33. Search in Rotated Sorted Array


Problem Description

In this problem, we have an integer array nums that is initially sorted in ascending order and contains distinct values. However, nums might have been rotated at some unknown pivot index k, causing the order of elements to change. After the rotation, the part of the array after the pivot index comes before the part of the array before the pivot index. Our goal is to find the index of a given integer target in the rotated array. If the target exists in the array, we should return its index; otherwise, return -1.

Since the array is rotated, a standard binary search won't immediately work. We need to find a way to adapt binary search to work under these new conditions. The key observation is that although the entire array isn't sorted, one half of the array around the middle is guaranteed to be sorted.

The challenge is to perform this search efficiently, achieving a time complexity of O(log n), which strongly suggests that we should use binary search or a variation of it.

Intuition

The intuition behind the solution is to modify the binary search algorithm to handle the rotated array. We should focus on the property that the rotation splits the array into two sorted subarrays. When we calculate the middle element during the binary search, we can determine which part of the array is sorted: the part from the start to the middle or the part from the middle to the end. Once we know which part is sorted, we can see if the target lies in that range. If it does, we adjust our search to stay within the sorted part. If not, we search in the other half.

To implement this, we have two pointers left and right that define the bounds of our search space. At each step, we compare the target with the midpoint to decide on which half to continue our search. There are four cases to consider:

  1. If the target and the middle element both lie on the same side of the pivot (either before or after), we perform the standard binary search operation.
  2. If the target is on the sorted side, but the middle element isn't, we search on the sorted side.
  3. If the target is not on the sorted side, but the middle element is, we search on the side that includes the pivot.
  4. If neither the target nor the middle element is on the sorted side, we again search on the side including the pivot.

By repeatedly narrowing down the search space and focusing on either the sorted subarray or the subarray containing the pivot, we can find the target or conclude it's not present in O(log n) time.

Learn more about Binary Search patterns.

Solution Approach

The solution implements a modified binary search algorithm to account for the rotated sorted array. Below is a step-by-step walkthrough of the algorithm as shown in the provided code snippet:

  1. Initialize two pointers left and right to represent the search space's bounds. left starts at 0, and right starts at n - 1, where n is the length of the input array nums.

  2. The binary search begins by entering a while loop that continues as long as left < right, meaning there is more than one element in the search space.

  3. Calculate mid using (left + right) >> 1. The expression >> 1 is equivalent to dividing by 2 but is faster, as it is a bitwise right shift operation.

  4. Determine which part of the array is sorted by checking if nums[0] <= nums[mid]. This condition shows that the elements from nums[0] to nums[mid] are sorted in ascending order.

  5. Check if target is between nums[0] and nums[mid]. If it is, that means target must be within this sorted portion, so adjust right to mid to narrow the search space to this sorted part.

  6. If the target is not in the sorted portion, it must be in the other half containing the rotation point. Update left to mid + 1 to shift the search space to the right half of the array.

  7. If the sorted portion was the right half (nums[mid] < nums[n - 1]), check if target is between nums[mid] and nums[n - 1]. Adjust the search space accordingly by either moving left or right depending if the target is in the sorted portion or not.

  8. This loop continues until the search space is narrowed down to a single element.

  9. After exiting the loop, check if nums[left] is the target and return left if that's the case, indicating that the target is found at that index in the array.

  10. If the element at nums[left] is not the target, return -1 to indicate that target is not found in the array.

This binary search modification cleverly handles the rotation aspect by focusing on which part of the search space is sorted and adjusting the search bounds accordingly. No additional data structures are used, and the algorithm strictly follows an O(log n) runtime complexity as required by the problem statement.

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Example Walkthrough

Let's illustrate the solution approach with a small example. Assume we have the rotated sorted array nums = [6, 7, 0, 1, 2, 4, 5] and we are looking for the target value 1. The original sorted array before rotation might look something like [0, 1, 2, 4, 5, 6, 7], and in this case, the pivot is at index 2, where the value 0 is placed in the rotated array.

  1. We initialize our search bounds, so left = 0 and right = 6 (since there are 7 elements in the array).

  2. Start the binary search loop. Since left < right (0 < 6), we continue.

  3. Calculate the middle index, mid = (left + right) >> 1. This gives us mid = 3.

  4. Check if the left side is sorted by checking if nums[left] <= nums[mid] (comparing 6 with 1). It is false, so the left-half is not sorted, the right-half must be sorted.

  5. Since 1 (our target) is less than 6 (the value at nums[left]), we know the target is not in the left side. We now look into the right half since the rotation must have happened there.

  6. Update left to mid + 1, which gives left = 4.

  7. Re-evaluate mid in the next iteration of the loop. Now mid = (left + right) >> 1 = (4 + 6) >> 1 = 5. The middle element at index 5 is 4.

  8. Check again where the sorted part is. Now nums[left] <= nums[mid] (1 <= 4) is true, which means we are now looking at the sorted part of the array.

  9. Check if the target is within the range of nums[left] and nums[mid]. Here, 1 is within 1 to 4, so it must be within this range.

  10. Now we update right to mid, setting right = 5.

  11. Continue the loop. At this point, left = 4 and right = 5, mid = (4 + 5) >> 1 = 4. The value at nums[mid] is the target we are looking for (1).

