner

Why can't we use simple binary search directly on a rotated sorted array?

Because the array is not fully sorted in ascending order due to rotation, simple binary search may fail. We need to identify which part of the array is sorted to decide where to search next.

intermediate

In the search algorithm, how do we decide which half to search next?

We check if the left half is sorted by comparing the start and middle elements. If sorted, we check if the target lies in this range. If yes, search left half; else, search right half. If left half is not sorted, right half must be sorted, so we do the same check there.

beginner

What is the time complexity of searching in a rotated sorted array using the modified binary search?

The time complexity is O(log n) because each step halves the search space, similar to normal binary search.

intermediate

Show the key steps of the binary search in rotated sorted array with an example array [6,7,0,1,2,4,5] searching for 2.

1. Start with low=0, high=6, mid=3 (value=1).
2. Left half [6,7,0,1] is not sorted, right half [1,2,4,5] is sorted.
3. Target 2 lies in right half range (1 to 5), so search right half.
4. Update low=4, high=6, mid=5 (value=4).
5. Left half [2,4] is sorted, target 2 lies in this range.
6. Update high=4, mid=4 (value=2), found target at index 4.

What is the main challenge when searching in a rotated sorted array?

A The array contains duplicates

B The array is not fully sorted in ascending order

C The array is too large

D The array is sorted in descending order

The rotation breaks the full ascending order, so normal binary search needs adjustment.

If the left half of the array is sorted, what should you check next?

A If the right half is sorted

B If the target is greater than the end element

C If the target is less than the start element

D If the target lies between the start and mid elements

If left half is sorted, check if target lies in that range to decide where to search.

What is the time complexity of the search in a rotated sorted array using binary search?

A O(log n)

B O(n)

C O(n log n)

D O(1)

The search halves the search space each time, so it is O(log n).

In the array [4,5,6,7,0,1,2], what is the pivot point where rotation happened?

A Between 7 and 0

B Between 2 and 4

C Between 5 and 6

D Between 1 and 2

Rotation happened between the largest element 7 and smallest element 0.

If the left half is not sorted, what can we conclude about the right half?

A Right half is empty

B Right half is also not sorted

C Right half must be sorted

D Right half contains duplicates

If left half is not sorted, the right half must be sorted in a rotated sorted array.

Explain how to search for a target in a rotated sorted array using binary search.

Think about how rotation affects sorting and how to use that to guide search.

Describe the difference between normal binary search and binary search in a rotated sorted array.

Focus on how rotation changes the array order.

Practice

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

easy

A. The array contains only negative numbers

B. The array elements are all the same

C. The array is sorted in descending order

D. The array is split into two sorted parts, so normal binary search needs adjustment

Solution

1. Step 1: Understand the array structure

   A rotated sorted array is created by taking a sorted array and rotating it at some pivot, splitting it into two sorted parts.

2. Step 2: Reason about search difficulty

   Normal binary search assumes the whole array is sorted, but here the array is split, so we must adjust the search to handle two sorted parts.

3. Final Answer:

   The array is split into two sorted parts, so normal binary search needs adjustment -> Option D

4. Quick Check:

   Rotated array = split sorted parts [OK]

Hint: Look for the pivot splitting the array into two sorted halves [OK]

Common Mistakes:

• Assuming the entire array is sorted
• Ignoring the pivot point
• Using linear search instead of binary search

2. Which of the following is the correct syntax to start a binary search in a rotated sorted array in JavaScript?

easy

A. let left = 0, right = arr.length - 1;

B. let left = 1, right = arr.length;

C. let left = -1, right = arr.length;

D. let left = 0, right = arr.length;

Solution

1. Step 1: Identify valid index range for binary search

   Binary search uses zero-based indexing, so left starts at 0 and right at last index (length - 1).

2. Step 2: Check each option

   let left = 0, right = arr.length - 1; correctly sets left=0 and right=arr.length-1. Others have invalid ranges or off-by-one errors.