  12. Since nums[mid] is equal to the target we return mid, which is 4.

Using this approach, we successfully found the target 1 in the rotated sorted array and return the index 4. This example demonstrates the modified binary search algorithm used within the rotated array context.

Solution Implementation

1class Solution:
2    def search(self, nums: List[int], target: int) -> int:
3        # Initialize the boundary indexes for the search
4        left, right = 0, len(nums) - 1
5      
6        # Use binary search to find the target
7        while left < right:
8            # Calculate the middle index
9            mid = (left + right) // 2  # Updated to use floor division for clarity
10
11            # Determine if the mid element is in the rotated or sorted part
12            if nums[0] <= nums[mid]:
13                # If target is between the first element and mid, go left
14                if nums[0] <= target <= nums[mid]:
15                    right = mid
16                # Else, go right
17                else:
18                    left = mid + 1
19            else:
20                # If target is between mid and last element, go right
21                if nums[mid] < target <= nums[-1]:  # Use -1 for last element index
22                    left = mid + 1
23                # Else, go left
24                else:
25                    right = mid
26      
27        # Check if the left index matches the target, otherwise return -1
28        return left if nums[left] == target else -1
29
1class Solution {
2
3    public int search(int[] nums, int target) {
4        // Length of the array.
5        int arrayLength = nums.length;
6      
7        // Initialize start and end pointers.
8        int start = 0, end = arrayLength - 1;
9      
10        // Binary search algorithm to find target.
11        while (start < end) {
12            // Calculate middle index of the current segment.
13            int mid = (start + end) / 2;
14          
15            // When middle element is on the non-rotated portion of the array.
16            if (nums[0] <= nums[mid]) {
17              
18                // Check if the target is also on the non-rotated portion and adjust end accordingly.
19                if (nums[0] <= target && target <= nums[mid]) {
20                    end = mid;
21                } else {
22                    start = mid + 1;
23                }
24              
25            // When middle element is on the rotated portion of the array.
26            } else {
27              
28                // Check if the target is also on the rotated portion and adjust start accordingly.
29                if (nums[mid] < target && target <= nums[arrayLength - 1]) {
30                    start = mid + 1;
31                } else {
32                    end = mid;
33                }
34            }
35        }
36      
37        // After narrowing down to one element, check if it's the target.
38        // If nums[start] is the target, return its index, otherwise return -1.
39        return nums[start] == target ? start : -1;
40    }
41  
42}
43
1class Solution {
2public:
3    int search(vector<int>& nums, int target) {
4        // Initialize the size of the input vector
5        int size = nums.size();
6        // Define the initial search range
7        int left = 0, right = size - 1;
8
9        // Perform binary search
10        while (left < right) {
11            // Find the middle index of the current search range
12            int mid = left + (right - left) / 2; // Avoids potential overflow compared to (left + right) >> 1
13
14            // Determine the side of the rotated sequence 'mid' is on
15            if (nums[0] <= nums[mid]) {
16                // 'mid' is in the left (non-rotated) part of the array
17                if (nums[0] <= target && target <= nums[mid]) {
18                    // Target is within the left (non-rotated) range, search left side
19                    right = mid;
20                } else {
21                    // Search right side
22                    left = mid + 1;
23                }
24            } else {
25                // 'mid' is in the right (rotated) part of the array
26                if (nums[mid] < target && target <= nums[size - 1]) {
27                    // Target is within the right (rotated) range, search right side
28                    left = mid + 1;
29                } else {
30                    // Search left side
31                    right = mid;
32                }
33            }
34        }
35
36        // The final check to see if the target is found at 'left' index
37        return (left == right && nums[left] == target) ? left : -1;
38    }
39};
40
1function search(nums: number[], target: number): number {
2    const length = nums.length;
3    let leftIndex = 0;
4    let rightIndex = length - 1;
5  
6    // Use binary search to find the target
7    while (leftIndex < rightIndex) {
8        // Calculate the middle index
9        const midIndex = Math.floor((leftIndex + rightIndex) / 2); // shifted to use Math.floor for clarity
10
11        // Check if the first element is less than or equal to the middle element
12        if (nums[0] <= nums[midIndex]) {
13            // If target is between the first element and middle element
14            if (nums[0] <= target && target <= nums[midIndex]) {
15                // Narrow down the right bound
16                rightIndex = midIndex;
17            } else {
18                // Target must be in the second half
19                leftIndex = midIndex + 1;
20            }
21        } else {
22            // If target is between the middle element and the last element
23            if (nums[midIndex] < target && target <= nums[length - 1]) {
24                // Narrow down the left bound
25                leftIndex = midIndex + 1;
26            } else {
27                // Target must be in the first half
28                rightIndex = midIndex;
29            }
30        }
31    }
32  
33    // Check if we have found the target
34    return nums[leftIndex] == target ? leftIndex : -1;
35}
36

Time and Space Complexity

Time Complexity

The given code performs a binary search over an array. In each iteration of the while loop, the algorithm splits the array into half, investigating either the left or the right side. Since the size of the searchable section of the array is halved with each iteration of the loop, the time complexity of this operation is O(log n), where n is the number of elements in the array nums.

Space Complexity

The space complexity of the algorithm is O(1). The search is conducted in place, with only a constant amount of additional space needed for variables left, right, mid, and n, regardless of the size of the input array nums.

Learn more about how to find time and space complexity quickly using problem constraints.


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