3. Final Answer:

   let left = 0, right = arr.length - 1; -> Option A

4. Quick Check:

   Index range = 0 to length-1 [OK]

Hint: Start binary search with left=0 and right=length-1 [OK]

Common Mistakes:

• Using right = arr.length (out of bounds)
• Starting left at 1 instead of 0
• Negative indices for left or right

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

const arr = [4,5,6,7,0,1,2];
function search(nums, target) {
  let left = 0, right = nums.length - 1;
  while (left <= right) {
    let mid = Math.floor((left + right) / 2);
    if (nums[mid] === target) return mid;
    if (nums[left] <= nums[mid]) {
      if (target >= nums[left] && target < nums[mid]) right = mid - 1;
      else left = mid + 1;
    } else {
      if (target > nums[mid] && target <= nums[right]) left = mid + 1;
      else right = mid - 1;
    }
  }
  return -1;
}
console.log(search(arr, 0));

medium

A. 6

B. 0

C. 4

D. -1

Solution

1. Step 1: Understand the array and target

   The array is rotated sorted: [4,5,6,7,0,1,2], target is 0.

2. Step 2: Trace the binary search steps

   Initial left=0, right=6, mid=3 (value 7). Left half is sorted (4 to 7). Target 0 is not in left half, so search right half: left=4, right=6.
   New mid=5 (value 1). Left half (0 to 1) is sorted. Target 0 is in left half, so right=4.
   New mid=4 (value 0), found target at index 4.

3. Final Answer:

   4 -> Option C

4. Quick Check:

   Target 0 found at index 4 [OK]

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

Common Mistakes:

• Not updating left/right correctly
• Confusing sorted half
• Returning wrong index or -1

4. Identify the bug in this search function for rotated sorted array:

function search(nums, target) {
  let left = 0, right = nums.length - 1;
  while (left < right) {
    let mid = Math.floor((left + right) / 2);
    if (nums[mid] === target) return mid;
    if (nums[left] <= nums[mid]) {
      if (target >= nums[left] && target < nums[mid]) right = mid - 1;
      else left = mid + 1;
    } else {
      if (target > nums[mid] && target <= nums[right]) left = mid + 1;
      else right = mid - 1;
    }
  }
  return -1;
}

medium

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

B. The mid calculation is incorrect

C. The target comparison uses wrong operators

D. The function should return left instead of -1

Solution

1. Step 1: Analyze the loop condition

   The loop uses while(left < right), which misses checking when left equals right, possibly skipping the last element.

2. Step 2: Correct the loop condition

   Changing to while(left <= right) ensures all elements are checked, including when left == right.

3. Final Answer:

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

4. Quick Check:

   Loop must cover all indices [OK]

Hint: Use while(left <= right) to cover all elements [OK]

Common Mistakes:

• Using while(left < right) misses last element
• Incorrect mid calculation (not here though)
• Returning wrong index when not found

5. Given a rotated sorted array with possible duplicates, which modification is needed in the binary search to correctly find the target?

const arr = [2,5,6,0,0,1,2];
function search(nums, target) {
  let left = 0, right = nums.length - 1;
  while (left <= right) {
    let mid = Math.floor((left + right) / 2);
    if (nums[mid] === target) return true;
    if (nums[left] === nums[mid] && nums[mid] === nums[right]) {
      left++;
      right--;
    } else if (nums[left] <= nums[mid]) {
      if (target >= nums[left] && target < nums[mid]) right = mid - 1;
      else left = mid + 1;
    } else {
      if (target > nums[mid] && target <= nums[right]) left = mid + 1;
      else right = mid - 1;
    }
  }
  return false;
}

hard

A. Remove the duplicate elements before searching

B. Add checks to skip duplicates when left, mid, right are equal

C. Use linear search instead of binary search

D. Ignore duplicates and proceed as normal

Solution

1. Step 1: Understand duplicates impact

   Duplicates can cause ambiguity in deciding which half is sorted, breaking normal binary search logic.

2. Step 2: Skip duplicates when left, mid, right are equal

   When nums[left] == nums[mid] == nums[right], increment left and decrement right to skip duplicates and reduce search space.

3. Final Answer:

   Add checks to skip duplicates when left, mid, right are equal -> Option B

4. Quick Check:

   Skip duplicates to maintain binary search correctness [OK]

Hint: Skip duplicates when left, mid, right values match [OK]

Common Mistakes:

• Ignoring duplicates causing infinite loops
• Trying to remove duplicates before search
• Switching to linear search unnecessarily